Question

Two concave mirrors are placed $40\text{cm}$ apart and are facing each other. A point object lies between them at a distance of $12\text{cm}$ from the mirror of focal length $10\text{cm}$. The other mirror has focal length $15\text{cm}$. Find the location of final image formed after two reflections, first at the mirror nearer to the object and $2^{nd}$ at the other mirror.

• A. $17\text{cm}$ from the $2^{nd}$ mirror
• B. $8.57\text{cm}$ from the $1^{st}$ mirror
• C. $8.57\text{cm}$ from the $2^{nd}$ mirror
• D. $17\text{cm}$ from the $1^{st}$ mirror

Answer 1

The correct option is C $8.57\text{cm}$ from the $2^{nd}$ mirror


For first reflection

$u_1=-12\text{cm};f_1=-10\text{cm}$

using mirror formula

$\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f_1}$

$\Rightarrow \frac{1}{v_1}=\frac{-1}{10}+\frac{1}{12}$

$\Rightarrow v_1=-60\text{cm}$

The rays after reflection are going to converge at a point $I_1=60\text{cm}$ from the first mirror i.e, $60-40=20\text{cm}$ behind the mirror.

For second reflection

As incident rays are directed toward right, all quantities towards right will be taken $+ve$.

$u_2=+20\text{cm};f_2=-15\text{cm}$

Again using mirror formula,

$\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f_2}$

$\Rightarrow \frac{1}{v_2}=\frac{-1}{15}+\frac{-1}{20}$

$\Rightarrow v_2=\frac{-300}{35}=-8.57\text{cm}$

Hence correct option is (c).