An object of height $4 c m$ is placed at a distance of 20 $c m$ from a converging lens of focal length 30 $c m$ in an upright position. Then, height and nature of image is

c m

and virtual

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c m

and virtual

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c m

and virtual

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c m

and real

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Solution

The correct option is B 12 $c m$ and virtual
Using lens formula,
$\frac{1}{v} + \frac{1}{20} = \frac{1}{30}$
$\frac{1}{v} = \frac{1}{30} - \frac{1}{20} = \frac{2}{60} - \frac{3}{60} = \frac{- 1}{60}$
$v = - 60 c m$
$m = \frac{h_{i}}{h_{o}} = \frac{- 60}{- 20} = + 3$
$h_{o}$ = 4 $c m$
$\frac{h_{i}}{4} = 3$
$h_{i} = 3 \times 4 = 12 c m$
Nature of image is magnified.

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An object of height 4 cm is placed at a distance of 20 cm from a converging lens of focal length 30 cm in an upright position. Then, height and nature of image is

The correct option is B 12 cm and virtual Using lens formula, 1v+120=130 1v=130−120=260−360=−160 v=−60 cm m=hiho=−60−20=+3 ho = 4 cm hi4=3 hi=3×4=12 cm Nature of image is magnified.

An object of height $4 c m$ is placed at a distance of 20 $c m$ from a converging lens of focal length 30 $c m$ in an upright position. Then, height and nature of image is