A convex lens of focal length 20,cm is placed | vedclass

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(D) For the lens,using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given $u = -60\,cm$ and $f = +20\,cm$,we have $\frac{1}{v} - \frac{

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$10$

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(D) For the lens,using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given $u = -60\,cm$ and $f = +20\,cm$,we have $\frac{1}{v} - \frac{1}{-60} = \frac{1}{20}$.$\frac{1}{v} + \frac{1}{60} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$.So,$v = 30\,cm$. This image is formed $30\,cm$ behind the lens.For the final image to coincide with the object,the rays must strike the mirror normally (perpendicularly). This happens if the rays are directed towards the center of curvature $(C)$ of the mirror.The distance of the image from the mirror is $30\,cm - 10\,cm = 20\,cm$. Thus,the radius of curvature $R = 20\,cm$.The focal length of the mirror is $f_m = \frac{R}{2} = \frac{20}{2} = 10\,cm$.

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