A thin cylindrical rod of length 10 cm is placed horizontally on the principle axis of a concave mir

Question

A thin cylindrical rod of length 10 cm is placed horizontally on the principle axis of a concave mirror of focal length 20 cm . The rod is placed in a 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 $x / 3 \mathrm{~cm}$. The value of $x$ is _______ .\

Professor PC Thomas Classes · Answer

(32) : $f=-20 \mathrm{~cm}$ Length of object $=10 \mathrm{~cm}$\ $u_{A}=-45 \mathrm{~cm}$\ $\frac{1}{u_{A}}+\frac{1}{v_{A}}=\frac{1}{f}-\frac{1}{45}+\frac{1}{v_{A}}=-\frac{1}{20}$\ $\Rightarrow \quad v_{A}=-36 \mathrm{~cm}$\ As, $u_{B}=-35 \mathrm{~cm}, f=-20 \mathrm{~cm}$\ $\frac{1}{-35}+\frac{1}{v_{B}}=-\frac{1}{20} ; \frac{1}{v_{B}}=\frac{1}{35}-\frac{1}{20}=\frac{20-35}{700}$\ $v_{B}=\frac{700}{-15} \mathrm{~cm}$\ Length of image $=\left|v_{A}-v_{B}ight|=\left|36-\frac{700}{15}ight|$ $$ =\frac{160}{15}=\frac{32}{3} \mathrm{~cm} $$ On comparing with given value with $\frac{x}{3}$, we get $x=32$\