# NCERT Solutions For Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

## NCERT Solved Exercise Questions – Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

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?

Ans – Size of the candle, h= 2.5 cm

Image size = h’

Object distance, u= −27 cm

Keep the object between and C. Therefore, the image should be accurate, inverted, and outside of C. The screen should be positioned at the location of the image in order to find the sharp image.

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.

Ans – Height of the needle, h1 = 4.5 cm

Object distance, u = −12 cm

A convex mirror always creates a virtual picture of an item placed in front of it that is upright and small in size. Convex mirror focal length: +15 cm Object separation u = -12 cm with the mirror formula

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?

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

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

The apparent depth will alter if the water is replaced with another liquid, and the microscope will need to be pushed further to focus the image. Using fresh liquid

9.4 Figures 9.31(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 45° with the normal to a water-glass interface [Fig. 9.31(c)].

Ans – Angle of incidence, i = 60°

Angle of refraction, r = 35°

use Snell’s law to account for air-to-glass refraction. Glass’s refractive index with respect to air

9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. 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.)

Ans – depth of the bulb in water,

d1 = 80 cm

= 0.8 m

When light rays from bulb B strike the water’s surface at an angle equal to the critical angle (C), they barely scrape the surface,

while light rays that strike the water’s surface at an angle greater than C are completely internally reflected back into the water.

Through a circular patch with a radius of r, the light image rays emerge from the water.

9.6 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.

Ans – Angle of the prism, A = 60°

Refractive index of water, µ = 1.33

The angle of lowest deviation is 40° when the light beam strikes the glass prism from the air. Glass’s air-reflective index.

new angle of minimum deviation can be calculated.

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 20cm?

Ans –  Double-convex lens’s focal length is 20 cm.

R1 is the radius of curvature of one lens face.

The other face of the lens’s radius of curvature is equal to R2.

Double-convex lens radius of curvature = R

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

Ans – (a) The light beam converges at Pt when a convex lens is put in the direction of the light converging at P. Point P serves as the convex lens’s virtual object as a result.

(b) In order to further diverge the light, a concave lens is put in the path of the convergent beam.

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

Ans – Size of the object, h1 = 3 cm

Object distance, u = −14 cm

The picture is therefore virtual, upright, 1.8 cm in size, and 8.4 cm away from the lens on the same side as the item. The virtual image travels toward the lens’s focus but never beyond it as the object is moved away from the lens. As the focus moves in, the size of the image also decreases.

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

Ans – Focal length of the convex lens, f1 = 30 cm

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

The lens system’s negative focal length causes the combination to function as a diverging lens.

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 15cm. 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 (25cm), and
(b) at infinity? What is the magnifying power of the microscope in each case?

Ans –The objective lens’s focal length is f1 = 2.0 cm.

Eyepiece focal length, f2 = 6.25 cm

d = 15 cm is the distance between the objective lens and the eyepiece.

(A) The final image should be within the range of distinct vision. Aim to have the object in front of the objective at a distance of zero.

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

Ans – 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

The image is created at a minimum distance that allows for sharp focus. The distance between two lenses will be equal to 0 + |e|.

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

Ans – The objective lens’s focal length is 144 cm.

Eyepiece focal length, fe = 6.0 cm

The telescope’s magnification power is indicated as follows:

9.14 (a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm 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 × 106m, and the radius of lunar orbit is 3.8 × 108m.

Ans – The objective lens’s focal length is 15 metres, or 15 x 102 cm.

Eyepiece focal length, fe = 1.0 cm

• A telescope’s angular magnification is given as

9.15 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 deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Ans –  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

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

Ans – Actual depth of the pin, d = 15 cm

It is possible to calculate the picture shift caused by the thick glass slab. Here, shift is solely dependent on the glass’s refractive index and slab thickness.

The lateral displacement

9.17 (a) Figure 9.32 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?

Ans – Angle of incidence equals i.

Refraction angle = r

Angle of incidence = I at the interface

The inner core – outer core interface’s refractive index () is given as:

(a) You have learnt 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 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 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?

Ans -(a) Convex and plane mirrors can both create actual images. A true image will be created if the item is virtual, that is, if light rays converge at a point behind a plane mirror (or a convex mirror) and reflect to a point on a screen placed in front of the mirror.

(B) In this instance, there is no contradiction. The virtual representation of the object serves as an object for our convex lens, which then creates a real image of the object on the retina.

(c) Assume AB is the lake’s bank-standing fisherman. The light coming from the fisherman’s head is refracted at the water/air interface, where it bends toward the normal. For the fish, point B’ rather than point B appears to be where the refracted rays are coming from. As a result, the fisherman’s height for a diver is AB’, which is higher than his actual height AB.

(d) Decrease; Yes

When viewed from an angle, a water tank’s apparent depth changes. This occurs because light bends when it transitions between different media. When seen from an angle, the tank appears to be deeper than it actually is.

(e) Yes

Diamond has a higher refractive index (2.42) than regular glass (1.5). Compared to glass, diamond has a smaller critical angle. To ensure that all of the light entering the diamond is completely reflected from its faces, a diamond cutter will utilise a wide angle of incidence. This is the cause of a diamond’s brilliant appearance.

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

Ans – L = 4f is the formula for the minimal separation between a real item and its real image created by a convex lens with a focal length.

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

Ans – D = 90 cm is the distance between the object and the image (screen).

d = 20 cm is the distance between the convex lens’s two sites.

The lens’s focal length is f.

9.21 (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm 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.

Ans – Convex lens’s f1 equivalent focal length is 30 cm.

Concave lens’s f2 focal length is approximately 20 cm.

d = 8.0 cm is the distance between the two lenses.

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.

Ans – Refractive index of the prism, µ = 1.524 = Incident angle = Refracted angle = Angle of incidence at the face AC

e = Emergent angle = 90°

9.23 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.

Ans –

(c) Unless the image is situated close to the shortest distance of distinct vision, e. v = D, it is obvious that magnification and power magnification are not equivalent to one another.

9.24 (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.

Ans – (A) The image is created at the close point (d = 25 cm), where the highest magnification is reached.

Image separation, v = d = 25 cm

Focusing distance, f = 10 cm

Distance to object = u

We have the following using the lens formula:

9.25 What should be the distance between the object in Exercise 9.24 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 9.23 to 9.25 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Ans – Each square’s virtual image area is 6.25 mm2 in size.

A0 is the area per square millimetre.

As a result, the object’s linear magnification can be determined as follows:

(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?

Ans – (A) When not using a magnifying glass, the object that has to be viewed clearly should be set at a distance of 25 cm. However, an object can be placed closer to the eye than at 25 cm when using a magnifying lens. Compared to an identical object placed 25 cm away, the closer object has a larger angular dimension. An angular amplification is what a magnifying glass offers in this sense.

(b) Yes. Because the angle subtended at the eye is a little less than the angle subtended at the lens, the angular magnification somewhat lowers.

(c) If a convex lens with a decreasing focal length is created, aberrations including spherical and chromatic aberrations begin to appear.

(d) The eyepiece’s angular magnification is given by (1+D/fe) while the objective’s angular magnification is roughly given by /f0. It is obvious that for improved magnification, the focal lengths of the eyepiece and the objective should be reduced.

(e) The field of vision will be reduced if we place our eye very close to the eyepiece since the entire light will not shine on it. In order to increase the field of view, we move our eye a small distance away from the eyepiece to capture the significant amount of light that is refracted through the eyepiece.

9.27 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope? 9.28 A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. 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 (25cm)?

Ans – For the desired magnification of 30, we want to know the distance between the provided objective and eye lens in this case. Make sure that the final image is produced at least a clear distance from the eyepiece.

9.29 (a) For the telescope described in Exercise 9.28 (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 25cm?

Ans – Objective lens focal length, NCERT Solutions for Class 12 Physics = 140 cm

Eyepiece focal length, fe = 5 cm

The shortest distance at which vision is distinct is 25 cm.

9.30 A Cassegrain telescope uses two mirrors as shown in Fig.9.30. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?

Ans – The objective lens’s focal length is 140 cm.

Eyepiece focal length, fe = 5 cm

• The distance between the objective lens and the eyepiece when adjusted normally.

f0 + fe = 140 + 5 = 145 cm

9.31 Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.33. A current in the coil produces a deflection of 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Ans – d = 20 mm is the distance between the secondary mirror and the objective mirror.

R1 = 220 mm is the objective mirror’s radius of curvature.

9.32 Figure 9.34 shows an equiconvex 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 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?

Ans – The distance of the needle from the lens when there is liquid is equal to the focal length f of the convex lens and piano concave lens created by the liquid below it, or f = 45 cm.

Also n = 1.5

The distance between the needle and the lens is only equal to the convex lens’s focal length, or f = 30 cm, when there is no liquid present.

If f2 is the plane concave lens’s produced focal length

## Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Solved Exercise Questions Free PDF Download #### Robin Singh

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