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Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Ray Optics and Optical Instruments

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.

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Payal Gupta | Contributor-Level 10

4 months ago

9.23 Area of each square, A = 1 ${m m}^{2}$

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

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

$\frac{1}{10}$ = $\frac{1}{v}$ - $\frac{1}{- 9}$

$\frac{1}{v}$ = $\frac{1}{10} - \frac{1}{9}$

v = -90 cm

Magnification, m = $\frac{v}{u}$ = $\frac{- 90}{- 9}$ = 10

Therefore the area of each square of the virtual image

= 10 $* 10 {m m}^{2}$ = 100 ${m m}^{2} = 1 {c m}^{2}$

Magnifying power of the lens = $\frac{d}{|u|}$ = $\frac{25}{9}$ = 2.8

The magnification in (a) is not the same as the magnifying power in (b). The magnification magnitude is = $\frac{v}{u}$ and the magnifying power is = $\frac{d}{|u|}$ . The two quantities will be equal when the image is for

...more

9.23 Area of each square, A = 1 ${m m}^{2}$

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

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

$\frac{1}{10}$ = $\frac{1}{v}$ - $\frac{1}{- 9}$

$\frac{1}{v}$ = $\frac{1}{10} - \frac{1}{9}$

v = -90 cm

Magnification, m = $\frac{v}{u}$ = $\frac{- 90}{- 9}$ = 10

Therefore the area of each square of the virtual image

= 10 $* 10 {m m}^{2}$ = 100 ${m m}^{2} = 1 {c m}^{2}$

Magnifying power of the lens = $\frac{d}{|u|}$ = $\frac{25}{9}$ = 2.8

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