A thin cylindrical rod of length 10cm is placed horizontally on the principle axis of a concave mirror of focal length 10cm. The rod is placed in such a way that midpoint of the rod is at 20cm from the pole of mirror. If the length of the image formed by the mirror will be $\frac{x}{3}$ cm, then the value of x is …

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answer is 40.

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Detailed Solution

$\begin{array}{l} \begin{matrix} F = - 10 c m & u_{A} = - 25 c m \\ \frac{1}{F} = \frac{1}{v_{A}} + \frac{1}{μ_{A}} & v_{A} = - \frac{250}{15} c m \end{matrix} \\ u_{B} = - 15 c m \\ \frac{1}{f} = \frac{1}{V_{B}} + \frac{1}{u_{B}} V_{B} = - \frac{150}{5} \\ L_{I} = V_{B} - V_{A} = \frac{40}{3} c m \end{array}$

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