Question

A square $ABCD$ of side $1mm$ is kept at distance $15cm$ in front of the concave mirror as shown in figure. The focal length of the mirror is $10cm$. The length of the perimeter of its image will be

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

Answer 1

Answer: (A)

$8mm$

From the question we have

$u=-15cm$ and $f=-10cm$

The mirror formula is given as

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}$

$\Rightarrow \frac{1}{v}=\frac{2-3}{30}=-\frac{1}{30}$

$\Rightarrow v=-30cm$

Magnification $m=\frac{v}{u}=\frac{30}{15}$

Also magnification $m=\frac{h_i}{h_o}=\frac{h_i}{0.1}$

$\therefore \frac{30}{15}=\frac{h_i}{0.1}$

$h_i=\frac{30\times 0.1}{15}=0.2$

Since the object is a cube, the perimeter can be given as, $4\times \text{side}$

$\therefore p=4\times 0.2=0.8cm=8mm$

Hence the correct answer is option (A).