When an object of height 3 cm is placed at a distance of 30 cm from a lens, a real image is formed at a distance of 60 cm. Find out the height of the object.

-6 cm

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+6 cm

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-3 cm

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+3 cm

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Solution

The correct option is A
-6 cm

Here,

Image distance, u = -30 cm

Object distance, v = +60 cm

Height of object $h_{o}$ = 3 cm

Height of image $h_{i}$ = ?

Using magnification

$m = \frac{u}{u}$ $= \frac{+ 60}{- 30} = - 2$

Also, $m = \frac{h_{i}}{h_{o}} = \frac{h_{i}}{3}$

$\Rightarrow \frac{h_{i}}{3} = - 2$

$\Rightarrow h_{i} = - 6 c m$

The negative sign shows that the image is real and inverted.

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When an object of height 3 cm is placed at a distance of 30 cm from a lens, a real image is formed at a distance of 60 cm. Find out the height of the object.

The correct option is A -6 cm Here, Image distance, u = -30 cm Object distance, v = +60 cm Height of object ho= 3 cm Height of image hi= ? Using magnification m=uu =+60−30=−2 Also, m=hiho=hi3 ⇒hi3=−2 ⇒hi=−6 cm The negative sign shows that the image is real and inverted.