• # question_answer 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 independent of $x$ then its value, is A) $\left[ 1-\frac{1}{\mu t} \right]$                     B) $\left[ t-\frac{\mu }{t} \right]$ C) $t\left[ 1-\frac{t}{\mu } \right]$                      D) $[\mu t-1]$

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 refraction, 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 Oi acts as the object for the refraction 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         $\frac{AB+B{{O}_{1}}}{BI}=\mu$         or $\frac{t+\mu x}{BI}=\mu$ or         $BI=x+\frac{t}{\mu }$ The net shift is $OI=OB=\,(x+t)-\,\left( x-\frac{t}{\mu } \right)$ $=t\,\left( 1-\frac{1}{\mu } \right)$ Which is independent of $x$.