Question

A point object O is placed in front of a transparent slab at a distance x from its closer surface. It is seen from the other side of the slab by light incident nearly normally to the slab. The thickness of the slab is t and its refractive index is $\mu .$ The apparent shift in the position of the object is

• A. $t(1+\frac{t}{\mu })$
• B. $t(1-\frac{t}{\mu })$
• C. $t(2-\frac{t}{\mu })$
• D. $t(2+\frac{t}{\mu })$

Answer 1

The correct option is B

$t(1-\frac{t}{\mu })$

The situation is shown in figure. Because of the refraction at the first surface, the image of O is formed at $O_1.$ For this refractiion, the real depth is AO = x and the apparent depth is $A{O}_1.$ Also the first medium is air and the second is the slab. Thus,
$\frac{x}{A{O}_1}=\frac{1}{\mu }or,A{O}_1=\mu x.$
The point $O_1$ acts as the object for the refractiion at the second surface. Due to this refraction the image of $O_1$ is formed at I. Thus,
$\frac{B{O}_1}{BI}=\mu$
or, $\frac{AB+A{O}_1}{BI}=\mu or\frac{t+\mu x}{BI}=\mu$
or $BI=x+\frac{t}{\mu }.$
The net shift in $OI=OB-BI=(x+t)-(x+\frac{t}{\mu })$
$=t(1-\frac{t}{\mu })$
which is independent of x.