Question

An object is placed in front of a slab $(\mu =1.5)$ of thickness $6\text{cm}$ at a distance of $28\text{cm}$ from it, while the other face of the slab is silvered. Find the position of the final image.

• A. $54\text{cm}$ from surface $1$
• B. $54\text{cm}$ from surface $2$
• C. $36\text{cm}$ from surface $1$
• D. $36\text{cm}$ from surface $2$

Answer 1

The correct option is C $36\text{cm}$ from surface $1$

Refraction at surface :-

${\mu }_1=1,d_1=28\text{cm},{\mu }_2=1.5$

$d_2=\frac{d_1}{{\mu }_{relative}}=\frac{d_1}{1/\mu }$

$\Rightarrow d_2=\mu d_1=1.5\times 28=42\text{cm}$

Therefore, 1st image $I_1$ will be formed at $42\text{cm}$ from the 1st surface.


Reflection at surface $2:-$

For silvered surface object distance will be

$\therefore u=42+6=48\text{cm}$

$\therefore$ Image by the second surface will be $I_1$ at $48\text{cm}$ behind the mirror.

$I_1$ will act like an object for second surface.

Second refraction at surface $1:-$

Object distance from surface $1$ will be

$d_1^{\prime }=48+6=54\text{cm}$

So, image distance from surface $1$ will be

$d_2^{\prime }=\frac{d_1^{\prime }}{{\mu }_{relative}}=\frac{d_1^{\prime }}{\left(n_{incident}/n_{refraction}\right)}$

$\Rightarrow d_2^{\prime }=\frac{d_1^{\prime }}{1.5/1}=\frac{54}{1.5}$

$\therefore d_2^{\prime }=36\text{cm}$

Hence, option $\left(c\right)$ is correct.