Bihar Board Class 12 Physics Solutions Chapter 9 Ray Optics and Optical Instruments Textbook Questions and Answers, Additional Important Questions, Notes.

## BSEB Bihar Board Class 12 Physics Solutions Chapter 9 Ray Optics and Optical Instruments

**Bihar Board Class 12 Physics **Ray Optics and Optical Instruments Textbook Questions and Answers

Question 1

A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Answer:

Here, O = size of the object = 2.5 cm

u = distance of the object from the concave mirror = – 27 cm

R = radius of curvature of concave mirror = – 36 cm

∴ f = \(\frac {R}{2}\) = –\(\frac {-36}{2}\) = -18 cm

(i) υ – distance of the image from the mirror = ?

Using the relation,

\(\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}\) , we get

\(\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}=\frac{1}{-18}-\frac{1}{(-27)}\)

= \(\frac{-3+2}{54}=-\frac{1}{54}\)

υ = – 54 cm.

The negative sign shows that the image is formed in front of the mirror i.e., on the side of the object itself. Thus the screen must be placed at a distance of 54 cm in front of the mirror.

(ii) Nature and size of the image = ?, I = ?

Using the relation,

m = \(\frac {I}{O}\) = –\(\frac {υ}{u}\)

We get,

\(\frac{\mathrm{I}}{2.5}=-\frac{(-54)}{(-27)}\)

or I = (2.5) x (- 2) = – 5 cm.

Thus clearly the image is real, inverted and magnified.

(iii) If the candle is moved closer to the mirror, then the screen is to be moved farther and farther. But when the distance of the candle is lesser than the focal length of the mirror, as u → f, υ → ∞, for u < f, then the image would be virtual, so it cannot be collected on the screen i.e., no screen is required.

Question 2.

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Answer:

Here, u = distance of the needle from the convex mirror = – 12 cm

f = focal length of the mirror – + 15 cm.

O = size of the object = 4.5 cm

v = location of the image = ?

Using the relation,

\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) , we get

\(\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}} \cdot \frac{1}{\mathrm{u}}=\frac{1}{15}-\left(-\frac{1}{-12}\right)\)

= \(\frac{4+5}{60}=+\frac{9}{60}\)

∴ υ = \(\frac{20}{3}\) cm = + 6.67 cm. = 6.7 cm

The positive sign shows that the image is formed behind the mirror.

I = size of image = ?

using the relation,

m = \(\frac{20}{3}\) = \(\frac{I}{O}\) , we get

The positive sign shows that the image formed is erect, virtual and of reduced size.

Also, magnification m is given by

m = \(\frac{20}{3}\)

As the needle is moved farther from the mirror, the image moves towards the focus (upto the focus only) and gets smaller and smaller in size i.e. as u → ∞, υ →f (but never beyond) while m → O.

Question 3

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Answer:

Case I: When tank is filled with water

Here, real depth = 12.5 cm

Apparent depth = 9.4 cm

μ = Refractive index of water = ?

We know that

Case II. When the tank is rilled with the liquid :

Here, μ = R.I of the liquid = 1.63.

real depth = 12.5 cm

apparent depth = ?

real depth

∴ Using the relation,

the distance through which the microscope has to be moved up = 9.4 – 7.67 = 1.73 cm = 1.70 cm.

Question 4.

Figures (a) and (b) show refraction of an incident ray in air at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water at 45° with the normal to a water-glass interface [Fig. (c)].

Answer:

i = angle of incidence = 60°

r = angle of refraction = 35°

^{a}μ_{g}= R.I. of glass w.r.t. air =?

Using the relation,

From fig. (b), here, i = 60°, r = 47°

^{a}μ_{w} = R.I. of water w.r.t. air = ?

We know that

From fig. (c), i = angLe of incidence = 45°

Let r = angie of refraction = ?

^{w}μ_{g} = R.I. of glass w.r.t. water =?

We know that

Question 5.

A small bulb is placed at the bottom of tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Answer:

Here, source of light (O) is 80 cm below the surface of water i.e., OO’ = 80 cm = 0.80 m.

^{a}μ_{2} = 1.33

The ray of light emitted from O will be refracted into air only if the angle of incidence is less than the critical angle C. When the angle of incidence is equal to critical angle C, the light will not be refracted along water surface into air, but it will graze the air water interface. Thus, the light will appear to come out of a cone having vertex angle 2C. If the angle of incidence at water-air interface is more than C, then the rays of light will be totally reflecfed.

∴When i = C, r = 90°

∴ Area of the surface of water through which light from the bulb can emerge out is the area of the circle of radius O’A = (O’B) i.e., r’ = \(\frac {AB}{2}\) = O’A = O’B

Now in ∆ OO’B,

tan C = \(\frac{\mathrm{O}^{\prime} \mathrm{B}}{\mathrm{O}^{\prime} \mathrm{O}}\)

or O’B = OO’ tan C

= 0.80 x tan 48. 6°

= 0.8 x 1.1345 = 0.907 m = 90.7 cm

∴ Area of the surface of water through which light will emerge

= πr^{a} = 3.14 x (0.907)^{2}

= 2.584 m^{2} = 2.6 m^{2}

Question 6.

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Answer:

Here, A = angle of prism = 60°

δ_{m} = angle of minimum deviation = 40°

μ = R.I. of the material of prism = ?

Using the relation,

Let δ_{m} be the new angle of minimum deviation of a parallel beam of light when the prism is placed in water. Here ^{a}μ_{w} = 1.33, ^{a}μ_{g} = 1.532

Question 7.

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Answer:

μ = R.I of glass = 1.55

f = focal length of the lens = 20 cm

R_{1} = R, R_{2} = – R where R_{1} and R_{2} are the radii of curvatures of the two faces forming the double convex lens. R = ?

Using the relation

or R = 1.10 x 20 = 22 cm.

Question 8.

A beam of light converges to a point P. Now a lens is placed in the path of the convergent beams 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm?

Answer:

Here, the point P on the right of the lens acts as a virtual object,

∴ u = + 12 cm, v = ?

(a) For convex lens, f = + 20 cm.

Using lens formula,

i.e. the image is formed on the right of the lens and is real,

(b) for concave lens, f = – 16 cm

∴ Using the relation,

The image formed is 48 cm on the right of the lens where the beam would converge and is real.

Question 9.

An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Answer:

Here, O = size of object = 3.0 cm.

v = distance of object from concave lens = -14 cm.

f = focal length of the concave lens = – 21 cm v = ?, I = ?

Using the relation

∴the image is virtual, erect and located at 8.4 cm from the lens on the same side as the object.

Also we know that

i.e., the image is of diminished size.

As the object is moved away from the lens, the virtual image moves towards the focus of the lens (but never beyond focus). The size of the image progressively goes on decreasing i.e. as u → ∞, v → f but never beyond f while m → 0.6

Question 10.

What is the focal length of a convex lens of focal length 30 cm in contact with a concave length of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Answer:

Here, f_{1} = focal length of the convex lens = + 30 cm.

f_{2} = focal length of concave lens = – 20 cm

Let f = focal length of the combination of two lenses in contact = ?

We know that

\(\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}\)

As the focal length of the combination of two lenses is – ve, so the combination behaves as a diverging lens i.e., as a concave lens.

Question 11.

A compound microscope consists of an objective lens of focal length 2.0 cm and an eye-piece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer:

Here, f_{0} = focal length of the objective lens = 2.0 cm

f_{e} = focal length of the eye-piece = 6.25 cm

L = distance between two lenses = 15 cm.

(a) v_{e} = – 25 cm = distance of the final image from E- lens

u_{0} = distance of the object from O-lens = ?

Using lens formula,

\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{u}}\) we get for eve-piece,

∴ Now L = 15 cm = v_{0} + u_{e}

∴ v_{0} = L – |u_{e} = 15 – 5 = 10 cm.

∴ For objective lens,

We know that the magniíyiig power of the compound microscope is given by

(b) The final image will be formed at infinity only if the image formed by the objective is in the focal plane of the eye piece i.e., at principal focus of the eye piece.

Thus here, v_{e} = – u_{e} = f_{e}= 6.25 cm.

|v_{0}| = L – |u_{e}| = 15- 6.25 = 8.75 cm

Question 12.

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eye-piece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Answer:

Here, D = 25 cm

f_{0} = focal length of objective = 8.0 mm = 0.8 cm.

f_{e} = focal length of eye-piece = 2.5 cm

u_{0} = – 9.0 mm = – 0.9 cm.

L = separation between two lenses = ?

M = ?

For eye piece, using lens formula, we get

for objective,

Now magnifying power of the compound microscope is given by

Question 13.

A small telescope has an objective lens of focal length 144 cm and an eye-piece of local length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?

Answer:

Here, f_{0} = focal length of the objective of the telescope = 144 cm.

f_{e} = focal length of the objective of the telescope = 6.0 cm.

M = magnifying power of telescope = ?

L = separation between the objective and the eye-piece = ?

(i) In normal adjustment (i.e. when the final image is formed at co)

M = – \(\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\) = \(\frac {144}{6}\) = -24 ,M = 24

∴ L = f_{0} + f_{e} = 144 + 6 = 150 cm.

Question 14.

(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye-piece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 x 10^{8} m, and the radius of lunar orbit is 3.8 x 10^{6} m.

Answer:

Here, f_{0} = focal length of the objective of the telescope = 15 m.

f_{e} = focal length of the eye piece = 1.0 cm = 10^{-2} m.

(a) Angular magnification = ?

We know that Angular magnification = – \(\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\)

= – \(\frac{15}{10^{-2}}\) = – 1500

(b) Let d be the diameter of the image of the moon formed by the objective lens.

∴ angle subtended by the image = \(\frac{\mathrm{d}}{\mathrm{f}_{0}}=\frac{\mathrm{d}}{15}\) ….(i)

Also we know that angle subtended by the diameter of the moon

from (i) and (ii), we get

Question 15.

Use the mirror equation to deduce that :

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note : This exercise helps you to deduce algebraically properties of images that one obtains from explicit ray diagram.]

Answer:

(a) Here, f < u < 2f (given), υ = ?

Using mirror formula, \(\frac{1}{\mathrm{u}}+\frac{1}{v}=\frac{1}{\mathrm{f}}\), we get [w and f both – ve and |u| > f]

Suppose u here is \(\frac {3}{2}\) f

This image is real because υ’ is – ve.

Aliter :

\(\frac{1}{v}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}\) For f

∴ 2f < υ < ∞ ⇒ υ is between 2f and infinity which means that the image is formed beyond 2f.

(b) For convex mirror :

∴ uf is – ve and u – f is also – ve.

∴ υ is always + ve or image formed by convex mirror is always virtual and thus is independent of the location of the object. Now for convex mirror, f > 0. Also we have u < 0 (object lies on left)

∴ \(\frac {1}{υ}\) < 0 or υ > 0 or υ = + ve image is formed on right and is virtual whatever may be the value of u.

(c) From (b) υ = \(\frac{\mathrm{uf}}{\mathrm{u}-f}\)

and m = \(\frac{-v}{u}\left(\frac{-f}{u-f}\right)\)

|u – f| is always grater than f ( ∵ u is – ve ) and \(\left|\frac{f}{u-f}\right|\) < 1 and \(\frac{f}{\mathrm{u}-f}\) is -ve is sign. i.e., as f > 0 for convex mirror and u < 0, ∴ \(\frac{1}{υ}\) > f i.e., υ < f (image located between the pole and the focus)

Also υ < |u| (image diminished).

∴ m is + ve and less than one.

Therefore image produced by convex mirror is virtual and diminshed in size.

(d) From (a) for concave mirror

\(\frac{1}{υ}\) = \(\frac{u – f}{uf}\)

|u and f both – ve| i.e. f < 0 for concave mirror.

Here, u is less than f and is equal to f/2.

∴ u = \(\frac{- f}{2}\)

or υ = f

∴ υ > |u| and υ is + ve. or \(\frac{υ}{u}\) > 1

∴ m = – \(\frac{υ}{u}\) = + ve and > 1

Thus the image formed is virtual and magnified.

Aliter :

f < u < 0 implies \(\frac{1}{f}\) – \(\frac{1}{u}\) > 0 or – \(\frac{1}{υ}\) > 0, or υ > 0 image is 1 1

formed on right and is virtual. Also \(\frac{1}{u}\) < \(\frac{1}{|\mathrm{u}|}\), ∴υ > |u| ⇒ image is enlarged.

Question 16.

A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Answer:

Here, t = thickness of glass slab = 15 cm.

p = R.I of glass = 1.5.

The normal shift in the position of the pin is given by

d = t \(\left(1-\frac{1}{\mu}\right)\)

where t = real depth of pin = thickness of slab

Let x = apparent depth of pin

∴ d = Normal shift = t – x = ?

Also µ = \(\frac{t}{x}\) or x = \(\frac{t}{µ}\)

∴ d = normal shift = t – \(\frac{t}{u}\) = t \(\left(1-\frac{1}{\mu}\right)\)

= 15 \(\left(1-\frac{1}{1.5}\right)\) = 15 \(\left(1-\frac{2}{3}\right)\)

= 15 x \(\frac{1}{3}\) = 5 cm,

i.e. the pin appears raised by 5 cm.

No, the answer does not depend upon the location of the slab for small angles of incidence,

Question 17.

(a) Figure shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place as shown in the figure,

(b) What is the answer if there is no outer covering of the pipe?

Answer:

(a) Here ^{a}µ_{g} = Refractive index of the glass fibre w.r.t. air = 1.68.

^{a}µ_{c} = Refractive index of the outer coating material w.r.t. air = 1.44.

If ^{c}µ_{g} be the Refractive index of the glass w.r.t. outer coating, then ^{g}µ_{c} = \(\frac{{ }^{a} \mu_{c}}{{ }^{a} \mu_{g}}=\frac{1.44}{1.68}\)

∴ ^{c}µ_{g} = \(\frac{1}{g_{\mu_{c}}}=\frac{1.68}{1.44}\) = 1.67

Let C be the critical angle for the fibre material w.r.t. the material of the outer coating.

∴ sin C = \(\frac{1}{{ }^{c} \mu_{g}}=\frac{1}{1.167}\) = 0.8569

∴ C = sin^{-1} (0.8569) = 58.97° = 59°

Total internal reflection will take place when i > C i.e. i > 58.97° = 59°

or when r < r_{max}

Where r_{max} = 90 – C = 90° – 59° = 31°.

Now refractive index of glass-fibre w.r.t. air is given by .

Thus all the rays which are incident in the range 0 < i < 60° will suffer total internal reflection in the pipe.

(b) If there is no outer coating of pipe, then refraction inside the pipe shall take place from glass to air.

∴ ^{c}µ_{g} = 1.68, ^{a}µ_{a} = 1.

∴ Sin C’ = \(\frac{1}{\mu}=\frac{{ }^{a} \mu_{\mathrm{a}}}{{ }^{a} \mu_{\mathrm{g}}}=\frac{1}{1.68}\) = 0.5952

∴ C’ = 36.5°.

(∵ \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) = µ = 1.68).

∴ r_{max} = 90° – 36.5° = 53.5°

which is greater than C’.

Thus all rays incident at angles in the range O to 900 with the axis

will suffer total internal reflection from inside the pipe.

Question 18.

Answer the following questions:

(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real ¡mages under some

circumstances? Explain.

(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?

(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisher man look taller or shorter to the diver than what he actually is?

(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?

(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

Answer:

(a) Yes, plane and convex mirrors can produce real images. When, the rays incident on them (plane or convex mirrors) are converging to a point behind the mirror, they are reflected to a point on a screen infront of the mirror. In other words, a plane or convex mirror can produce a real image if the object is virtual.

(b) No, there is no contradiction. It is due to the fact that the virtual image formed by a spherical mirror acts as virtual object for eye lens which in turn forms real image on the retina of the eye because eye lens is convergent lens.

(c) The man would look taller to the diver than what he actually is. As the man is in air, so light travels from rarer to denser medium and bends towards the normal and it appears to be coming from a larger distance as shown in the figure here. The size of light BP and BQ from the head (B) of the fisherman AB, on refraction at the water-air interface, bend towards the normal at points P and Q and appear to be coming from B’ to the diver. Clearly AB’ > AB. Where AB’ is the image of fisherman AB.

(d) Yes, the apparent depth of a tank of water changes if viewed obliquely. The apparent depth decreases when water tank is viewed obliquely as compared to the depth of the tank when seen normally.

(e) Yes. Refractive index of diamond is 2.4 which is much larger than the refractive index of ordinary glass (p = 1.5). The critical angle for diamond is 24.4° which is much smaller than for ordinary glass (= 90°), A skilled diamond cutter exploits the large range of angles of incidence in diamond 24° to 90° to ensure that light entering the diamond on any face at an angle of incidence more than 24°, it suffers total reflection from many faces before getting out. This produces sparkling effect in diamond.

Question 19.

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Answer:

Here, u + v = 3m, ∴ υ = 3 – u.

From lens formula,

For real solution, 9 -12 f should be positive i.e., 9 -12 f shouldbe positive

or 9 – 12 f > 0

9 > 12 f.

or f < \(\frac {9}{12}\) < \(\frac {3}{4}\) m.

∴ The maximum focal length of the lens required for the purpose is \(\frac {3}{4}\) m i.e. f_{max} = 0.75 m.

Aliter :

u + v = 3 or v = 3 – u = 3- (-v) = 3 + u ……(1) (using sign conventions)

From lens formula,

For to be maxium

Question 20.

A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

Answer:

Here, O is position of object and I is position of image (screen). Distance Ol = 90 cm.

L_{1} and L_{2} are the two positions of the lens.

∴ Distance between L_{1} and L_{2} = O_{1} O_{2} = 20 cm.

For position L_{1} of the lens—Let x be the distance of the object from the lens.

∴ u_{1} = – x

∴ distance of the image from the lens, v_{1} = + (90 – x)

if be the focal length of the lens, then using lens formula,

For position L_{2} of the lens—Let u_{2} and v_{2} be the distances of the object and image from the lens in this position.

∴ u_{2} = – (x + 20),

∴ υ_{2} = + [90 – (x + 20)] = + (70 – x)

Using lens formula,

From (1) and (2), we get

Question 21.

(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10 if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?

(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangement (a) above. The’ distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Answer:

Here, as per Q. 9.10,

f_{1} = + 30 cm (focal length of convex lens)

f_{2} = – 20 cm (focal length of concave lens)

d = separation between the lenses = 8 cm

Let F = effective focal length of the combination = ?

Using the relation,

Thus as F is – ve, so the combination of these lenses acts as a diverging lens.

(i) Now let a parallel beam of light be incident on the lens of the combination.

∴ f_{1} = + 30 cm, u_{1} = – ∞, υ_{1} = ? (position of the image formed by convex lens).

Using lens formula,

This image (I_{1}, ray) acts as a virtual object for the second lens having

f_{2} = – 20 cm,

u_{2} = distance of the object,

υ_{2} = distance of final image I = ?

= u_{1} – d = 30 – 8 = 22 cm.

∴ From lens formula, we get

Thus the parallel incident seam appears to diverge from a point (220 cm – 4) = 216 cm from the centre of two lens system.

(ii) Now Let the parallel beam be incident from Left on the concave Lens first,

∴ u_{1} = – oo, f_{1} = – 20 cm, υ_{1} = ?

∴ \(-\frac{1}{\mathrm{u}}+\frac{1}{v}=\frac{1}{\mathrm{f}}\) gives

This image acts as a real object for the convex lens.

∴u_{2} = υ_{1} – d = – 20 – 8 = – 28 cm

f_{2} = + 30 cm, υ_{2} = ?

∴Using lens formula, we get

a distance of (420 – 4) = 416 cm from the centre of two lens system.

Thus we conclude that the answer depends on which side a parallel beam of light is incident as the distances are different in the two cases. Also the notion of effective focal length is not useful in this case.

(b) Here O = 1.5 cm, 1 = size of image = ?

u_{1} = distance between the object and convex lens = – 40 cm

M = magnification produced by the combination = ?

f_{1} = + 30 cm, υ_{1} = ?

∴ Using the relation, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\), we get

If m_{1} be the magnification produced by the convex lens, then m_{1}.

\(\frac{v_{1}}{\left|\mathrm{u}_{1}\right|}=\frac{120}{40}\) = 3

For concave lens :

If m_{2} be the linear magnification produced by concave lens, then

Also we know that

M = m_{1} x m_{2}

= 3 x (\(\frac{5}{23}\)) = \(\frac{15}{23}\) = 0.652

i.e. net magnitude of magnification = 0.652

Also M = \(\frac{I}{O}\)

or I = M x O = \(\frac{15}{23}\) x 1.5 = \(\frac{45}{46}\) cm

= 0.98 cm.

Question 22.

At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Answer:

Here, A = angle of prism = 60°

μ = refractive index of the material of prism = 1.524.

A ray of light PQ incident on the face AB will suffer total internal reflection at the other face AC when it gets incident on the face AC at an angle of incidence equal to the critical angle for the material of the prism. Let C = critical angle for the material of the prism.

∴ sin C = \(\frac{1}{μ}\) = \(\frac{1}{1.524}\) = 0.6562

∴ C = sin^{-1} (0.6562) = 41°.

We know that for a prism,

A = r_{1} + r_{2} ”

Here r_{2} = C = 41°, A = 60°

∴ 60° = r_{1} + 41° or r_{1} = 60 – 41 = 19°.

For the face AB,

i = angle of incidence = ?

r_{1} = 19°, m = 1.524 .

According to Snells’ law,

Question 23.

You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion, (b) disperse (and displace) a pencil of white light without much deviation.

Answer:

When a beam of white light is incident on a prism, the emergent beam is dispersed and is deviated from its original path. When two prisms of different material and angles are combined so as to produce dispersion without deviation then such a combination is called direct vision prism. On the other hand if the two prisms combined produce deviation without dispersion, then such a combination is called achromatic prism.

(a) Condition of deviation without dispersion for prism combination:

We know that θ = δ_{v} – δ_{r} = (μ_{v} – μ_{r}) A …..(i)

and θ’ = δ’_{v} – δ’_{r} = (μ’_{v} – μ’_{r}) A’ …..(ii)

For no dispersion produced by the combination θ + θ’ = 0 i.e., dispersion produced by the two prisms are equal and opposite.

or (μ_{v} – μ_{r}) A+ (μ’_{v} – μ’_{r}) A’ = 0.

or \(\frac{\mathrm{A}^{\prime}}{\mathrm{A}}=-\frac{\mu_{\mathrm{v}}-\mu_{\mathrm{r}}}{\mu_{\mathrm{v}}-\mu_{\mathrm{r}}}\) ………(iii)

– ve sign shows that the two prisms have to be placed in opposite manner. Eqn. (iii) is the condition for no dispersion or condition for achromatism.

Eqn. (iii) can also be expressed as :

(b) The condition of dispersion without deviation by a combination of two prisms of different materials and different angles:

Let A, μ be the angle of prism and R. I. for mean colour of the crown glass.

Also A’ be the angle of prism for flint glass having R.I. μ’ for mean colour.

If δ, δ’ be the deviation suffered by mean light through crown and flint glass prisms respectively.

∴ δ = (μ – 1) A …(i)

and δ’ = (μ’ – 1) A’ …(ii)

If the combination does not produce any deviation, then δ + δ’ = 0 i.e., the deviations produced by the two prisms are equal and opposite,

or (μ – l)A + (μ’- l)A’= 0

\(\frac{\mathbf{A}^{\prime}}{\mathbf{A}}=-\frac{\mu-\mathbf{1}}{\mu^{\prime}-\mathbf{1}}\)]

The negative sign shows that the two prisms have to be placed in opposite manner.

eqn. (iii) is the required condition for no deviation.

Let θ and θ’ be the angular dispersions produced by crown and flint glasses respectively.

θ = δ_{v} – δ_{r} = (μ_{v} – μ_{r}) A

and θ’ = δ’_{v} – δ’_{r} =(μ’_{v} – μ’_{r}) A’

If θ + θ’ be the net dispersion produced, then

Question 24.

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accomodation (i.e. the range of converging power of the eye-lens) of a normal eye.

Answer:

Power of cornea = 40 dioptre

And least converging power of eye lens = 20 dioptre

For far points situated at infinity least converging power = 40 + 20 = 60 dioptre.

f = focal length = \(\frac {5}{3}\) cm [∵f_{m} = \(\frac{1}{P}=\frac{1}{60}\) m = \(\frac{100}{60}\)

Here u = – ∞

By Lens formula – \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

υ = \(\frac{5}{3}\)

To focus the object at the near point :

Power of combination = ?

∴ Power of eye lens = 64 – 40 = 24 dioptre.

Range of accommodation of eye lens = 20 dioptre to 24 dioptre.

Question 25.

Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accomodation? If not, what might cause these defects of vision?

Answer:

No, short – sightedness (Myopia) or long-sightedness (Hypermetropia) do not necessarily mean that eye has partially lost its

ability of accommodation. Myopia arises when the eye can see near by objects clearly but distant objects are not seen clearly due to the shift of the farthest point towards the eye and thus it becomes difficult to see beyond certain limit. This defect may be due to either the eye ball has become large or focal length of the eye lens has become too small. So the ray coming from infinity are focussed at a near point. In long sightedness, one cannot see near by objects clearly. This defects may be due to either the eye ball has become short or the focal length of the lens has become too large.

Question 26.

A myopic person has been using spectacles of power – 1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.

Answer:

Here, P = -1D, ∴ f = \(\frac{1}{p}\) = \(\frac{1}{-1}\) = – 100 cm

For normal vision, far point is at infinity.

∴ u = – ∞, υ = ?

Using lens formula,

\(\frac{1}{\mathrm{u}}+\frac{1}{v}=\frac{1}{f}\)

we get

\(-\frac{1}{-\infty}+\frac{1}{v}=-\frac{1}{100}\)

or \(\frac{1}{υ}\) = – \(\frac{1}{100}\)

∴ υ = -100 cm

Thus virtual image of the object at infinity is produced at 100 cm distance using spectacles.

To view objects at distance 25 cm to 100 cm, the person using ability of accomodation of eye which is partially lost in old age for which he needs another spectacles having power.

P = + 2 D,

∴ f = \(\frac{1}{p}\) = \(\frac{1}{2}\) = 0.5 m = 50 cm, v = 25 cm, u = ?

Using eqn. (1), we get

His near point shifts to 50 cm.

Question 27.

A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Answer:

The person looking at a mesh of crossed wires is able to see vertical wire more clearly than horizontal wires due to imperfect spherical nature of eye lens. As a result the focal length of this lens become different in two mutually perpendicular directions and eye cannot see objects in these two directions clearly simultaneously. This defect is directional and is removed by using cylindrical lenses in a particular direction. This defects is called astigmatism. The defect (called astigmatism) arises because the curvature of the cornea plus eye-lens refracting system is not the same in different planes.

[The eye- lens is usually spherical i.e., has the same curvature on different planes but the cornea is not spherical in case of an astigmatic eye.] In the present case, the curvature in the vertical plane is enough, so sharp images of vertical lines can be formed on the retina. But the curvature is insufficient in the horizontal plane, so horizontal lines appear blurred. The defect can be corrected by using a cylindrical lens with its axis along the vertical. Clearly, parallel rays in the vertical plane will suffer no extra refraction, but those in the horizontal plane can get the required extra convergence due to refraction by the curved surface of the cylindrical lens if the curvature of the cylindrical surface is chosen appropriately.

Question 28.

A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Answer:

(a) u = ?

v = – 25 cm

f = 5 cm

Using lens formula, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) , we get

\(-\frac{1}{\mathrm{u}}+\frac{1}{25}+\frac{1}{5}=\frac{1+5}{25}=\frac{6}{25}\)

∴ u = – \(\frac {25}{6}\) = – 4.167 cm. = – 4.2 cm

Thus closest distance at which he can read the book is – 4.2 cm. For the farthest distance,

u’ = ? v’ – oo, f = 5 cm

Using lens formula, we get’

which is the farthest distance.

(b) Maximum angular magnification (magnifying)

\(\frac{v}{u}=\frac{d}{|u|}=\frac{25}{\frac{25}{6}}\) = 6

(∴ υ = d = 25 cm = distance of distinct vision). Minimum angular magnification is at the farthest distance.

= \(\frac{v}{\left|v^{\prime}\right|}=\frac{25}{5}\)

Question 29.

A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(c) Is the magnification in (a) equal to the magnifying power in

(b) Explain.

Answer:

Here, size of each square in the figure, = 1 mm^{2} = area of object

∴ O = size of object = 1mm

u = distance of object from convex lens = -9 cm

f = focal length of magnifying glass = +10 cm:

(a) m = magnification produced by the lens = ?

A = area of each square in virtual image = ?

For a lens,

M = \(\frac{υ}{u}\) ……..(1)

Now v can be calculated using lens formula,

∴ from (i),

m = \(\frac{-90}{-9}\) = 10

Now each square in the figure has area 1 mm2 i.e., 1 mm x 1 mm. Since the lens produces a linear magnification of 10, thus size of each square in the virtual image will appear as (10 x 1 mm) x (10 x 1 mm)

or A = 100 mm^{2}.

Aliter :

m = \(\frac{I}{O}\),

∴ I = m x O = 10 x 1 mm^{2} = 10 mm.

∴ Area of each square in virtual image = (10 mm)^{2} = 100 mm^{2}.

(b) Least distance of distinct vision, D = – 25 cm.

∴ M = Angular magnification – magnifying power = \(\frac{D}{u}=\frac{-25}{. .9}\)

or M = \(\frac{25}{9}\) = 2.8

(c) No, the magnification produced in (a) \(\left(=\frac{v}{\mathrm{u}}\right)\) cannot be equal to the magnifying power \(\left(=\frac{D}{\mathrm{u}}\right)\)unless υ = D i.e., the image is located at the least distance of distinct vision.

Question 30.

(a)At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this case?

(c) Is the magnification equal to the magnifying power in this case? Explain.

Answer:

(a) Here, maximum magnifying power is obtained when the virtual image is formed at the least distance of distinct vision.

υ = – 25 cm, f = + 10 cm, u = ?

Using lens formula \(\frac{1}{\mathrm{u}}+\frac{1}{v}=\frac{1}{f}\) , we get

(b) m – magnification = ?

Using the relation, m = \(\frac{υ}{u}\), we get

m = \(\frac{-25}{\left(-\frac{50}{7}\right)}=25 \times \frac{7}{50}\)

or m = \(\frac{7}{2}\) = 3.5

(c) Magnifying power, M = ?

When the object is placed, so that the image is formed at the least distance of distinct vision, then M is given by

m = 1 + \(\frac{D}{f}\) = \(\left(1+\frac{25}{10}\right)\) = 3.5

Also M = \(\frac{\mathrm{D}}{|\mathrm{u}|}=\frac{25}{\left(\frac{50}{7}\right)}\) = 3.5

Yes, the magnification is equal to magnifying power in this case because image is formed at the least distance of distinct vision.

Question 31.

What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

[Note : Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Answer:

Here, area of object = 1 mm^{2}, O = size of object = 1 mm.

Area of image = 6.25 mm^{2},

∴ I size of image = – \(\sqrt{6.25}\) mm.

m = linear magnification is given by,

m = \(\frac{\mathrm{I}}{\mathrm{O}}=\frac{\sqrt{6.25}}{1}\) = 2.5

Also we know that m = \(\frac {υ}{u}\)

∴ 2.5 = \(\frac {υ}{u}\)

or υ = 2.5 u; f = 10 cm.

Using lens formula,

Hie image will be formed at 15 cm from the lens. Since the least distance of distinct vision is 25 cm so the square will not be seen distinctly by the eye when the eye is held close to the magnifier.

Question 32.

Answer the following questions :

(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?

(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?

(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?

(d) Why must both the objective and the eye piece of a compound microscope have short focal lengths?

(e) When viewing through a compound microscope, our eyes should be positioned not on the eye piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eye-piece?

Answer:

(a) It is true that the angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass i.e., angular size of the image is equal to the angular size of the object. By using magnifying glass, we keep the object far more closer to the eye than at 25 cm. The closer object has larger angular size than the same object at 25 cm. It is in this sense that a magnifying lens produces angular magnification.

(b) Yes, the angular magnification changes if the eye is moved back and it (angular magnification) decreases a little. It is because, now the angle subtended at the eye by the image is less than the angle subtended by the image at the lens. Tire angle subtended by the object at the eye is also less than that subtended at the lens but the difference is very small. Of course this effect is negligible when the image is at very-very large distance.

(c) Theoretically it is true, but when we decrease focal length, abberations both spherical as well as chromatic become larger. Moreover it is difficult to manufacture lenses of very-very short focal length.

(d) We know that the magnifying power of a compound microscope is given by

M = \(\frac{\mathrm{L}}{\mathrm{f}_{0}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)\)

where 1 + \(\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\) = magnification of eye lens

and \(\frac{\mathrm{L}}{f_{0}}=\frac{-v_{0}}{\mathrm{u}_{0}}\) = magnification of objective.

D = least distance of distinct vision.

L = Distance between objective and eye-piece i.e., length of microscope tube f0 and f are the focal lengths of objective and eye piece. Thus for M to be large, both f0 and fe must be very small.

(e) The eye-piece produces the image of the object lens itself in the eye and is called eye-ring. An important point about the eye ring is that all the rays from the object refracted by the objective pass through it. Thus ideal position for our eyes for viewing is this eye ring only. The eye will receive all the rays from the object if it is placed at the position of the eye ring, provided area of the pupil of the eye is greater than or atleast equal to the area of the eye ring. In case eye is placed closed to the eye piece, then it would not collect all the rays from the object as the field of view will be reduced. The precise i.e. exact location of the eye ring would depend upon the sepration between the objective and eye piece and also on the focal length of the eye piece

Question 33.

An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eye-piece of focal length 5 cm. How will you set up the compound microscope?

Answer:

Here, f, = focal length of the objective = 1.25 cm.

f_{e}, = focal length of the eye piece – 5 cm

M = angular magnification of the compound microscope = 30

In normal adjustment, the final image is formed at the least distance of distinct vision i.e., D = 25 cm.

∴ For eye piece, M_{e} = 1 + \(\frac{D}{\mathrm{f}_{\mathrm{e}}}\) = 1 + \(\frac{25}{5}\) = 1 + 5 = 6.

Let M_{0} = angular magnification of the objective.

∴ Using the relation, M = M_{0} x M_{e} we get

M_{0} = \(\frac{\mathrm{M}}{\mathrm{M}_{\mathrm{e}}}=\frac{30}{6}\) = 5

∴ M_{0} = \(-\frac{v_{0}}{-\mathrm{u}_{0}}\) gives \(-\frac{v_{0}}{-\mathrm{u}_{0}}\) = 5 or υ_{0} = – 5 υ_{0}

Using lens formula for objective,

i.e. the object should be placed at a distance of 1 cm in front of the objective lens.

∴ υ_{0} = – 5(u_{0}) = – 5 x (-.1.5) = 7.5 cm.

For eye piece, υ_{e} = -25 cm, f_{e} = 5 cm

∴ Using the reltion,

– \(\frac{1}{u_{e}}+\frac{1}{v_{e}}=\frac{1}{f_{e}}\), we get

∴ The compound microscope must be set up such that distance between objective and eyepiece is

Question 34.

A small telescope has an objective lens of focal length 140 cm and an eye-piece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) the final image is formed at the least distance of distinct vision (25 cm)?

Answer:

Here, f_{0} = focal length of objective = 140 cm.

f_{e} = focal length of eyepiece = 5.0 cm

(a) When the telescope is in normal adjustment, the magnifying power is given by :

M = \(\frac{\mathrm{f}_{0}}{\left|\mathrm{f}_{\mathrm{e}}\right|}=+\frac{140}{5}\) = 28

(b) When the final image is formed at the least distance of distinct vision, then M is given by :

Question 35.

(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eye-piece?

(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

(c) What is the height of the final image of the tower if it is formed at 25 cm?

Answer:

Here f_{0} = 140 cm

f_{e} = 5 cm

(a) Separation between the objective and eye-lens is given by

L = f_{0} + f_{e} = 140 + 5 = 145 cm.

(b) Let 0 be the angle made by 100 m tall tower AB at the point of observation O.

Here h = 100 m.

b = 3 km = 3,000 m.

∴ Using the relation, θ = \(\frac{100}{3,000}\) = \(\frac{1}{30}\) radian. ……….(1)

Let y be the height of the image formed by the objective.

∴ AngIe subtended by the image produced by the objective is

\(\frac{y}{f_{0}}=\frac{y}{140}\) ………(ii)

Equating (i) and (ii), we get

\(\frac{y}{f_{0}}=\frac{y}{140}\)

or y = \(\frac{140}{30}=\frac{14}{3}\) cm = 4.7 cm

(c) υ_{e} = D = position of the final image formed by the telescope = 25 cm

if m_{e} be the magnification produced by the eye piece, then

m_{e} = 1 + \(\frac{y}{f_{0}}\) = 1 + \(\frac{25}{5}\) = 1 + 5 = 6

Let I be the height of the final image of the tower =?

O = size of the object for eye piece

= size of the image formed by objective

= \(\frac{14}{3}\) cm.

∴ Using the relation,

m = \(\frac{I}{O}\), weget

m_{e} = \(\frac{I}{O}\)

or I = m_{e} x O = 6 x \(\frac{14}{3}\) = 28cm,

Question 36.

A carsegrain telescope uses two mirrors as shown in Fig. given here. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of larger mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Answer:

Here, R_{1} = radius of curvature of large mirror i.e. concave mirror = – 220 mm = – 22 cm

R_{2} = radius of curvature of small i.e. convex mirror = 140 mm = 14 cm

If f_{1}, and f_{2} be the focal lengths of large and small mirror, then

f_{1} = \(\frac{\mathrm{R}_{1}}{2}=-\frac{22}{2}\) = – 11cm and f_{2} = \(\frac{\mathrm{R}_{2}}{2}=\frac{14}{2}\) = 7 cm

d = spacing between the mirrors = 20 mm = 2 cm

Using the sign conventions, f_{1}, R_{1} are taken as – ve.

u – ∞ as object lies at infinity.

As per ray diagram final image of the object is formed at the back of objective mirror which is seen through eye piece.

Hie rays coming from object at ∞ tend to meet at principal focus of objective but before that they are intervened by the concave mirror of smaller focal length.

For objective :

u = -∞, f = f_{1} = 11 cm υ = ?

Using the relation,

= distance from objective

∴ Distance from convexjnirror

= – (υ + d) = – (-11 + 2) = + 9 cm

= This acts as virtual object for convex mirror

i.e u’ = + 9 cm

f’ = f_{2} = 7 cm

υ’ = distance of final image = ?

Question 37.

Light incident normally on a plance mirror attached to a galvanometer coil retraces backwards as shown in Fig. here. A current in the coil produces deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Answer:

We know that the reflected rays of light get deflected by twice the angle of rotation of the mirror.

From fig., we see that when the mirror is turned from position M

to M’ through angle θ = 3.5°, then the reflected ray OB turns through

∠ 2θ = 2 x 3.5° = 7° = ∠AOB .

Here, OA = 1.5m,

∴Displacement, AB = d = ?

Now in rt. ∠d ∆AOB,

tan 2θ = \(\frac {AB}{AO}\)

or tan 7° = \(\frac {d}{1.5}\)

or d = 1.5 x tan 7°

= 1.5 x 0.1228 = 0.1842 m

= 18.42 cm

Question 38.

Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a place mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Answer:

Here let ft = focal length of double convex lens L = 30 cm.

Let F = Focal length of the combination of double-convex lens L and plano-concave liquid lens L’

∴ F = 45.0 cm, μ = 1.5 = \(\frac {3}{2}\)

Also let f_{2} be the focal length of the plano-concave lens made of liquid between the convex lens and plane mirror.

∴ radii of curvature of the two surfaces of plano-concave lens of liquid formed between foci convex lens and plane mirror are -R and oo, where R is the radius of curvature of the surface of equi-convex lens.

Also, we know that the object uncoincide with its image only if the ray of light retrace their paths i.e. they fall normally on the plane mirror and it can be possible only if the object lies at the focus of the combined lens system.

For lens L,

∴ Using the relation,

∴ For lens L’ i.e. plano-concave liquid lens, let ^{a}μ_{e} be its refractive index = ?

∴ Using the relation,

**Bihar Board Class 12 Physics **Chapter 9 Ray Optics and Optical Instruments Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.

Define refraction of light.

Answer:

It is defined as the process of bending of light from its path when it travels from one medium to the another.

Question 2.

Why a glass sphere in air behaves as a convergent lens?

Answer:

The convergent lens is one which converges the rays of light. The glass sphere in air behaves as a convergent lens because the rays of light falling on the glass sphere converge at a point after refraction from both the surfaces.

Question 3.

State principle of reversibility of light.

Answer:

It states that if the final path of ray of light is reversed after suffering a number of reflections and refractions then it retraces its path i.e., travels back along the same path in the opposite direction. It is shown in the figure here.

Question 4.

Define photometry.

Answer:

It is defined as the branch of physics which deals with the measurement of the intensity of illumination of the source or the intensity of light emitted by a source of light.

Question 5.

Categorise the sources of fight.

Answer:

The sources of light are mainly of the following three types

depending on the wavelength of light they emit.

(a) Thermal sources : They emit light of all wavelength in the visible region, e.g. An incandescent bulb.

(b) Gas discharge tube : It emits light in few wavelength band, e.g., a neon tube light gives out its characteristics red colour while argon gas tube gives its characteristic blue light.

(c) Luminescent sources : Such sources produce light partly in visible region and partly in ultraviolet region.

Question 6.

(a) Define photoluminescence, (b) Name the colours visible to human eye.

Answer:

(a) It is defined as the process of emission of light after absorbing some e.m. radiations. e.g. when UV light falls on phosphorus, zinc sulphate etc., it gets absorbed and visible light is emitted.

(b) The colours which are visible to human eye are violet, indigo, blue, green, yellow, orange and red.

Question 7.

Define : (a) Luminous flux, (b) Luminous intensity, (c) Efficiency of light source.

Answer:

(a) It is defined as the quantity of energy in visible range passing per unit time across a given surface held perpendicular to the radiant energy. It is denoted by ø and its unit is Lumen (lm).

(b) It is defined as the luminous flux per unit solid angle of a source. It is also known as illuminating power and denoted by I Its unit is lumen per steradian (Lm Sr^{-1}) or Candela (C’d)

∴ I = \(\frac{\phi}{\Omega}=\frac{\phi}{4 \pi}\)

(c) It is defined as the ratio of the output power in visible region to the input electrical power.

Its unit is lm w-1 or /m per watt.

Question 8.

Define illuminance or intensity of illumination. On what factors does it depends?

Answer:

It is defined as the amount of visible radiations received per unit area per unit time falling normal to the surface. It is denoted by E.

E = \(\frac{\phi}{A}\) Its unit is lux.

It depends upon the following factors :

(i) luminous intensity of the source (I).

(ii) distance of the surface from the source of light.

(iii) area of the surface.

(iv) inclination of the surface w.r.t. the direction of rays incident on it.

Question 9.

State (a) Laws of reflection.

(b) Laws of refraction.

Answer:

(a) The following are the two laws of reflection :

(i) Angle of incidence is always equal to the angle of reflection.

(ii) The incident ray, reflected ray and normal to the surface at the point of incidence all lie in the same plane.

(b) The following are the two laws of refraction :

(i) The ratio of the sine of angle of incidence to the sine of the angle of refraction is always constant for a given pair of media.

where _{a}μ_{b} is called relative refractive index of medium b w.r.t. a.

(ii) The incident ray, refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane.

Question 10.

Define absolute refractive index of a medium.

Answer:

It is defined as the ratio of the velocity of light in vacuum to the velocity of light in a medium.

i.e., μ = \(\frac {c}{υ}\)

Note : “ii,, becomes air ^ or p for medium V when medium ‘a is air or vacuum.

Question 11.

State new cartesian sign conventions used for mirrors.

(i) All the distances are measured from the pole of the mirror.

(ii) All the distances measured in the direction of incident ray are taken as positive and the distances measured opposite to the incident ray are taken as negative.

(iii) All heights measured perpendicular to the principal axis in upward direction are taken as + ve and those measured in downward direction are taken as – ve.

Note : Direction of incident light is always to be shown falling from left to right.

So distance of the object and real image is always – ve while that of virtual image is always + ve, height of real image is always – ve while that of the virtual image and the size of real object are always + ve.

Question 12.

On what factors does the refractive index of a medium depend?

Answer:

It depends upon the following factors:

(i) The nature of the pair of media.

(ii) Direction of incident light e.g. \({ }^{a} \mu_{b} \neq^{b} \mu_{a}\)

(iii) The wavelength of light used.

Question 13.

Why there is apparent shift in the position of sun?

Answer:

There is an apparent shift in the position of sun due to the atmospheric refraction of light. It becomes visible nearly 2 minutes before actual sunrise and remains visible for nearly 2 minutes after actual sunset. The difference between actual and apparent timings of sunrise and sunset is called apparant shift in the position of sun and is nearly 4 minutes.

Question 14.

Define lateral shift. On what factors does it depend?

Answer:

It is defined as the perpendicular distance between the incident ray and the emergent ray, when the light incident obliquely on a parallel sided refracting slab.

It depends upon the following factors :

(i) thickness of the slab i.e.; it increases with the increase in the thickness of the slab.

(ii) angle of incidence i.e., it increases with the increase in the value of the angle of incidence.

(iii) It increases with the increase in the refractive index of the slab.

Question 15.

Define total internal reflection.

Answer:

It is defined as the process of reflection of light that takes place when a ray of light travelling from denser to rarer medium gets incident at the interface of the two media at an angle greater than the critical angle for the given air of media.

Question 16.

Define critical angle for a pair of media.

Answer:

It is defined as the angle of incidence in the denser medium for which the angle of refraction becomes 90″ in the rarer medium.

Question 17.

What are the conditions for the total internal reflection to take place?

Answer:

The following are the conditions for the total internal reflection to take place:

(i) The light should travel from a denser medium to a rarer medium.

(ii) The angle of incidence in the denser medium must be greater than the critical angle for the given pair of media in contact.

Question 18.

Define mirage.

Answer:

It is defined as an optical illusion that occurs in deserts and coal tarred roads appear to be covered.with water but on approaching at that place no water is obtained. In deserts thirsty animals observe virtual images of trees on hot sand so expecting a pond of water there but on reaching there, they do not get water pond and hence called optical illusion.

Question 19.

Explain twinkling of stars.

Answer:

Different layers of earth’s atmosphere move to and fro and are never at rest and their density changes. Thus the light coming from stars suffers atmospheric refractions and thus their image is formed at different places. Hence they look twinkling.

Question 20.

Why diamond sparkles?

Answer:

Tire critical angle for diamond is low i.e., 23° and its refractive index is 2.47. The faces of diamond are cut in such a way that when a ray of light entering from a face undergoes multiple total internal reflections from its different faces. Due to small value of the critical angle, almost all light rays entering the diamond suffer multiple total internal reflection and thus it shines brilliantly.

Question 21.

Does the apparent depth of a water tank changes if viewed obliquely? If so does the apparent depth increases or decreases?

Answer:

Yes, the apparent depth of a water tank changes if viewed obliquely: The apparent depth for oblique view decreases from its value for near normal viewing.

Question 22.

Do the materials always have the same colour whether viewed by reflected light or through transmitted light?

Answer:

No, all the materials don’t have the same colour. Some materials reflect light of one colour strongly while transmit some other colour strongly. Hence they look different in colour when viewed on both sides.

Question 23.

For the same angle of incidence, the angle of refraction in three media A, B and C are r_{b}, r_{b} and r_{c} s.t. r_{a} > r_{b} > r_{c}. In which medium will the velocity of light be minimum ?

Answer:

We know that

The velocity of light is minimum for minimum value of angle of refraction. Thus as ra for medium A is minimum, so the velocity of light is minimum in medium A.

Question 24.

Mention new cartesian sign conventions for refraction of light from a spherical refracting surface.

Answer:

(i) All the distances are measured from the pole of the spherical refracting surface. .

(ii) All the distances measured in the direction of incident ray are taken as + ve and those measured opposite to the direction of incident ray are taken as – ve.

Question 25.

What the assumptions for refraction of light from spherical refracting surface?

Answer:

(i) The aperture of the refracting surface is small.

(ii) The object is a point object lying on the principal axis.

(iii) The incident and refracted rays make small angles with the principal axis.

Question 26.

Define achromatic combination of lenses.

Answer:

Achromatic aberration for a convex lens is positive while for a concave lens, it is negative. If we combine a concave lens and a convex lens in such a way that the combination is free from chromatic aberration than such a combination is known as achromatic combination of lenses.

Question 27.

What is the condition to obtain achromatism in two tenses?

Answer:

To obtain achromatism in two lens system, fo should be equal to ft so that red and violet rays are focused at the same point, f and f are the focal length of red and violet colours respectively.

Question 28.

On what factors does the focal length of a lens depend?

Answer:

The focal length of a lens depends on the:

(i) radii of curvature of the two surfaces of the lens.

(ii) refractive index of the material of the lens.

Question 29.

A concave mirror and a convex lens are held in water.

What changes, if any, do you expect to find the focal length of the either?

Answer:

Focal length of the mirror will remain same as the laws of reflection are same in water and air but focal length of lens will increase due to change in refractive index.

Question 30.

Why does the goggle lens not have poweralthough surface of the lens are curved?

Answer:

The outer and inner surfaces of the goggle lens are parallel and they have same radius of curvature. As the focal length is given by

\(\frac {1}{f}\) = (μ – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

when R_{1} = R_{2}, \(\frac {1}{f}\) = 0 or f = ∞

Also we know that P = \(\frac {1}{f(m)}\), so P = 0.

Question 31.

Does the focal length of a thin convex lens remain the same when red monochromatic light is used instead of blue light ?

Answer:

No, it will not remain same.

We know that \(\frac {1}{f}\) = (μ – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) and according to Cauchy’s formula, μ = A + \(\frac{B}{\lambda^{2}}\) or μ ∝ \(\frac{1}{\lambda^{2}}\) Also λ_{r} > λ_{b}

∴ µ_{r}< µ_{b}, thus f_{b} < f_{r}. So the focal length of the lens will reduce when we use blue light.

Question 32.

What happens when the angle of incidence is changed by rotating the prism?

Answer:

When the direction of an incident ray of light is kept fixed and the prism is rotated in clockwise direction, then the angle of incidence increases and thus the angle of deviation decreases as shown below. If we go on rotating the prism, the angle of incidence increases’ and the angle of deviation goes on decreasing. For a particular valye of i the angle of deviation 8 becomes minimum and is called angle of minimum deviation (δ_{m}). When the angle of incidence is further increased, the angle of deviation also starts increasing.

Question 33.

Define dispersion of light. What is its cause?

Answer:

It is defined as the process of splitting up of white light into its constituent colours on passing through a prism.

We know that for small angled prism,

δ = (µ – 1)A. ‘

Also according to Cauchy’s formula, we know that µ ∝ \(\frac{1}{\lambda^{2}}\)

Thus µ of the material of prism is different for different colours, so δ is also different for different incident colours. Thus due to different values of angle of deviation, each colour occupies different direction in emergent beam of light and thus constituent colours of white light get dispersed. λ_{v}< λ_{t}, so δ_{v} > δ_{r}.

The violet colour deviates more than the red colour.

Question 34.

Define angular dispersion of light. On what factors does it depend?

Answer:

The angular dispersion for any two colours is defined as the difference in the deviations suffered by the two colours in passing through a prism. It is denoted by θ.

for extreme colours, θ = δ_{v} – δ_{r} = (µ_{v} < µ_{r}) A

It depends on the angle of prism and the refractive index for the two colours.

Question 35.

Define dispersive power of a prism. On what factors does it depend?

Answer:

It is defined as the ratio of the angular dispersion for extreme colours to the deviation suffered by the mean light. It is denoted by ω.

∴ ω = \(\frac{\theta}{\delta}=\frac{\delta_{\mathrm{v}}-\delta_{\mathrm{I}}}{\delta}=\frac{\mu_{\mathrm{v}}-\mu_{\mathrm{r}}}{\mu-1}\)

a depends upon the refractive index of the prism material for the extreme and mean colours and is independent of the angle of prism. Yellow light is called mean colours.

Question 36.

How can we obtain pure spectrum?

Answer:

The following conditions must be satisfied so as to obtain pure spectrum:

(i) A narrow source of light should be used so that minimum number of rays strike the prism.

(ii) Prism must be placed in minimum deviation condition.

(iii) A convex lens L_{1} must be placed in between the source of light and the prism so as to make the incident beam falling on the prism as parallel.

(iv) Another convex lens L_{2} is placed between the prism and the screen so that the rays of same colour converge at the same point after refraction. It is shown in figure given below :

Question 37.

Why is the colour of sky blue?

Answer:

It is due to scattering of blue light. According to Ray leigh’s law of scattering, the intensity of light scattered is inversely proportional to the fourth power of the wavelength of light.

i.e I ∝\(\frac{1}{\lambda^{4}}\)

The light from sun traveling through earth’s atmosphere gets scattered by large number of gaseous molecules. As wavlength of blue light is lesser, so it is scattered more and hence sky appears blue.

Question 38.

Why danger signals are of red light?

Answer:

Scattering of light is inversely proportional to the fourth power of wavelength of incident light. As red light has longer wavelength as compared to other visible colours, so its scattering is least and thus red light signals can be seen from a longer distance.

Question 39.

Define angular vision. On what factors does it depends?

Answer:

It is defined as the angle subtended at the eye by the size of the object seen by the eye. It depends upon :

(i) the size of the object.

(ii) distance of the object from the eye.

Question 40.

Define magnifying power of an optical instrument.

Answer:

It is defined as the ratio of the angle made by the image at the eye to the angle made by the object at eye when both are placed at the least distance of distinct vision.

i.e., M = \(\frac{β}{α}\)

Question 41.

How magnifying power of a simple microscope can be increased?

Answer:

We know that M = 1 + \(\frac{D}{f}\)

Thus M can be increased if (i) focal length (f) of the convex lens is shorter.

(ii) image of the object is formed at the least distance of distinct vision which is possible when it is placed between the focus and pole of lens.

Question 42.

Define an astronomical telescope.

Answer:

It is defined as an optical instrument used to see heavenly bodies such as stars, planets etc clearly.

Question 43.

How magnifying power of an astronomical telescope can be increased for normal adjustment? What is its length in this case?

Answer:

We know that M = \(\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\) for normal adjustment. Thus M will be large if f is smaller and f_{e} is larger where f_{0} and fe are the focal lengths of the objective and eye piece respectively.

The length of astronomical telescope for this case is given by L = f_{0} + f_{e} = focal length of objective + focal length of eye piece.

Question 44.

What are the requirements for a good telescope?

Answer:

The following are the requirements for a good telescope :

- It should have high magnifying power.
- It should have high resolving power.
- It should have large light gathering power.

Question 45.

What are the differences between a real and a virtual image?

Answer:

Real image

- An image formed is said to be real if two rays from a given object after being reflected or refracted from an optical surface actually meet at a point.
- It is always inverted.
- It may be enlarged, diminished or of same size depending upon the position of the object.

Virtual image :

- An image formed is said to be virtual if two rays from a given object after being reflected or refracted from the optical surface appear to meet or come from a point.
- It is always erect.
- It is always of bigger (enlarged) size.

Question 46.

Explain the physical meaning of the power of a lens.

Answer:

By the term, power of a lens we mean converging or diverging power of the lens. When parallel rays are incident on a convex lens or concave lens, then they converge or diverge these rays respectively . Thus convex lens has converging power while the concave lens has diverging power.

Question 47.

Define an equivalent lens.

Answer:

It is defined as a single lens which will form the image of a given object at the same point as is formed by each of the lenses used in the combination.

Question 48.

On what factors does the deviation produced by a prism depend?

Answer:

The deviation produced by the prism in the path of a ray of light depends upon the following factors :

- angle of prism.
- angle of incidence.
- material of the prism i.e., refractive index of the material of the prism.

Question 49.

Under what conditions, the magnifying power of astronomical telescope is :

(a) minimum

(b) maximum.

Answer:

The magnifying power of astronomical telescope is :

(a) minimum when the final image is formed at infinity.

M = – \(\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\)

(b) maximum when the final image is fonned at the least distance of distinct vision and is given by :

M = \(\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{D}}\right)\)

Short Answer Type Questions

Question 1.

Prove that ^{a}μ_{b} = \(\frac{1}{{ }^{b} \mu_{\mathrm{a}}}\)

Answer:

Let OA be an incident ray, incident at angle i on the interface at point A and gets refracted at angle r along AB in water.

Let a plane mirror be placed normal to the ray AB at point B. So, it will be reflected back along BA and it suffers refraction at point A along AO. Thus, the ray retraces its original path in opposite direction.

Applying Snell’s law at point A when the refraction takes place from medium a to b, we get

Again applying Snell’s law at A when refraction takes place from b to a, we get

Multiplying (i) and (ii), we get

Hence, proved.

Question 2.

Derive the relation between refractive index of three media when light is passing through their combination.

Answer:

Let a ray of light OA passes from air medium (a) to the combination of two media water (b) and glass (c). It gets refracted along AB and BC respectively in medium (b) and (c) and finally emerges out along CD in air s.t. CD||OA.

Let i = angle of incidence at A. r_{1} and r_{2} are the angles of refraction at point A and B respectively. As CD||OA, so ∠NCD = i.

Applying Snell’s law at points A, B and C, we get

Multiplying (i), (ii) and (iii), we get

Question 3.

Derive the relation

Answer:

Let O be an object placed at the bottom of a beaker full of water. A ray GA incident normalLy on the interface XY between water and air from the object O travels along A A’ without refraction as ∠i = 0, so ∠r = 0. Another ray OB incident at an angle of incidence i on XY gets refracted at angle r along BC away from normal NN’. When BC is produced backward, it meets OA at point I. Thus 1 is the virtual image of the point object O.

∠AIB = r and ∠AOB = i

∴AO = real depth of the object

Al = apparent depth of the object.

Applying Snell’s law at point B when refraction takes place from b to a, we get

Also we know that

from (i) and (ii), we get

Now in rt. ∠d ∆s AOB and AIB,

from (iii) and (iv)

Also as the point B is close to A, so BO ≈ AO and BI ≈ AI

Question 4.

Prove that ^{a}μ_{b} = \(\frac{1}{\sin C}\) where C is the critical angle.

Answer:

We know that when a ray of light travels from denser to rarer medium, it bends away from the normal and thus angle of refraction r is always greater than angle of incidence. As per definition of critical angle, when i = C, r becomes 90° in the rarer medium and the refracted ray travels along the interface between two media.

In the figure given here, when ∠OBN = C, ∠NBC = r = 90° and the ray OB travels from water (b) to air (a), so applying Snell’s law at point B, we get

Also we know that

∴ from (1) and (2), we get

where ^{a}μ_{b} = ^{a}μ_{w} = R.I. of water w.r.t. air.

Question 5.

Derive the expression for the lateral shift produced by a * glass slab in the path of the light.

Answer:

The ray diagram is shown below :

Let t = thickness of the glass slab having R.I. p.

OA = incident ray travelling from air to glass.

AB = refracted ray.

BC = emergent ray which is parallel to the ray OA.

Draw BK normal to the initial path OAL from point B. If d = BK, then d is called lateral shift.

Now in rt. ∠d ∆AKB,

Also in rt. ∠d ∆AN’B,

∴ from (i) and (ii), we get

d = \(\frac{\mathbf{t}}{\operatorname{cosr}}\). sin (i – r)

which is the required expression for ‘d’.

Question 6.

Define lens maker’s formula. State assumptions and new cartesian sign conventions used to derive it.

Answer:

Defintion : It is defined as the formula used to manufacture a lens of a particular focal length from the glass of given refractive index.

Assumptions :

- The lens is thin.
- All the distances are measured from the optical centre of the lens.
- The aparture of the lens is small.
- The object is a point object lying on principal axis of the lens.
- The incident ray, refracted ray and the normal all make small angles with the principal axis.

Sign. Conventions : (i) All the distances measured in the direction of incident ray are taken as positive and those measured in the direction opposite to the incident ray are taken as negative.

(ii) The size of the object or image measured in the upward direction is taken as positive while those measured in downward direction are taken as negative.

Question 7.

Derive lens maker’s formula.

Answer:

Consider a concave lens made of a material hav ing absolute refractive index µ_{2} placed in an optically rarer medium of refractive index µ_{1}and bound by two surfaces XP_{1}, Y and XP_{2} Y having radii of curvatures R_{1} and R_{2} respectively.

Let O be the point object lying on the principal axis. A ray OA incident on the spherical surface XP_{1}Y gets refracted along AI_{1} in the absence of XP_{2}Y and thus I_{1} is the real image of O. Here refraction takes place from rarer to denser medium, so using refraction formula, we get

where R_{1} = radius of curvature of surface XP_{1}Y = P_{1}C_{1} = CC_{1}

u = P_{1}O = CO

and υ’ = P_{1}l_{1} ≈ CI_{1}.

Refraction through XP_{2}Y :

For the spherical surface XP_{2}Y, the acts as the virtual object placed in a denser medium of refractive index µ_{2} and the ray AB gets refracted along BI. Thus I is the final image formed by the lens in the rarer medium of refractive index µ_{1}

Applying the formula for refraction through concave surface XP_{2}Y from denser to rarer medium, we get

Adding (i) and (ii), we get

Also from lens formula, we know that

– \(\frac{1}{\mathrm{u}}+\frac{1}{v}=\frac{1}{\mathrm{f}}\) ………..(iv)

∴From (i) and (ii), we get

\(\frac {1}{f}\) = (µ – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ……….(v)

Where µ = \(\frac{\mu_{2}}{\mu_{1}}\) is called absolute refractive index of the refracting media Equestion (4) is known as lens maker’s formula

Question 8.

Define and derive expressions for first and second principal focus of a concave lens.

Answer:

First principal focus :

It is defined as the position of the object on the principal axis of a lens for which the image is formed at infinity. It is denoted by F_{1}.

∴ u = CF_{1} = -f_{1}, and v = ∞

∴ From equation – \(\frac {1}{u}\) + \(\frac {1}{v}\) = (µ – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ……..(i)

we get ,

Second principal focus : It is defined as the position of the image on the principal axis of a lens for which the object lies at infinity. It is denoted by F_{2}

i.e., Here, u = ∞, υ = + f_{2}

∴ From equation (i), we get

∴ from (ii) and (iii), we get

or f_{1} = f_{2} = f (say)

Thus two focal length of the lens are same.

Question 9.

Derive the expression for linear magnification of a lens.

Answer:

Consider a convex lens of focal length f. Let AB be an object of size O placed on its principal axis at a distance u and A’B’ be its image of size I formed at a distance t’ from its optical centre as shown in the figure. Now triangles A’B’C and ABC are similar.

∴\(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{CA}^{\prime}}{\mathrm{CA}}\) …….(i) Applying new cartesian sign conventions, we get

A’B’ = -I

AB = +O

CA = -υ

CA’ = +υ ……(ii)

∴ From (i) and (ii), we get

\(\frac{-1}{+O}=\frac{+v}{-u}\) ……..(iii)

If m be the linear magnification, then by definition,

m = \(\frac{\mathrm{I}}{\mathrm{O}}=\frac{v}{\mathrm{u}}\) ……..(iv)

m in terms of u and f :

From lens formula,we know that

∴ \(\frac{1}{\mathrm{u}}+\frac{1}{v}=\frac{1}{\mathrm{f}}\) …….(v)

Multiplying on both sides of (y) by u, we get

m in terms of υ and f :

(V) x υ gives

Question 10.

Prove that \(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) for two thin lenses placed in contact with each other.

Answer:

Let f_{1} and f_{2} be focal lengths of the two thin lenses L_{1} and L_{2} respectively placed in contact with each other.

Final image l of the point object O lying on principal axis of the combination of lenses is formed in two steps :

(i) L_{1} produces a real image V of the object O.

∴ Using lens formula for lens L_{1}, we get

– \(\frac{1}{u}+\frac{1}{v^{\prime}}=\frac{1}{f_{2}}\) ……..(i)

(ii) The lens L_{2} produces the image I of the virtual object Fata distance v from the combination i.e, υ = C_{2}I ≈ C_{1}I

∴ using lens formula for lens L_{2}, we get,

– \(\frac{1}{v^{\prime}}+\frac{1}{v}=\frac{1}{\mathrm{f}_{2}}\) ……(ii)

Adding (i) and (ii), we get

\(\frac{1}{\mathrm{u}}+\frac{1}{v}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}\) …….(iii)

Question 11.

Prove that A + δ = i + e for a prism.

Answer:

Let a ray of light PQ be incident on the face AB of the prism at angle of incidence i. Let it gets refracted along QR at angle rr The refracted ray QR suffers refraction along RS at face AC. So RS is called emergent ray.

Let e = angle of emergence.

r_{2} = angles of refraction of face AC = ∠QRO when RS is produced backward it meets PQ at point T at angle δ. So δ = angle of deviation = ∠KTS.

r_{1} = ∠OQR.

∴ ∠TQR = i – r_{1} and ∠TRQ = e – r_{2}.

Now in ∆TQR,

∠KT’R = ∠TQR + ∠TRQ

or δ = i – r_{1} + e – r_{2} = i + e – (r_{1} + r_{2}) …….(i)

Also in ∆QRO,

∠QOR + r_{1} + r_{2} = 180° ……(ii)

and in quadrilateral ∆QOR,

∠A + ∠O + ∠AQO + ∠ARO = 360°

or ∠A + ∠O + 90° + 90° = 360°

∠A + ∠O = 180° …..(iii)

∴ from (ii) and (iii), we get

∠r_{1} + ∠r_{2} + ∠O = ∠A +∠O

or A = r_{1} + r_{2} …….(iv)

∴from (i) and (iv), we get

δ = i + e – A

or A + δ = i + e

Hence, proved.

Question 12.

Derive the relation for refractive index of the prism in terms of angle of minimum deviation.

Answer:

When the prism is set in the position of minimum deviation,

i.e., when δ – δ_{m}, i = e, r_{1}; = r_{2} = r (say)

Also we know that

A = r_{1} + r_{2} = r + r = 2r

∴ r = \(\frac {A}{2}\) ……….(i)

and A + δ = i + e

or A + δ_{m} = i + i = 2i

∴ i = \(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\) ……..(ii)

If μ be the refractive index of the material of the prism, then according to Snell’s law,

Question 13.

Derive the expression for the deviation through a small angled prism.

Answer:

Same diagram of Q. 11.

Applying Snell’s law at face AB, we get

As the angle of prism is small, so the ray of light is incident at a small angle i, r_{1} will also be small. So sin i ≈ i and sin r_{1} = r_{1}

from (i), μ = \(\frac{\mathrm{i}}{\mathrm{r}_{1}}\) …(ii)

or i = μr_{1}

For refraction at face AC,

As the angle of prism is small, so e and r_{2} are also small, thus sine ≈ e and sin r_{2} ≈ r_{2}.

∴ from (iii), μ = \(\frac{\mathrm{e}}{\mathrm{r}_{2}}\), or e = μr_{2} ……(iv)

Adding (ii) and (iv), we get

i + e = μ(r_{1} + r_{2} ) …….(v)

Also we know that for a prism,

A = r_{1} + r_{2} ……(vi)

and i + e = A + δ ……(vii)

∴ from (v), (vi) and (vii), we get

A + δ = μA

or δ = μA – A = (μ – 1) A.

∴ δ = (μ -1) A

which is the required expression for 8.

Question 14.

Explain why a ray of light travelling from air to glass does not suffer total internal reflection.

Answer:

For light to suffer total internal reflection, it should travel from denser to rarer medium. But here light is travelling from rarer to denser medium so total internal reflection is not possible.

Question 15.

If a lens is used in a medium of which it is made up, how will it behave and what will be value of its power and focal length?

Answer:

In this case refractive index of lens w.r.t. medium is 1.

[μ_{1} = μ_{2}

∴ μ = [\(\frac{\mu_{2}}{\mu_{1}}\) = 1]

By lens-maker formula we know that

\(\frac{1}{f}\) = (μ – 1) \(\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\)

putting μ = 1, we get

\(\frac{1}{f}\) = 0

or f = \(\frac{1}{0}\) = ∞

or lens behaves as a plate having zero power and infinite focal length.

Question 16.

A lamp is placed at a certain distance from screen. If the source is made 4 times stronger and its distance is doubled from screen. What will happen to intensity?

Answer :

We know that E = \(\frac{\mathrm{I}}{r^{2}}\)

From this formula E ∝ I

If I is made four times, E also becomes four times.

Also E ∝ \(\frac{\mathrm{I}}{r^{2}}\)

If r is doubled, E becomes \(\frac{1}{4}\) times.

Combining both these facts, E remains same.

Question 17.

Explain by sketches the formation of (i) real enlarged image and (ii) a virtual image in case of concave mirror.

Answer:

In Figure (I) image is real, inverted and magnified.

En figure (ii) image is virtual, erect and magnified

Question 18.

Name the five features of the image formed by a plane mirror.

Answer:

- It is always virtual.
- It is always erect.
- It is of the same size as the object.
- It is formed as far behind the mirror as the object is infront of the mirror.
- It is laterally inverted.

Question 19.

Define critical angle and calculate its value for light going from a medium of p = 4l into air.

Ans. It is that angle of incidence in denser medium for which angle of refraction in rarer medium is 90”.

We know that μ = \(\frac{1}{\sin C}\)

Here μ = \(\sqrt{2}\)

∴ \(\sqrt{2}\) = \(\frac{1}{\sin C}\)

sin C = \(\frac{1}{\sqrt{2}}\)

or C = 45°.

Question 20.

What are optical fibres? Give their one use.

Answer:

Optical fibres are thousands of very fine quality fibres of glass or quartz. The diameter of each fibre is of the order of 10 4 cm having refractive index of material equal to 1.7. These are coated with a thin layer of material having μ = 1.5.

They are used in transmission and reception of electrical signals by converting them first into light signals.

Question 21.

Prove that M = \(\left(1+\frac{D}{f}\right)\) for simple microscope.

Answer:

The ray diagram of simple microscope is shown here. We know that the magnifying power of the simple microscope is given by

M = \(\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}\) ……..(i)

Now in rt. ∠d ∆s, CA’Q and CAB, we get

tan α = \(\frac{\mathrm{AQ}}{\mathrm{CA}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{CA}^{\prime}}\) …….(2)

and tan β = \(\frac{\mathrm{AB}}{\mathrm{CA}}\) ………….(3)

∴ from (1), (2) and (3), we get

M = \(\frac{\mathrm{AB} / \mathrm{CA}}{\mathrm{AB} / \mathrm{CA}^{\prime}}=\frac{\mathrm{CA}^{\prime}}{\mathrm{CA}}\)

= \(\frac{-D}{-\mathbf{u}}=\frac{D}{u}\) ……….(4)

From lens formula,

or M = 1 + \(\frac{D}{f}\) ………..(5)

which is the required expression for M, here D is the least distance of distinct vision, f = focal length of the simple microscope.

Question 22.

Prove that M = – \(\frac{\mathbf{L}}{\mathbf{f}_{0}}\left(\mathbf{1}+\frac{\mathbf{D}}{\mathbf{f}_{\mathrm{e}}}\right)\) for a compound microscope.

Answer:

The ray diagram of the compound microscope is shown here

Its magnifying power is given by :

M = \(\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}\)

Now in rt. ∠d ∆ C’A” Q, tan α = \(\frac{\mathrm{A}^{\prime \prime} \mathrm{Q}}{\mathrm{C}^{\prime} \mathrm{A}^{\prime \prime}}=\frac{\mathrm{AB}}{\mathrm{C}^{\prime} \mathrm{A}^{\prime \prime}}\)

Also in rt. ∠d ∆ C’A” B”,tan β = \(\frac{\mathrm{A}^{\prime \prime} \mathrm{B}^{\prime \prime}}{\mathrm{C}^{\prime} \mathrm{A}^{\prime \prime}}\)

M = \(\frac{\mathrm{A}^{\prime \prime} \mathrm{B}^{\prime \prime} / \mathrm{C}^{\prime} \mathrm{A}^{\prime \prime}}{\mathrm{AB} / \mathrm{C}^{\prime} \mathrm{A}^{\prime \prime}}\)

= A’B”/AB = \(\frac{A^{\prime \prime} B^{\prime \prime}}{A^{\prime} B^{\prime}} \times \frac{A^{\prime} B}{A B}\)

= m_{e} x m_{0} ……..(1)

where m_{e} = \(\frac{\mathrm{A}^{\prime \prime} \mathrm{B}^{\prime \prime}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}\) is the linear magniíication produced by eyepiece.

and m_{0} = \(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}\) is the linear magnification produced by objective.

Now m_{0} = \(\frac{v_{0}}{\cdots u_{0}}=-\frac{v_{0}}{u_{0}}\) ……….(2)

and m_{e} can be found as follows :

Using lens formula for eyepiece, we get

As υ_{e} = – D ∴m_{e} = \(\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)\) ……..(3)

∴ From (1), (2) and (3), we get

M = – \(\frac{v_{0}}{u_{0}} \cdot\left(1+\frac{D}{f_{i}}\right)\)

Now the focal length of the objective is very small, so u_{0} = f_{0}, and the image A’B’ must be formed near the eyepiece i.e., v_{0} = L = length of microscope tube.

∴ M = \(\frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)\)

When the final image is formed at infinity, then m_{e} = 1 + \(\frac{\mathrm{D}}{\mathrm{t}_{\mathrm{e}}}=\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{C}}}\)

as f_{e} < < D.

∴ M = \(\frac{v_{0}}{\mathrm{u}_{0}} \times \frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}=\frac{\mathrm{L}}{\mathrm{f}_{0}} \times \frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\)

Question 23.

What conclusions you draw from the equivalent focal length of the combination of two lenses?

Answer:

We know that the focal length of the combination of two lenses is given by

\(\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}\)

We conclude from this expression that :

(a) The power of the combination is given by.

P = P_{1} + P_{2} .

When P_{1} and P_{2} are the powers of lenses having focal length f_{1} and f_{2} respectively, P is the power of combination of lenses. When the two lenses are placed in contact, then the image produced by first iens acts as the object for the second lens and thus produces its final image. So the equivalent magnification produced by the combination is given by

m = m_{1} x m_{2}

(b) If two lenses are combined by placing them at finite distance d apart, then F is given by :

\(\frac{1}{\mathrm{~F}}:=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}} \cdot \frac{\mathrm{d}}{\mathrm{f}_{1} \mathrm{f}_{2}}\)

(c) Two lens combined may :

(i) produce magnified image of the object

(ii) produce an erect image.

(iii) minimise the lens defects such as chromatic aberration or spherical aberration.

Question 24.

Derive the refraction formula for (he convex spherical surface when the refraction takes place from rarer to denser medium.

Answer:

Let O be a point object lying in the rarer medium of R.I. μ_{1} on the principal axis of the convex spherical surface. Let I be its real image formed in the denser medium of R.I. μ_{2}. The various angles are shown in the figure.

Draw AN perpendicular to the principal axis from point A. Now in ∆ AOC,

i = ∝ + γ ……..(i)

Now as the aperture of the spherical refracting surface is small so a, y can be replaced bys their tangents.

i.e., i = tan α + tan γ …….(ii)

From right angled ∆s ANO and ANC, we get

∴ from (ii) and (iii), we get

i = \(\frac{\mathrm{AN}}{\mathrm{PO}}+\frac{\mathrm{AN}}{\mathrm{PC}}\) ……….(iv)

Also in ∆ AIC,

γ = r + β

or r = γ – β

= tan γ – tan β

Also in rt. angled triangle ANC and ANI,

∴ From (v) and (vi), we get

r = \(\frac{\mathrm{AN}}{\mathrm{PC}}-\frac{\mathrm{AN}}{\mathrm{PI}}\) ……(vii)

Let μ be the relative refractive index of the denser medium w.r.t. the rarer medium.

For small angles î and r,

sin i ≈ i and sin r ≈ r.

∴ From (viii),

Using new cartesian sign conventions,

P0 = -u

PC = + R

PI = + υ, we get

Dividing on both sides of (x) by i, we get

which is the required refraction formula for convex spherical surface when the refraction takes place from rarer to denser medium.

Question 25.

Derive refraction formula for convex spherical surface for a virtual image when the refraction takes place from rarer (µ_{1}) to denser medium (µ_{2})

Answer:

The refraction formula is

Prove yourself.

[Hint: both the object and virtual image lie in the rarer medium].

Question 26.

Derive the refraction formula for concave spherical surface for an object lying in rarer medium.

Answer:

In this case, the image is formed in rarer medium as shown in the ray diagram. I is the virtual image.

The refraction formula is

– \(\frac{\mu_{1}}{\mathrm{u}}+\frac{\mu_{2}}{v}=\frac{\mu_{2} \cdot \mu_{1}}{\mathrm{R}}\)

Derive yourself.

Question 27.

Write a short note on the common defects of vision. Ans. The following are the common defects of vision :

Answer:

(a) Short sightedness or myopia :

If is the defect when the eye can see the nearby objects clearly but the distant objects are not clearly seen. For normal eye, the farthest point lies at infinity. But due to this defect, this point shifts towards the eye and the ray coming from infinity are focussed at point R’ before the retina as shown in the ray diagram So the object does not appear clear. This defect can be removed by using spectacles of diverging nature or concave lens of focal length equal to the distance of the far point of the myopic eye as shown in the diagram.

(b) Long sightedness or Hypermetropia :

It is the defect when the eye can see the distant objects clearly but the near objects are not clearly seen.

The object placed at a distance of 25 cm is not clearly visible as its image is formed at a point T, behind the retina which can be possible when the focal length of the eye lens becomes large or eye ba i I becomes short. This defect may be corrected by using a convex lens of focal length /given by

\(\frac{1}{f}=\frac{1}{\mathrm{~d}}-\frac{1}{\mathrm{D}}\)

where D = 25 cm, d = least distance of distinct vision for defective eye as shown in the ray diagram.

(c) Presbyopia : The power of accomodation of the eye lens for old persons decreases, so that neither near not distant objects are clearly visible. This defect may be avoided by using bifocal lenses.

Question 28.

(a) List some advantages of a reflecting telescope, especially for high resolution astronomy.

(b) A reflecting telescope has a large mirror for its objective with radius of curvature equal to 80 cm. What is the magnifying power of the telescope if the eye-piece used has a focal length of 1.6 cm?

Answer:

(a) The following are some of the advantages of a reflecting type telescope especially for high resolution astronomy.

(i) The objective is a concave spherical mirror and not a lens, so it is free from chromatic aberration.

(ii) The spherical aberration is minimised by using a parabolic mirror in place of the concave spherical mirror as objective.

(iii) The image is brighter compared to that in a refracting type telescope.

(iv) A concave spherical mirror of large aperture can be easily made. Such a telescope will possess high light gathering power and hence can be used to see even faint stars as it has high resolution.

(v) Mirror requires grinding and polishing of only one side.

(vi) There is no absorption of light as it is in case of lenses.

(b) Here, R = radius of curvature of objective of the reflecting telescope = .80 cm.

f_{e} = focal length of its eye piece = 1.6 cm.

M = magnifying power of telescope = ?

We know that f = \(\frac {R}{2}\) ,

∴ f_{0} = \(\frac{\mathrm{R}}{2}=\frac{80}{2}\) = – 40 cm (by signconventions, f, R are – ve)

∴ Using the relation, M = \(\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\) we get

M = \(\frac{40}{1.6}\) = 25

Question 29.

Explain why

(a) A diamond glitters in a brightly lit room but not in a dark room.

(b) A crack in window-pane appears silvery.

(c) The bubbles of air rising up in a water tank appear silvery when viewed from top.

Answer:

(a) The glittering of diamond is due to total internal reflection of light in a brightly lit room. The refractive index for diamond is 2.42 so that critical angle for diamond air interface is 24.4°. The diamond i cut suitably so that light entering the diamond from any face at angle > 24.4° suffers multiple total internal reflections at various faces of the diamond and thus it remains with in it. But in a dark room, no light ray is there to undergo multiple total internal reflections. Hence the diamond does not shine in a dark room.

(b) A crack in window-pane appears silvery due to total internal reflection of light. When light is incident on the crack at an angle greater than critical angle, then it suffers total internal reflection i.e., gets reflected back as if the crack is a mirror.

(c) The bubbles of air rising up in a water tank appear silvery i.e., shine brightly due to total internal reflection from them. When light rays travel from water to air bubble i.e., from denser to rarer medium, light suffers total internal reflection at the surface of the air bubble for angle of incidence for than critical angle and reflects back as if air bubble is a miror. Hence bubbles appear silvery when viewed from the top.

Long Answer Type Questions

Question 1.

Derive the expression for the magnifying power of a telescope for the normal adjustment and when the final image is formed at the least distance of distinct vision.

Answer:

(a) When the final image is formed at infinity (normal adjustment)

When a parallel beam of light rays falls on the objective lens from a distant object, its real and inverted image is formed at A’B’ at a distance f_{0} from it. If the position of eye piece is so adjusted that A’B’ lies at its focus, then the final image will be formed at infinity and the telescope is said to be in normal adjustment. As the object is at infinity, so the angle made by it at the eye i.e. at C_{2} is equal to the angle made by it at the objective i.e. at C_{1}, because C_{1} is close to C_{2}.

∴ ∠A’C_{1}B’ = α and ∠A’C_{2}B’ = β

By definition, the magnifying power is given by

M = \(\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}\) …….(1)

Now in rt. ∠d ∆s A’B’C_{1} and A’B’C_{2}.

tan α = \(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{C}_{1} \mathrm{~A}}\) …….(2)

and tan β = \(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{C}_{2} \mathrm{~A}^{\prime}}\) ……(3)

∴ from (1), (2) and (3), we get

Where C_{1} A’ = f_{0} = focal length of objective lens,

and C_{1}A’ = – f_{e} = focal length of eye lens.

– ve signs shows that the final image is inverted.

(b) When the final image is formed at least distance of distinct vision. The ray diagram is shown below :

Here, the final image A”B’ is formed at the least distance of distinct vision i.e., D.

M = \(\frac{B}{\alpha}=\frac{\tan \beta}{\tan \alpha}\)

= \(\frac{A^{\prime} B^{\prime} / C_{2} A^{\prime}}{A^{\prime} B^{\prime} / C_{1} A^{\prime}}=\frac{C_{1} A^{\prime}}{C_{2} A^{\prime}}=\frac{+f_{0}}{-u_{e}}\) …….(i)

where C_{1}A’ = f_{0} = focal length of the objective lens,

and C_{2}A’ = – u_{e} = distance of A’B’ from the eye lens, here A’B’ acts as an object for eye lens.

Now for eye lens, the lens formula is given by

\(\frac{1}{\mathrm{u}_{\mathrm{e}}}+\frac{1}{v_{\mathrm{e}}}=\frac{1}{\mathrm{f}_{\mathrm{e}}}\) …….(2)

Using new cartesian sign conventions,

u_{e} = – u_{e} v_{e} = – D, f_{e} = + f_{e}

∴ from (2) and (3), we get

∴ from (1) and (4), we get

M = – \(\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{D}}\right)\)

which is the required expression.

Numerical Problems

Question 1.

Where should an object be placed in front of concave mirror of focal length f so that the image to be of the same size as that of the object?

Answer:

Let O and I be the size of the object and image resepetively.

∴ O = I and M = linear magnification = \(\frac {I}{O}\) = 1 = – \(\frac {υ}{u}\).

or υ = – u

where υ and u are the distances of the image and object respectively from the concave mirror.

Let f be the focal length of the concave mirror.

∴ Using sign conventions, for concave mirror, u, fare – ve and υ = +ve.

Using mirror formula, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) , we get

where R = radius of curvature of the mirror.

Thus the object must be placed at the centre of curvature of the mirror.

Question 2.

A square wire of side 3.0 cm is placed 25 cm away from the concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire?

Answer:

Here, O = size of the object = 3.0 cm

u = distance of the object = – 25 cm .

f = focal length of mirror = – 10 cm (for concave mirror)

A = Area enclosed by the image = ?

Using mirror formula,

Here -ve sign shows that th the mirror is formed to the left of the mirror

Also we know that

– ve sign shows that the image is inverted.

∴ A = t^{2} = 4 cm^{2}.

Question 3.

A rod of length 5 cm lies along the principal axis of the concave mirror of focal length 15 cm in such a way that the end closer to the pole is 30 cm away from it. Find the length of the image.

Answer:

Here, f = focal length of concave mirror = – 15 cm.

∴ R = 2f = 2 x (- 15) = – 30 cm.

Also u’ = distance of the end of the rod closer to the mirror i.e., of end B = – 30 cm

which means that the closer end B of the rod AB lies at the centre of curvature C of the mirror.

Let u = distance of the end A from mirror = u’ + BA = -30 + (-5) = -35 cm

Let υ be the distance of the image A’ of the end A, l = A’B = length of image = ?

∴ Using mirror formula, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) we get

i.e., A’ lies between C and P of the mirror.

∴ l = PB – PA’ = PC – PA’ = 30 – 26.25 = 3.75 cm.

Question 4.

A object is kept in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate the possible distances of the object from the mirror.

Answer:

Here, f = focal length of the mirror = – 20 cm.

Let I, O be the size of the image and object respectively.

∴ I = 3 x O = 3O

u = distance of the object from the mirror = ?

The following two possible cases arise :

Case I :

When the image formed is real :

Using mirror formula,

Case II. When the image formed is virtual.

∴ From eq (1) we get

Thus the possible distances of the object are 13.33 cm and 26.67 cm towards left of the mirror i.e. both these distances are – ve.

Question 5.

A fish at a depth of \(\sqrt{7}\) cm below the surface of water sees the outside world through a circular horizon. Calculate the radius of

this circular horizon. Refractive index of water w.r.t. air is \(\frac{4}{3}\)

Answer:

Here, O = Position of the fish.

OP = \(\sqrt{7}\) cm = distance of fish below water = real depth.

u = \(\frac{4}{3}\) = R.I. of water

Diamter of circular horizon through which the fish sees the outside world = QR.

If r’ be its radius = ?

Then r’ = PR

C = Critical angle

r = angle of refraction = 90°

∴ Using the relation, μ = \(\frac{1}{sin C}\), we get

sin C = \(\frac{1}{μ}\) = \(\frac{3}{4}\) ……….(1)

Also in rt. angled ∆ OPR,

sin C = \(\frac{PR}{OR}\) ………(2)

∴ from (1) and (2), we get

\(\frac{\mathrm{PR}}{\mathrm{OR}}=\frac{3}{4}\)

Question 6.

A beam of light strikes a glass sphere of 20 cm diameter converging towards a point 40 cm behind the pole of the spherical surface. Find the position of the image if the refractive index of glass is 1.5.

Answer:

Diameter of glass sphere = 20 cm

R = radius of the glass sphere = \(\frac{D}{2}\) = \(\frac{20}{2}\) = 10 cm

μ_{1} = 1, μ_{2} = 1.5

In the absence of the glass sphere, the rays of light converge at O at a distance of 40 cm from the pole P of the sphere which are focussed at point I in its presence. Thus l is the real image of the virtual object O and refraction takes place from rarer to denser medium.

u = PO = + 40 cm

υ = distance of image I from P = ?

∴ Using the relation,

Thus the image is formed at the surface of the glass sphere as shown in the figure drawn.

Question 7.

The radii of curvature of two surfaces of a double convex lens are 10 cm and 20 cm respectively. Calculate the power of the lens if the refractive index of the material of the lens be 1.5.

Answer:

Here, R_{1} = 10 cm, R_{2} = – 20 cm.

μ = 1.5

P = Power of lens = ?

Using lens maker’s formula,

Question 8.

A convex lens gives a real image 4 times as large as an object when the object is at certain distance from the lens. When the object is moved away from the lens by 2 cm from its original position, the magnification is only 2 times. Find

(a) the focal length of the lens power of the lens.

(b) original position of the object (C).

Answer :

Here, m linear magnification produced by lens = 4.

Let u be the distance of the object and υ be the distance of its image from the lens.

∴Using the relation, m = \(\frac {υ}{u}\) ,we get

4 = \(\frac {υ}{u}\) or υ = 4u. ……….(1)

(a) f = focal length of the lens = ?

Using lens formula,

– \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) ………(2)

We get,

(b) Let u’ be the new distance of the object from lens when moved 2 cm away.

∴ u’ = u + 2

and m = 2

If υ’ be the distance of the image produced in this case, then

∴Using eq (2), we get \(\frac{1}{\mathrm{f}}=-\frac{1}{\mathrm{u}^{\prime}}+\frac{1}{v^{\prime}}\)

∴ From (3) and (5), we get

∴ From (3)

f = – \(\frac {4}{3}\) x (-6) = 8 cm

(c) let P be the power of the lens

Here f = 8 cm = \(\frac {8}{100}\) m

∴Using the relation,

Question 9.

Two lens of focal lengths 20 cm and – 25 cm are placed in contact. Calculate the power of the combination.

Answer:

Here, f_{1} – 20 cm, f_{2} = – 25 cm

If P_{1} and P_{2} be the powers of the two lenses respectively, then

P_{1} = \(\frac{100}{\mathrm{f}_{1}}\) D and P_{2} = \(\frac{100}{\mathrm{f}_{2}}\) D

or P_{1} = and \(\frac{100}{20}\) = 5D, and P_{2} = \(\frac{100}{-25}\) = – 4D

Let P be the power of the combination, then

Using the relation P = P_{1} + P_{2}, we get

P = 5 + (-4) = 5 – 4 = +1D

Question 10.

Two plano-concave lenses of glass of refractive index 1.5, each of radii of curvature 15 cm and 20 cm respectively placed in contact with the curved faces towards each other. The space between the curved faces is filled with a liqud of refractive index 1.65. Find the focal length of the combination.

Answer:

Here for first plano-concave lens, R_{1} = ∞ R_{2} = 15 cm, μ = 1.5

If f_{1} be its focal length, then using the relation,

f_{1} = – 30 cm.

For second plano-concave lens, R_{1} = – 20 cm, R_{2} = ∞, μ = 1.5

If f_{2} be its focal length, then using, \(\frac{1}{\mathrm{f}_{2}}\) = (μ -1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) , we get

f_{2} = – 40 cm.

When the two lenses are placed in contact with both curved surfaces in contact and liquid is filled in the space then the lens becomes as shown in the figure here.

Now μ_{1} = 1.65, R_{1} = 15 cm, R_{2} = – 20 cm.

If f_{1}, be the focal length of the lens made of liquid, then using

If F be the focal length of the combination, then using

Question 11.

A convex lens of focal length 10 cm and a concave lens of focal length 20 cm are kept 5 cm apart. Find the focal length of the equivalent lens.

Answer:

Here, f_{1} = 10 cm, f_{2} = – 20 cm, d = 5 cm.

F = focal length of the combination =?

Using the relation,

Question 12.

A convex lens of focal length 0.2 m and made of glass (μ = 1.5) is immersed in water (μ = 1.33). Find the change in focal length of the lens.

Answer:

Here, f_{a} = focal Length of lens placed in air = 0.2 m = 20 cm

^{a}μ_{g}= 1.5, ^{a}μ_{w} = 1.33

If ^{w}μ_{g} be the refractive index of glass w.r.t. water, when the lens is immersed in water, then

Using the relation, ^{w}μ_{g} = \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) we get

^{w}μ_{g} = \(\frac{1.5}{1.33}\)

∆f = change in focal length of the lens = f_{w} – f_{a} = ?

Let R_{1} and R_{2} be the radii of curvatures of the two surfaces of the convex lens.

∴ Using the relation, \(\frac{1}{f}\) = (μ – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)of lens in air f_{a} given as

Also let f_{w} be the focal length of the lens placed in water

Af = f_{w} – f_{a} = 78.2 – 20 = 58.2 cm.

Question 13.

The rays incident at 60° on one refracting face of a prism having angle 30° suffered deviation of 30°. Calculate the angle of emergence. Draw the ray diagram. Also calculate the R.I. of the material of the prism.

Answer:

Here, i = angle of incidence = 60°

A = angle of prism = 30°

δ = angle of deviation = 30°

e = angle of emergence = ?

Using the relation, A + δ = i + e, we get

e = A + δ – i = 30 + 30 – 60 = 0°

Thus the emergent ray is normal to the face AC as shown in the figure.

We know that R.I. is given by, u = \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\)

Here, i = 60°, f_{1} = i – δ = 60 – 30° = 30°

∴ μ = \(\frac{\sin 60^{\circ}}{\sin 30^{\circ}}\) = \(\frac{\sqrt{3} / 2}{\frac{1}{2}}\)

= 1.732

Question 14.

A glass prism of angle 72° and refractive index 1.66 is immersed in a liquid of refractive index 1.33. Find the angle of minimum deviation for a parallel beam of light passing through the prism.

Answer:

Here, A = 72°, ^{w}μ_{g} =1.66

^{w}μ_{l} = 1.33, δ_{m} = angle of minimum deviation = ?

When the prism is immersed in the liquid, then the refractive index of glass w.r.t. liquid is given by

^{l}μ_{g} = \(\frac{{ }^{a} \mu_{g}}{{ }^{a} \mu_{l}}=\frac{1.66}{1.33}\) = 1.248

We know that

Question 15.

The refractive indices of crown glass and flint glass for violet and red light are 1.523, 1.513 and 1.773, 1.743 respectively. Calculate the dispersive power of these glasses and tell which one has more dispersive power?

Answer:

Here, for crown glass

μ_{v}= R.I. for violet colour = 1.523

μ_{r} = R.I. for red colour = 1.513.

If p be the R.I. for mean colour, Then

W_{c} disersive power of crown glass = ?

We know that

For flint glass,

μ_{v} = 1.773 = R.I. for violet colour

μ_{r} = 1.743 = R.I. for red colour Pr

If μ’ = R.I. for mean colour, then

μ = \(\frac{\mu_{\mathrm{v}}+\mu_{\mathrm{r}}}{2}=\frac{1.773+1.743}{2}\) = 0.0193

If w_{f} be the dispersive power of flint glass, then

w_{f} = \(\frac{\mu_{\mathrm{v}}-\mu_{\mathrm{r}}}{\mu^{\prime}-1}\)

= \(\frac{1.773-1.743}{1.758-1}=\frac{0.030}{0.758}\) = 0.0396

Clearly w_{f} w_{c} i.e., disersive power of flint glass is more than that of the crown glass.

Question 16.

An achromatic lens combination of focal length 50 cm consists of a crown glass convex lens and flint glass concave lens. Find the focal length of each lens. Given μ_{b} = 1.520, μ_{r} = 1.510 for crown glass and μ_{b} = 1.660, μ_{r} = 1.650 for flint glass.

Answer:

μ_{b} = 1.520

μ_{r} = 1.510 for crown glass.

μ_{b} = 1-660

μ_{r}= 1.650 for flint glass.

F = focal length of achromatic combination = 50 cm.

f_{1}, f_{2} = focal length of each lens = ?

If μ, μ’ be R.I. for mean colour for crown and flint glass respectively.

then μ = \(\frac{\mu_{b}+\mu_{r}}{2}\) = \(\frac{1.520+1.510}{2}\) = 1.515

and μ’ = \(\frac{\mu_{\mathrm{b}}+\mu_{\mathrm{r}}}{2}=\frac{1.660+1.650}{2}\) = 1.655

Let w_{1} and w_{2} be the dispersive powers for crown and flint glasses respectively.

Let f_{1} and f_{2} be the focal lengths of crown and flint glasses respectively.

Now for an achromatic combination,

Also we know that

Question 17.

A man wants to see two poles separately situated at 11 km. Calculate the minimum approximate distance between the poles.

Answer:

We know that resolving power of eye = 1’ = \(\left(\frac{1}{80}\right)^{\circ}\)

= \(\frac{1}{60} \times \frac{\pi}{180}\) raciian ………(1)

Let d be the minimum distance between the two poles to see them as just separate.

Then, resolving power = \(\frac{d}{R}\) = \(\frac{d}{11,000}\) ……….(ii)

∴ from (i) and (ii), we get

Question 18.

(a) A combination of two thin lenses in contact is to be made which has the same focal length for blue and red light. (Such a combination is known as an ‘achromatic doublet’. Show that the ratio of their focal lengths (for yellow light( must be equal in magnitude and opposite in sign to the ratio of the dispersive powers of the materials of the two lenses.

(b) Use the results in (a) to suggest a way of removing chromatic aberration of the lens which made on flint glass. You are given convex and concave lenses (made of crown glass) of various focal length. The ratio of the dispersive powers of the flint glass to crown glass is about 1.5, focal length for flint glass = 15 cm.

Answer:

(a) For combination F_{v} = F_{r}

F_{v} = focal length of combination for violet light (the combination is of two lenses of different material in contact)

∴ \(\frac{1}{\mathrm{~F}_{\mathrm{V}}}=\frac{1}{\mathrm{f}_{\mathrm{v}}}+\frac{1}{\mathrm{f}_{\mathrm{v}}^{\mathrm{t}}}\)

Similarly \(\frac{1}{F_{r}}=\frac{1}{f_{r}}+\frac{1}{f_{r}^{\prime}}\) ……(ii)

where f and f’ are given by relations,

\(\frac {1}{f}\) = (μ – 1) \(\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\) (for crown glass)

\(\frac {1}{f}\) = (μ – 1) \(\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\) (for flint glass)

The relations for f_{v}, f_{v}, f_{r} and f_{r}can be written by replacing μ

with μ_{v}, μ_{v}, μ_{r} and μ_{r}

Putting these results in (i) and (ii) we have

For the combination to behave as achromatic,

Hence proved

where f ∝ f’ are focal lengths for yellow light w and w’ are the respective dispersive powers.

(b) Here, \(\frac{w}{w^{\prime}}=\frac{3}{2}\) (given)

∴ \(\frac{f^{\prime}}{f}=-\frac{w^{\prime}}{w}=-\frac{3}{2}\)

Also f’ = 15 cm (given).

∴ f = \(\frac{2}{3}\) x f’ = \(\frac{-2}{3}\) x 15 = – 10 cm

Thus to remove the chromatic aberration of the flint glass lens, it must be combined with a crown glass lens of focal length – 10 cm i.e., a concave lens.

Question 19.

You are given a double-convex lens made of crown glass with each surface of radius of curvature 15 cm. A flint glass lens is grafted on to one of the surfaces of this lens. What is the radius of curvature of the second surface of the flint glass lens for the combination to be an ‘achromatic doublet’ for blue and red light? Data on refractive indices required is μ_{v} = 1.520, μ_{r} = 1.509 for crown

glass and μ_{v} = 166, μ_{r} = 1.644 for flint glass.

Answer:

For combination to behave as achromatic, crown glass lens should be double convex, flint glass lens should be double concave and both placed in contact.

For crown glass lens :

R_{1} = R and R_{2} = – R = 15 cm

For flint glass lens :

R_{1} = – R = – 15 cm and R_{2} = ?

For crown glass lens :

μ_{v} = 1.520, μ_{r} = 1.509 and for flint glass μ_{v} = 1.660, μ_{r} = 1.644. Tire combination behaves as achromatic doublet for blue and red lights. Therefore chromatic abberation f_{r} – f_{v} should be equal and

opposite in two cases or \(\frac{1}{f_{v}}-\frac{1}{f_{r}}\) should be same for two lenses.

For crown glass

imf

For flint glass :

This value is – 0.001467 cm and is equal to (μ_{v} – μ_{r}) \(\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\)

or – 0.001467 = (1.660 – 1.644) \(\left[-\frac{1}{15}-\frac{1}{R_{2}}\right]\)

.’. R_{2} = ∞

The other surface of flint glass lens is plane.

Question 20.

Two light sources A and B placed on both sides of a screen at distances of 60 cm and 40 cm respectively produce photometric balance. If a glass plate is inserted between source A and screen then the source B is moved by 5 cm to obtain balance again. What percentrage of light is absorbed by glass?

Answer:

Figure (b)

E_{1} = E_{2}

E_{1} = \(\frac{\mathrm{I}_{1}^{\prime}}{(60)^{2}}\)

[due to insertion of glass plate, E of screen due to A changes]

Fraction of light absorbed by glass plate

= I_{1} – I_{1}

= \(\mathrm{I}_{1}-\frac{64}{81} \mathrm{I}_{\mathrm{1}}=\frac{17}{81} \mathrm{I}_{1}\)

% of light absorbed by glass plate

\(\frac{\frac{17}{81} \mathrm{I}_{1}}{\mathrm{I}_{1}} \times 100=\frac{1700}{81}\)

Question 21.

A mark placed on the surface of a sphere is viewed through glass from a position directly opposite. If the diameter of sphere is 10 cm and p. of glass is 1.5. Find the position of image.

Answer:

Here, u = – 10 cm, R = – 5 cm, μ_{1} = 1, μ_{2} = 1.5, υ = ?

Here refraction is from denser to rarer medium.

Question 22.

The refractive index of a prism with apex angle A is cot A/2. Prove that angle of minimum deviation is given by δ_{m} = 180 – 2A., m

Answer:

We know by prism formula

[We know sin (90 – θ) = cos θ]

Question 23.

A ray of monochromatic light strikes the right angled prism at P from air and at Q it just grazes the right handed prism surface as it emerges at Q.

(a) Calculate its refractive index relative to air in terms of angle of incidence θ_{1}

(b) Give a numerical upper limit for μ.

Answer:

(a) Angles are shown in the figure here.

For refraction at P,

For refraction at Q,

Also we know that for a prism, r_{1} + r_{2} = 90° ……….(3)

∴ sin (90 – r_{1}) = \(\frac {1}{μ}\)

cos r_{1} = \(\frac {1}{μ}\) …………(4)

∴ From (2) and (4), sin r2 = cos rt = —

Squaring and adding (1) and (5), we get

(b) As maximum value of sin θ_{1}, = 1,

∴ \(\left(\sin _{\theta_{1}}^{2}\right)_{\max }\) = 1

So the upper limit for p = maximum value of p is given by

∴ μ_{max} = \(\sqrt{1+1}=\sqrt{2}\)

Fill In The Blanks

Question 1.

A concave lens made of a material of refractive index 1.5 is immersed in a medium of refractive index 1.5. A parallel beam of light is incident on the lens. The beam will ………….

Answer:

remain parallel.

Question 2.

The focal length of a lens of focal power 100 D is ……….

Answer:

1 cm.

Question 3.

The image of an object formed by a concave lens is …………

Answer:

always virtual.

Question 4.

Two lenses of power +16 D and – 6 D are placed in contact. Then the power and focal length of the combination are given by …………. and ……….. respectively.

Answer:

+ 10 D, 10 cm.

Question 5.

A bird flying high in the air appears to be ………… as the refraction takes place from rarer to denser medium.

Answer:

higher than in reality.

Question 6.

The brightness of image produced by a lens which is half painted black reduces to and the size of the image

Answer:

half, remains the same.

[Hint: Intenstiy of the image & square of the aperture of the lens.]

Question 7.

A lens forms a sharp image on a screen. On inserting a parallel sided slab of refractive index p between the screen and lens, it is found necessary to move the screen through a distance’d’ in order for the image to be again sharply focused. The thickness of the slab is ……….

Answer:

d\(\left(1-\frac{1}{\mu}\right)\)

Question 8.

The dispersive power of for a given prism is independent of but depends

Answer:

angle of prism, refractive index for the extreme and mean colours.

Question 9.

When a ray of white light passes through a glass prism, red light is divided through a angle than violet light because the refractive index for violet light is …………. than that of red light.

Answer:

Smaller, greater.

Question 10.

………. In glass violet light travels than red light.

Answer:

Slower.

Question 11

……….. is the main source of light.

Answer:

Sun.

Question 12.

With the rise in temperature, the density and hence the refractive index of the medium

Answer:

decreases.

Question 13.

With an object placed in a denser medium and observed by an observer in the rarer medium, the image of the object shifts …………. and when the object placed in rarer medium is observed in denser medium, the image shifts ………..

Answer:

nearer the refracting surface, away from the refracting surface.

Question 14.

The sparkle of diamond, the shine of air bubbles in a glass, paper weight, the mirage in hot deserts and the looking in cold countries is due to

Answer:

the total internal reflection of light.

Question 15.

In the position of minimum deviation, the deviation is equally shared at the two plane surfaces enclosing the angle of prism and the refracting ray passes through the prism symmetrically with respect to the refracting edge. If the base angles of the prism are equal, then the refracting ray ………. in the position of minimum deviation.

Answer:

runs parallel to the base.

Question 16.

Compound microscope consists of an objective of ………. focal length and aperture and an eyepiece of comparatively ……… focal length and ………. aperture.

Answer:

small, small, larger, larger.

Question 17.

In reflecting astronomical telescope, parabolic mirrors are used because they are free from ………… and in an astronomical telescope f_{0} ……… f_{e} where f_{0} and f_{e}, are the focal lengths of objective and eye piece respectively.

Answer:

spherical aberration, greater than.

Question 18.

The length of the astronomical telescope for normal adjustment is …………..

Answer:

f_{0} + f_{e}.

Question 19.

Focal length of a concave mirror when it is dipped in water does …………

Answer:

not changes.

Question 20.

The magnifying power of the erecting lens in a terrestrial telescope is ……….

Answer:

One.

Question 21.

Red light is used for danger signals because it is ……….. and scattered …………

Answer:

having high wave lengths, least.

Question 22.

If f be the focal length of a convex lens used as a simple microscope and D is the least distance of distinct vision, then its magnifying power is ………..

Answer:

1 + \(\frac {D}{f}\)

Question 23.

Green colour when seen in sodium light appears and when red and blue colours are mixed …………. colour will be formed.

Answer:

black, magenta.

Question 24.

Light travels from denser to rarer medium and when it gets incident on the interface at angle 0, the reflected and refracted rays are perpendicular to each other. The critical angle is …………

Answer:

sin^{-1} (cot θ).

Question 25.

The distance between the nearest point and the farthest point is known as ………

Answer:

range of vision.

Question 26.

An eye which does not show this range of vision is called ………..

Answer:

defective or ametropic.