Question 296: Ray Optics

Question

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

Accepted Answer

$	ext { A: }$
$$
 \begin{aligned}
& \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\
& \Rightarrow \frac{1}{v}+\frac{1}{-45}=\frac{1}{-20} \\
& \Rightarrow \frac{1}{v}=\frac{1}{45}-\frac{1}{20}=\frac{4-9}{180}=-\frac{1}{36} \\
& \Rightarrow v=-36 \mathrm{~cm}
\end{aligned}
$$
B: $\frac{1}{v}+\frac{1}{-35}=\frac{1}{-20}$
$\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{20}=\frac{4-7}{140}$
$\Rightarrow \quad v=-\frac{140}{3}$
$\Rightarrow$ length of image $=\frac{140}{3}-36=\frac{32}{3} \mathrm{~cm}$
$\Rightarrow x=32$

Ray Optics

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