NCERT Solutions for Class 12 Physics Chapter 9 – Ray Optics And Optical Instruments

Question 9.1:

A small candle, 2.5 cm in size is placed at 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:

Size of the candle, h= 2.5 cm

Image size = h

Object distance, u= −27 cm

The radius of curvature of the concave mirror, R= −36 cm

The focal length of the concave mirror, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m69fe1cce

Image distance = v

The image distance can be obtained using the mirror formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 7bc23f89

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m65fc6fab

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.

If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

Question 9.2:

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:

Height of the needle, h1 = 4.5 cm

Object distance, = −12 cm

Focal length of the convex mirror, f = 15 cm

Image distance = v

The value of can be obtained using the mirror formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m59aa5f3d

Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.

The image size is given by the magnification formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m5aebd801

Hence, magnification of the image, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 8a34ea6

The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.

If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

Question 9.3:

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:

Actual depth of the needle in water, h1 = 12.5 cm

Apparent depth of the needle in water, h2 = 9.4 cm

Refractive index of water = μ

The value of μcan be obtained as follows:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1144c031

Hence, the refractive index of water is about 1.33.

Water is replaced by a liquid of refractive index, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3e93f0b3

The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the needle. Hence, we can write the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 79a00fe2

Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up.

∴Distance by which the microscope should be moved up = 9.4 − 7.67

= 1.73 cm

Question 9.4:

Figures 9.34(a) and (b) show refraction of a ray in air incident 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 is 45º with the normal to a water-glass interface [Fig. 9.34(c)].

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m70d49137

Answer:

As per the given figure, for the glass − air interface:

Angle of incidence, = 60°

Angle of refraction, r = 35°

The relative refractive index of glass with respect to air is given by Snell’s law as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m77e7de83

As per the given figure, for the air − water interface:

Angle of incidence, i = 60°

Angle of refraction, r = 47°

The relative refractive index of water with respect to air is given by Snell’s law as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 29796e0a

Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 7c77bc5d

The following figure shows the situation involving the glass − water interface.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1922c973

Angle of incidence, i = 45°

Angle of refraction = r

From Snell’s law, r can be calculated as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 468ab6e7

Hence, the angle of refraction at the water − glass interface is 38.68°.

Page No 346:

Question 9.5:

A small bulb is placed at the bottom of a 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:

Actual depth of the bulb in water, d1 = 80 cm = 0.8 m

Refractive index of water, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m49c60e78

The given situation is shown in the following figure:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4e9d684e

Where,

i = Angle of incidence

r = Angle of refraction = 90°

Since the bulb is a point source, the emergent light can be considered as a circle of radius, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 11094008

Using Snell’ law, we can write the relation for the refractive index of water as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m7a5d1317

Using the given figure, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m163c1a9d

= tan 48.75° × 0.8 = 0.91 m

∴Area of the surface of water = πR2 = π (0.91)= 2.61 m2

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2.

Question 9.6:

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on the 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:

The angle of minimum deviation, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html mff80655 = 40°

The angle of the prism, = 60°

Refractive index of water, µ = 1.33

Refractive index of the material of the prism = NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1c3d18cb

The angle of deviation is related to refractive indexNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3519d07eas:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m41d02e55

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water, let NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3a6833edbe the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 2ddfe49b

Hence, the new minimum angle of deviation is 10.32°.

Question 9.7:

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:

Refractive index of glass,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 32413c1d

The focal length of the double-convex lens, f = 20 cm

The radius of curvature of one face of the lens = R1

The radius of curvature of the other face of the lens = R2

The radius of curvature of the double-convex lens = R

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 50db3aaa

The value of R can be calculated as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1feb51ce

Hence, the radius of curvature of the double-convex lens is 22 cm.

Question 9.8:

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

Answer:

In the given situation, the object is virtual and the image formed is real.

Object distance, = +12 cm

(a) Focal length of the convex lens, f = 20 cm

Image distance = v

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 5a5195e7

Hence, the image is formed 7.5 cm away from the lens, toward its right.

(b) Focal length of the concave lens, f = −16 cm

Image distance = v

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m25941257

Hence, the image is formed 48 cm away from the lens, toward its right.

Question 9.9:

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:

Size of the object, h1 = 3 cm

Object distance, u = −14 cm

Focal length of the concave lens, = −21 cm

Image distance = v

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3d148f0e

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 63808562

Hence, the height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

Question 9.10:

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

Answer:

Focal length of the convex lens, f1 = 30 cm

The focal length of the concave lens, f2 = −20 cm

The focal length of the system of lenses = f

The equivalent focal length of a system of two lenses in contact is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m118cc788

Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.

Question 9.11:

A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece 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:

Focal length of the objective lens, f1 = 2.0 cm

The focal length of the eyepiece, f2 = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

(a) Least distance of distinct vision, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1fdcbbdb

∴Image distance for the eyepiece, v2 = −25 cm

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 72d02435

Image distance for the objective lens,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3d6a24f2

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m5d988168

The magnitude of the object distance, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m59dedbb0 = 2.5 cm

The magnifying power of a compound microscope is given by the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 2f6d18c8

Hence, the magnifying power of the microscope is 20.

(b) The final image is formed at infinity.

∴Image distance for the eyepiece,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 791f167d

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4fdf3afb

Image distance for the objective lens,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m5fc99ea8

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 32fac850

The magnitude of the object distance, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m59dedbb0 = 2.59 cm

The magnifying power of a compound microscope is given by the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m329766ae

Hence, the magnifying power of the microscope is 13.51.

Question 9.12:

A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 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:

Focal length of the objective lens, fo = 8 mm = 0.8 cm

Focal length of the eyepiece, fe = 2.5 cm

Object distance for the objective lens, uo = −9.0 mm = −0.9 cm

Least distance of distant vision, = 25 cm

Image distance for the eyepiece, ve = −d = −25 cm

Object distance for the eyepiece = NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 61e00d68

Using the lens formula, we can obtain the value ofNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 61e00d68as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 65b89783

We can also obtain the value of the image distance for the objective lensNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3f9dd31d using the lens formula.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4c9724db

The distance between the objective lens and the eyepieceNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 19dddcc1

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m54d86c2b

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 602bbe83

The magnifying power of the microscope is calculated as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4f323d43

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 35313299

Hence, the magnifying power of the microscope is 88.

Question 9.13:

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

Answer:

Focal length of the objective lens, fo = 144 cm

Focal length of the eyepiece, fe = 6.0 cm

The magnifying power of the telescope is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m7fd71ba6

The separation between the objective lens and the eyepiece is calculated as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m1e31fa67

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m41015021

Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

Question 9.14:

(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece 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 × 106 m, and the radius of lunar orbit is 3.8 × 108 m.

Answer:

Focal length of the objective lens, fo = 15 m = 15 × 102 cm

The focal length of the eyepiece, fe = 1.0 cm

(a) The angular magnification of a telescope is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m20bc3b30

Hence, the angular magnification of the given refracting telescope is 1500.

(b) Diameter of the moon, d = 3.48 × 106 m

The radius of the lunar orbit, r0 = 3.8 × 108 m

Let NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3ac886e2be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 731b0a98

Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm

Question 9.15:

Use the mirror equation to deduce that:

(a) an object placed between and 2of 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 the focus of a concave mirror produces a virtual and enlarged image.

[Note: This exercise helps you deduce algebraically properties of

images that one obtains from explicit ray diagrams.]

Answer:

(a) For a concave mirror, the focal length (f) is negative.

< 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

u < 0

For image distance v, we can write the lens formula as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 7409bc80

The object lies between f and 2f.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 78af109b

Using equation (1), we get:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4ea619ec

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4785698d is negative, i.e., v is negative.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m288b75a0

Therefore, the image lies beyond 2f.

(b) For a convex mirror, the focal length (f) is positive.

∴ > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

∴ < 0

For image distance v, we have the mirror formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 6ae2183d

Thus, the image is formed on the backside of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object’s distance.

(c) For a convex mirror, the focal length (f) is positive.

f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative,

u < 0

For image distance v, we have the mirror formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m5ab65bec

Hence, the image formed is diminished and is located between the focus (f) and the pole.

(d) For a concave mirror, the focal length (f) is negative.

f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

u < 0

It is placed between the focus (f) and the pole.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 7dde5969

For image distance v, we have the mirror formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 6c63e886

The image is formed on the right side of the mirror. Hence, it is a virtual image.

For u < 0 and v > 0, we can write:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3c597cc

Magnification, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3dea8cc> 1

Hence, the formed image is enlarged.

Page No 347:

Question 9.16:

A small pin fixed on a tabletop 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:

The actual depth of the pin, d = 15 cm

Apparent depth of the pin = NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3ac886e2

Refractive index of glass,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 5f140e85

The ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m1e77a362

The distance at which the pin appears to be raised = NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 12d3cb1a

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m585d988

For a small angle of incidence, this distance does not depend upon the location of the slab.

Question 9.17:

(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fiber 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?

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 51ff7718

Answer:

(a) Refractive index of the glass fiber, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m34104393

Refractive index of the outer covering of the pipe, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1353b91d = 1.44

The angle of incidence = i

The angle of refraction = r

The angle of incidence at the interface = i

The refractive index (μ) of the inner core − outer core interface is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m7c9c4c74

For the critical angle, total internal reflection (TIR) takes place only whenNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m1dddc7fa, i.e., i > 59°

Maximum angle of reflection,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m72164e00

Let,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 75c7a6b3be the maximum angle of incidence.

The refractive index at the air-glass interface,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m34104393

We have the relation for the maximum angles of incidence and reflection as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1e8bd628

Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.

(b) If the outer covering of the pipe is not present, then:

Refractive index of the outer pipe,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 61988a2a

For the angle of incidence i = 90°, we can write Snell’s law at the air-pipe interface as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 39b5ce39.

Question 9.18:

Answer the following questions:

(a) You have learned that plane and convex mirrors produce virtual images of objects. Can they produce real images 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 onto the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?

(c) A diver underwater, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman 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 as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

(b) No

A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

(c) The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are traveling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.

(d) Yes; Decrease

The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on traveling from one medium to another. The apparent depth of the tank, when viewed obliquely, is less than the near-normal viewing.

(e) Yes

The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its face. This is the reason for the sparkling effect of a diamond.

Question 9.19:

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:

Distance between the object and the image, d = 3 m

The maximum focal length of the convex lens =NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 2fef4580

For real images, the maximum focal length is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m2e550531

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Question 9.20:

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:

Distance between the image (screen) and the object, D = 90 cm

Distance between two locations of the convex lens, = 20 cm

Focal length of the lens = f

Focal length is related to d and D as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m623a3fcc

Therefore, the focal length of the convex lens is 21.39 cm.

Question 9.21:

(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 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:

Focal length of the convex lens, f1 = 30 cm

Focal length of the concave lens, f2 = −20 cm

Distance between the two lenses, d = 8.0 cm

(a) When the parallel beam of light is incident on the convex lens first:

According to the lens formula, we have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4c2ba316

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 74ffd4e9 = Object distance = ∞

v1 = Image distance

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 5a547984

The image will act as a virtual object for the concave lens.

Applying lens formula to the concave lens, we have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html c3901a

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 494c1227= Object distance

= (30 − d) = 30 − 8 = 22 cm

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 5df737cf= Image distance

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m229c052f

The parallel incident beam appears to diverge from a point that isNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 18bdc6ac from the centre of the combination of the two lenses.

(ii) When the parallel beam of light is incident, from the left, on the concave lens first:

According to the lens formula, we have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4450f707

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 494c1227 = Object distance = −∞

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 5df737cf = Image distance

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 29807bbf

The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4c2ba316

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 74ffd4e9 = Object distance

= −(20 + d) = −(20 + 8) = −28 cm

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 680b2290 = Image distance

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 2b3af62f

Hence, the parallel incident beam appear to diverge from a point that is (420 − 4) 416 cm from the left of the centre of the combination of the two lenses.

The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

(b) Height of the image, h1 = 1.5 cm

Object distance from the side of the convex lens, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4cc202e8

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m64ffe8f5

According to the lens formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4c2ba316

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 680b2290 = Image distance

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html cf118e0

Magnification, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 72c613ae

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m573f8181

Hence, the magnification due to the convex lens is 3.

The image formed by the convex lens acts as an object for the concave lens.

According to the lens formula:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html c3901a

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 494c1227= Object distance

= +(120 − 8) = 112 cm.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 5df737cf = Image distance

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3a282778

Magnification, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 69775a70

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m600a89b9

Hence, the magnification due to the concave lens isNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 67f3cd3a.

The magnification produced by the combination of the two lenses is calculated as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m2bf8f43e

The magnification of the combination is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 10b384bd

Where,

h1 = Object size = 1.5 cm

h2 = Size of the image

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 61d03371

Hence, the height of the image is 0.98 cm.

Page No 348:

Question 9.22:

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:

The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m6c222275

Angle of prism, ∠A = 60°

Refractive index of the prism, µ = 1.524

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m56d2238a = Incident angle

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 8c6bec3 = Refracted angle

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m5be939b1 = Angle of incidence at the face AC

e = Emergent angle = 90°

According to Snell’s law, for face AC, we can have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m265a03be

It is clear from the figure that the angleNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4190fa5c

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m173139fa

According to Snell’s law, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m109bba20

Hence, the angle of incidence is 29.75°.

Question 9.23:

You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms that will

(a) deviate a pencil of white light without much dispersion,

(b) disperse (and displace) a pencil of white light without much deviation.

Answer:

(a)Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and the white light will emerge from the combination of the two prisms.

(b)Take the system of the two prisms as suggested in answer (a). Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.

Question 9.24:

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm 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 accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

Answer:

Least distance of distinct vision, d = 25 cm

The far point of a normal eye, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 176b2fd2

Converging power of the cornea, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 65aa0658

Least converging power of the eye-lens, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3a224d81

To see the objects at infinity, the eye uses its least converging power.

Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 D

Power of the eye-lens is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 72cbeb5f

To focus an object at the near point, object distance (u) = −d = −25 cm

The focal length of the eye-lens = Distance between the cornea and the retina

= Image distance

Hence, image distance, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m18454e0c

According to the lens formula, we can write:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 18939909

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m77b2f56d = Focal length

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3b4d7938

∴Power of the eye-lens = 64 − 40 = 24 D

Hence, the range of accommodation of the eye-lens is from 20 D to 24 D.

Question 9.25:

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

Answer:

A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eyeballs get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye-lens lose their ability to accommodate, the defect is called presbyopia.

Question 9.26:

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

Answer:

The power of the spectacles used by the myopic person, P = −1.0 D

The focal length of the spectacles, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m108a5a8a

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.

During old age, the person uses reading glasses of power,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4b8626c

The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

Question 9.27:

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:

In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

Question 9.28:

A man with a normal near point (25 cm) reads a book with a 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) Focal length of the magnifying glass, f = 5 cm

Least distance of distance vision, d = 25 cm

Closest object distance = u

Image distance, v = −d = −25 cm

According to the lens formula, we have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4b126354

Hence, the closest distance at which the person can read the book is 4.167 cm.

For the object at the farthest distance (u’), the image distance NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 36db5e5d

According to the lens formula, we have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m9aedafc

Hence, the farthest distance at which the person can read the book is

5 cm.

(b) Maximum angular magnification is given by the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html df025e8

Minimum angular magnification is given by the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 50b0346c

Question 9.29:

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:

Note : Here we took focal Length as 10 cm because if we take it as 9 cm then the image distance will be zero , which does not make any sense. (a) Area of each square, A = 1 mm2

Object distance, u = −9 cm

The focal length of a converging lens, =  9 cm

For image distance v, the lens formula can be written as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html ma6a209e

Magnification, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 171b29d7

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m1f119f87

∴Area of each square in the virtual image = (10)2A

= 102 × 1 = 100 mm2

= 1 cm2

(b) Magnifying power of the lens NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 636efdb6

(c) The magnification in (a) is not the same as the magnifying power in (b).

The magnification magnitude is NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m15992158and the magnifying power isNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m6eadf46a.

The two quantities will be equal when the image is formed at the near point (25 cm).

Question 9.30:

(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) The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm).

Image distance, v = −d = −25 cm

Focal length, f = 10 cm

Object distance = u

According to the lens formula, we have:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3b084e6f

Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.

(b) Magnification =NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m3af599c6

(c) Magnifying power =NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m6ad87d71

Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.

Question 9.31:

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:

Area of the virtual image of each square, A = 6.25 mm2

Area of each square, A0 = 1 mm2

Hence, the linear magnification of the object can be calculated as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m6685fcb8

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m17b124c3

The focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m5624cc32

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

Page No 349:

Question 9.32:

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 eyepiece of a compound microscope have short focal lengths?

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

Answer:

(a)Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.

(b) Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.

(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.

(d) The angular magnification produced by the eyepiece of a compound microscope is NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m8e2dd7f

Where,

fe = Focal length of the eyepiece

It can be inferred that if fe is small, then angular magnification of the eyepiece will be large.

The angular magnification of the objective lens of a compound microscope is given as NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m25419d4e

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 6218b072 = Object distance for the objective lens

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m33fd9f88 = Focal length of the objective

The magnification is large when NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 6218b072>NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m33fd9f88. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 6218b072is small, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m33fd9f88 will be even smaller. Therefore, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html bfbe8feand NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m33fd9f88 are both small in the given condition.

(e)When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much-refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.

The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

Question 9.33:

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

Answer:

The focal length of the objective lens,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m33fd9f88= 1.25 cm

The focal length of the eyepiece, fe = 5 cm

Least distance of distinct vision, d = 25 cm

Angular magnification of the compound microscope = 30X

The total magnifying power of the compound microscope, m = 30

The angular magnification of the eyepiece is given by the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 47e86a94

The angular magnification of the objective lens (mo) is related to me as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 382908bfNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 74c4cfd = m

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3c3e3869

Applying the lens formula for the objective lens:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4f22ecb8

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Applying the lens formula for the eyepiece:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4c3f2bf4

Where,

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 86ae25b = Image distance for the eyepiece = −d = −25 cm

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 61e00d68 = Object distance for the eyepiece

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m65b37e53

The separation between the objective lens and the eyepiece NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 304b5484

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html mce45d9c

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

Question 9.34:

A small telescope has an objective lens of focal length 140 cm and an eyepiece 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:

The focal length of the objective lens,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m33fd9f88= 140 cm

The focal length of the eyepiece, fe = 5 cm

Least distance of distinct vision, = 25 cm

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

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4ab47975

(b) When the final image is formed at d, the magnifying power of the telescope is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m4ff7f9

Question 9.35:

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

(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:

The focal length of the objective lens, fo = 140 cm

The focal length of the eyepiece, fe = 5 cm

(a) In normal adjustment, the separation between the objective lens and the eyepiece NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 61dd201e

(b) Height of the tower, h1 = 100 m

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

The angle subtended by the tower at the telescope is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 371eea67

The angle subtended by the image produced by the objective lens is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m47daa595

Where,

h2 = Height of the image of the tower formed by the objective lens

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html bf69278

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

(c) Image is formed at a distance, d = 25 cm

The magnification of the eyepiece is given by the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m44a77ce9

Height of the final imageNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m142abff1

Hence, the height of the final image of the tower is 28.2 cm.

Question 9.36:

A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Answer:

The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html mc5bfec

Distance between the objective mirror and the secondary mirror, d = 20 mm

The radius of curvature of the objective mirror, R1 = 220 mm

Hence, the focal length of the objective mirror, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 3c4be0a6

The radius of curvature of the secondary mirror, R= 140 mm

Hence, the focal length of the secondary mirror, NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m6e148f61

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.

Hence, the virtual object distance for the secondary mirror,NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m7cf9fabb

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 4298acdc

Applying the mirror formula for the secondary mirror, we can calculate image distance (v)as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 7ec420e5

Hence, the final image will be formed 315 mm away from the secondary mirror.

Question 9.37:

Light incident normally on a plane mirror attached to a galvanometer coil retraces backward as shown in Fig. 9.36. A current in the coil produces a 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?

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 197ec1bb

Answer:

Angle of deflection, θ = 3.5°

Distance of the screen from the mirror, D = 1.5 m

The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ= 7.0°

The displacement (d) of the reflected spot of light on the screen is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m500ab14d

Hence, the displacement of the reflected spot of light is 18.4 cm.

Page No 350:

Question 9.38:

Figure 9.37 shows an equi-convex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane 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?

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m382b92d0

Answer:

Focal length of the convex lens, f1 = 30 cm

The liquid acts as a mirror. The focal length of the liquid = f2

Focal length of the system (convex lens + liquid), = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m154762da

Let the refractive index of the lens beNS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html ca29345 and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is −R.

can be obtained using the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m1865cd05

Let NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1353b91dbe the refractive index of the liquid.

The radius of curvature of the liquid on the side of the plane mirror =NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 9594975

The radius of curvature of the liquid on the side of the lens, R = −30 cm

The value of NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html 1353b91d can be calculated using the relation:

NS 3 11 08 Sravana 12 Physics 9 38 NRJ LVN html m2e65899a

Hence, the refractive index of the liquid is 1.33.


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