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 image?

-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, u=-30 cm, v=+60 cm, $h_{o} = 3 c m$ , $h_{i} = ?$

Using, magnification, $m = (\frac{v}{u}) = \frac{+ 60}{- 30} = - 2$

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

Now, $- 2 = \frac{h_{i}}{3}$

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

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 image?

The correct option is A -6 cm Here, u=-30 cm, v=+60 cm, ho=3 cm, hi=? Using, magnification, m=(vu)=+60−30=−2 Also, magnification, m=hiho=hi3 Now, −2=hi3 ⇒hi=−6 cm The negative sign shows that the image is real and inverted.