Question 32 of 295

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

Your answerCheck answer

Show answer and solution▾

Answer: 32

As per sign conventions, object distance ${u}_{A}=-45cm$, focal length $f=-20cm$.
From the mirror formula
$\dfrac{1}{{v}_{A}}+\dfrac{1}{{u}_{A}}=\dfrac{1}{f} \Rightarrow \dfrac{1}{{v}_{A}}+\dfrac{1}{-45}=\dfrac{1}{-20} \Rightarrow {v}_{A}=\dfrac{-45\times (-20)}{-45-(-20)}=\dfrac{-900}{25} \Rightarrow {v}_{A}=-36cm$
And ${u}_{B}=-35cm$
$\dfrac{1}{{v}_{B}}+\dfrac{1}{{u}_{B}}=\dfrac{1}{f} \Rightarrow \dfrac{1}{{v}_{B}}+\dfrac{1}{-35}=\dfrac{1}{-20} \Rightarrow {v}_{B}=\dfrac{-35\times (-20)}{-35-(-20)}=\dfrac{700}{-15} \Rightarrow {v}_{B}=-\dfrac{140}{3}cm$
Now, length of image$={v}_{A}-{v}_{B}$
$=\left(-36+\dfrac{140}{3}\right)cm$
$=\dfrac{-108+140}{3}cm$
$=\dfrac{32}{3}cm$
$∴x=32$