Question

Refractive index of air is ${\mu }_1=1$ (outside medium), refractive index of slab is ${\mu }_2=1.5$, thickness of slab $t=6\text{cm}$ . If the object is placed at $28\text{cm}$ from nearest surface of slab, find the final image position wrt mirror. ( Assume that mirror is very near to the slab)

• A. $30\text{cm}$ behind the mirror.
• B. $32\text{cm}$ behind the mirror.
• C. $36\text{cm}$ behind the mirror.
• D. $26\text{cm}$ behind the mirror.

Answer 1

The correct option is A $30\text{cm}$ behind the mirror.

A ray of light from the object first encounters a glass slab, then a mirror and finally a glass slab again A slab simply shifts the object along the axis by a distance
$s_1=t(1-\frac{1}{\mu })=2\text{cm}$
Direction of shift is towards left. Therefore the object appears to be at $I_1$ which is $28–2=26\text{cm}$ from the slab.
Mirror: The object for the mirror is the image $I_1$ formed after shift due to the slab.
Therefore object distance from the mirror is $26+6=32\text{cm}$. The image will now be formed $32\text{cm}$ behind the mirror.
Slab: The ray now travels through the slab again but this time from right to left. Therefore it is shifted again by a distance of cm, but towards the right. Thus final position the image is $32–2=30\text{cm}$ behind the mirror.
Conclusion: Final image is formed $30\text{cm}$ behind the mirror.