A virtual image twice as big as the object is formed by a convex 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 B $30 c m$
For a lens the magnification is given as:
$m = \frac{v}{u} For a virtual image m = \frac{v}{u} = + 2, w h e n u = - 10 c m v = 2 u From lens formula \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \frac{1}{2 u} - \frac{1}{u} = \frac{1}{f} f = 20 c m For real image v is positive and u is negative hence v = - 2 u \frac{1}{(- 2 u)} - \frac{1}{u} = \frac{1}{f} \frac{1}{f} = - \frac{3}{2 u} f = - \frac{2 u}{3} u = - 30 c m$

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A virtual image twice as big as the object is formed by a convex 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 B 30cmFor a lens the magnification is given as: m=vuFor a virtual imagem=vu=+2,when u=−10cmv=2uFrom lens formula1v−1u=1f12u−1u=1ff=20cmFor real image v is positive and u is negativehencev=−2u1(−2u)−1u=1f1f=−32uf=−2u3u=−30cm