Question

The bottom of a container is 14.0 cm thick glass $(\mu =1.5)$ slab.The container contains two immiscible liquids A and B of depths 6.0 cm and 8.0 cm respectively.What is the apparent position of a scratch on the outer surface of the bottom of the glass slab when viewed through the container ? Refractive indices of A and B are 1.4 and 1.3 respectively.

Answer 1

Given: The bottom of a container is 14.0 cm thick glass ($\mu =1.5$) slab.The container contains two immiscible liquids A and B of depths 6.0 cm and 8.0 cm respectively. Refractive indices of A and B are 1.4 and 1.3 respectively.

To find the apparent position of a scratch on the outer surface of the bottom of the glass slab when viewed through the container

Solution:

The total apparent shift of the scratch on the outer surface is obtained by considering refraction at all the three media.

I.e.,

$t_1(1-\frac{1}{{\mu }_1})+t_2(1-\frac{1}{{\mu }_2})+t_3(1-\frac{1}{{\mu }_3})$

where $t_1,t_2,t_3$ are the thickness of the glass, liquid A,liquid B respectively.

and ${\mu }_1,{\mu }_2,{\mu }_3$ are the refractive index of glass, liquid A, liquid B respectively.

$⟹=14(1-\frac{1}{1.5})+6(1-\frac{1}{1.4})+8(1-\frac{1}{1.3})⟹=14\left(0.33\right)+6\left(0.28\right)+8\left(0.23\right)⟹=4.62+1.68+1.84⟹=8.14$

is the apparent position of a scratch on the outer surface of the bottom of the glass slab when viewed through the container.