Question

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 $\frac{x}{3}cm$. The value of $x$ is ______

Answer 1

Answer: 32

$U_A=-45cm,f=-20cm$
$V_A=\frac{-45\times (-20)}{-45-(-20)}=\frac{-900}{25}=-36cm$
$\text{ And }{U}_B=-35cm$
$\therefore V_B=\frac{-35\times (-20)}{-35-(-20)}=\frac{700}{-15}$
$\therefore V_A-V_B=\text{ length of image }$
$=\left(-36+\frac{140}{3}\right)cm$
$=\frac{-108+140}{3}cm$
$=\frac{32}{3}cm$
$\therefore x=32$