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

Answer 1

$U_A=-45 \,cm , f =-20 \,cm $
$ V_A=\frac{-45 \times(-20)}{-45-(-20)}=\frac{-900}{25}=-36 \,cm $
$ \text { And } U_B=-35 \,cm $
$ \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$