A virtual image twice as big as the object is formed by a concave lens when the object is 10 cm away from it. A real image twice as big as the object will be formed when it is placed at a distance from the lens :

40 c m

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30 c m

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20 c m

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15 c m

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Solution

The correct option is D $30 c m$
From the formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
We have, $u = - \frac{f (m - 1)}{m}$
For virtual image, $m = + 2$ , and
for real image, $m = - 2$
$∴ - 10 = - \frac{f (2 - 1)}{2}$
(Real object distances are always negative.)
$\Rightarrow f = 20$
For the real image, $u = \frac{f (- 2, - 1)}{- 2} = \frac{- 20 (- 1)}{- 2} = 30$

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A virtual image twice as big as the object is formed by a concave lens when the object is 10 cm away from it. A real image twice as big as the object will be formed when it is placed at a distance from the lens :

The correct option is D 30cmFrom the formula, 1f=1v−1uWe have, u=−f(m−1)mFor virtual image, m=+2, andfor real image, m=−2∴−10=−f(2−1)2(Real object distances are always negative.)⇒f=20For the real image, u=f(−2,−1)−2=−20(−1)−2=30