# 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. - Physics

Numerical

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.

#### Solution

(a) Area of each square, A = 1 mm2

Object distance, u = −9 cm

Focal length of a converging lens, f =  9 cm

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

1/"f" = 1/"v" - 1/"u"

1/10 = 1/"v" + 1/9

1/"v" = -1/90

∴ v = −90 cm

Magnification, "m" = "v"/"u"

= (-90)/-9

= 10

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

= 102 × 1

= 100 mm2

= 1 cm2

(b) Magnifying power of the lens = "d"/|"u"| = 25/9 = 2.8

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

The magnification magnitude is (|"v"/"u"|) and the magnifying power is ("d"/|"u"|).

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

Concept: Refraction at Spherical Surfaces and Lenses - Refraction by a Lens
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 9.23 | Page 346
NCERT Class 12 Physics Textbook
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 29 | Page 348
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