A point source is placed at a distance of 15 | vedclass

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(C) For the lens: $u = -15 \, cm$,$f = +10 \, cm$. Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} - \frac{1}{-1

Vedclass · Accepted Answer

$6$

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(C) For the lens: $u = -15 \, cm$,$f = +10 \, cm$. Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} - \frac{1}{-15} = \frac{1}{10} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$. So,$v = 30 \, cm$. For the image to form on the object itself,the rays must strike the convex mirror normally. This happens if the rays are directed towards the center of curvature of the mirror. The distance of the image from the lens is $30 \, cm$. The mirror is placed at a distance $d$ from the lens. The distance of the mirror from the image formed by the lens is $(30 - d)$. For the rays to be normal to the mirror,this distance must be equal to the radius of curvature $R = 2f = 2 	imes 12 = 24 \, cm$. Therefore,$30 - d = 24 \implies d = 6 \, cm$.

$A$ point source is placed at a distance of $15 \, cm$ from a converging (convex) lens of focal length $10 \, cm$. At what distance (in $cm$) should a convex mirror of focal length $12 \, cm$ be placed so that the image is formed on the object itself?

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