# Physics Class 12 NCERT Solutions: Chapter 9 Ray Optics and Optical Instruments Part 13 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q: 29. A card sheet divided into squares each of size is being viewed at a distance of through a magnifying glass (a converging lens of focal length ) 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:

(a) Area of each square,

Object distance,

Focal length of a converging lens,

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

Magnification,

∴ Area of each square in the virtual image

(B) Magnifying power of the lens

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

The magnification magnitude is and the magnifying power is .

The two quantities will be equal when the image is formed at the near point .

Q: 30. (A) At what distance should the lens be held from the figure in

Exercise (Above exercise) 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 .

Image distance,

Focal length,

Object distance

According to the lens formula, we have:

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

(B) Magnification

(C) Magnifying power

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

Q: 31. What should be the distance between the object in Exercise 30 (Above exercise) and the magnifying glass if the virtual image of each square in the figure is to have an area of . Would you be able to see the squares distinctly with your eyes very close to the magnifier?

[Note: Exercises 29 to 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,

Area of each square,

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

But

Focal length of the magnifying glass,

According to the lens formula, we have the relation:

According to the lens formula, we have the relation:

And

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