Question

A square $ABCD$ of side $1mm$ is kept at a distance $!5cm$ in front of the concave mirror a shown in the figure. The focal length of the mirror is $10cm$. The length of the perimeter of its image will be (nearly):

• A. $8mm$
• B. $2mm$
• C. $12mm$
• D. $6mm$

Answer 1

The correct option is A $8mm$

Given,

$f=10cm$

$h_0=1mm=0.1cm$

$d_0=15cm$

$\frac{1}{d_0}+\frac{1}{d_i}=\frac{1}{f}$

$d_i=\frac{1}{\frac{1}{f}-\frac{1}{d_0}}=\frac{1}{1/10-1/15}=30cm$

So, $\frac{h_i}{h_0}=-\frac{d_i}{d_0}$

$h_i=-h_0\frac{d_i}{d_0}$

$=-1\frac{30}{15}=-2mm$

So, $p=4.h_i=4.d=8mm$ where p is perimeter.