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Q.

A thin cylindrical rod of length $10 cm$ is placed horizontally on the principal axis of a concave mirror of focal length $20 cm$ . The rod is placed in such a way that mid point of the rod is at $40 cm$ from the pole of mirror. The length of the image formed by the mirror will be $\frac{x}{3} cm$ . The value of $x$ is _______.

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answer is 32.

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Detailed Solution

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$\begin{array}{l} \frac{1}{v_{1}} - \frac{1}{35} = - \frac{1}{20} \\ v_{1} = - \frac{700}{15} cm = - \frac{140}{3} cm \\ \frac{1}{v_{2}} - \frac{1}{45} = - \frac{1}{20} \Rightarrow v_{2} = - \frac{900}{25} cm = - 36 cm \end{array}$
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Length $= \frac{140}{3} - 36 = \frac{32}{3} cm$

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