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 10 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.

116 Views|Posted 8 months ago
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1 Answer
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8 months ago

9.23 Area of each square, A = 1 mm2

Object distance, u = - 9 cm

Focal length of the converging lens, f = 10 cm

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

1f = 1v - 1u

110 = 1v - 1-9

1v = 110-19

v = -90 cm

Magnification, m = vu = -90-9 = 10

Therefore the area of each square of the virtual image

= 10 *10mm2 = 100 mm2=1cm2

Magnifying power of t

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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