NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments (Part IV)

NCERT Solutions for Class 12 Physics, Chapter 9: Ray Optics & Optical Instruments are available here. The solutions are also available for download in PDF format. Students can download these solutions with the help of download link given at the end. This chapter has a large number of problems, so, we have provided the solutions in several parts. In Part I, we have provided solutions from question number 9.1 to 9.9. Part II, has solutions from question number 9.9 to 9.20. Part III has solutions from question number 9.21 to 9.30. In this part, you will find solutions from question number 9.31 to 9.31. You can also download these solutions in the PDF format with the help of download link given at the end.

NCERT Solutions for Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments from question number 9.21 to 9.30 are given below:

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.]

[Questions number 9.29 and 9.30 are available in Part III]

Solution 9.31:


Area of virtual image of each square, A = 6.25 sq mm

Area of each image, A’ = 1 sq mm


Magnification = (A/A’)1/2 = 2.5

As, v = 2.5 u

So, (1/2.5u) – (1/u) = 1 / 10


u = ‒ 6 cm

|v| = 15 cm

Since, v is less than the normal near point of 25 cm, virtual image cannot be seen by the eye distinctly.

NCERT Exemplar: Class 12 Physics – Chapter 9,  Ray Optics and Optical Instruments

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?

Solution 9.32:

(a) Angular magnification can be achieved by placing object much closer than 25 cm. This is exactly what magnifying glasses do.

(b) Yes. This is so because the angle subtended at the eye is slightly less than the angle subtended at the lens.

(c) It is not easy to make lens of very small focal length. Moreover, the spherical and chromatic aberration increases with decreasing focal length.

(d) For small fe, angular magnification increases. It is given by, (25/fe) + 1

Magnification of the objective, = 1 / [|uo| / fo] – 1

For |uo| > fo, m is large. So fo is small.

(e) To maximize our field of view, the light refracted by objective should be collected. This is achieved at eye-ring, the correct location of which depends no the separation between the objective and the eye-piece.

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?

Solution 9.33:


fo = 1.25 cm

fe = 5 cm

Angular magnification of eye-piece at 25 cm = (25 / 5) + 1 = 6

Magnification of objective = 30 / 6 = 5.

Since, vo / uo = 5


(1/5uo)  – (1/uo) = 1/1.25

Which yields,

uo = ‒1.5 cm

And, vo = 7.5 cm

For eyepiece

|ue| = 25 / 6 = 4.17 cm

Total separation between objective and eye-piece = 7.5 + 4.17 = 11.67 cm

For the desired magnification the objective should be placed 1.5 cm away from the objective.

Question 9.34: 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)?

Solution 9.34:


fo = 140 cm

fe = 5 cm

(a) For normal adjustment, m = fo / fe = 140/5 = 28

(b) For disinct vision, m = [fo/fe].[1 + {fe/25}]

Substitution yields, m = 33.6.

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 25cm?

Solution 9.35:

(a) The separation will be = fo + fe = 145 cm

(b) Angle subtended by the tower = h / u = 100 / 3000 = 1 / 30 rad

Angle subtended by the image produced by the objective = h’ / fo = h’/140


h’/140 = 1/30 or h’ = 4.7 cm

(c) Magnification of the eye-piece = 1 + (d/fe) = 1 + (25 / 5) = 6

Height of the final image = m.h’ = 6 x 4.7 = 28 cm.

CBSE Class 12 Physics Notes: All Chapters

Question 9.36: A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. 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 140 mm, where will the final image of an object at infinity be?

Solution 9.36:


d = 20 mm

R1 = 220 mm

R2 = 140 mm

So, f1 = 220 / 2 = 110 mm

And f2 = 140 / 2 = 70 mm

Large mirror forms an image, which is virtual object for smaller mirror.


u = 110 – 20 = 90 mm

Using mirror formula

(1/v) + (1/90) = (1/70)

Which gives, v = 315 mm.

Question 9.37: Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. 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?

Solution 9.37:

The deflection of reflected rays is twice the angle of rotation of the mirror.

So, 2A = 7o


d = 1.5 × tan2A = 18.4 cm.

Question 9.38: Figure 9.37 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?

Solution 9.38:


n = 1.5

d = 45 cm

d’ = 30 cm

Rearranging the given information,

Focal length of equiconvex lens and liquid

f = 45 cm

Focal length of the equiconvex lens

f’ = 30 cm

Therefore, the equivalent focal length

(1/f) = (1/f’) + (1/f’’)

Or (1/45) = (1/30) + (1/f’’)

Or f’’ = ‒90 cm

Now, by the relation,

1/f’ = (n ‒ 1) [(1/R) + (1/R)]

Substituting the values,

1 / 30 = (0.5) × [2/R]

Or R = 30 cm

In case of liquid,

1/f’’ = (n’ – 1) × [(1/R) – (1/∞)]


n’ = refractive index of the liquid

R’ of plane mirror is infinite.

Solving the above expression,

(‒1/90) = (n’ – 1) × [(1/30) ‒ 0]

Or n’ – 1 = 1/3

Which gives, n’ = 4/3 = 1.33.

Download NCERT Solutions for Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments

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