Question

An object is seen through a glass slab of thickness $36\text{cm}$ and refractive index $3/2$. If the system of object, observer and the glass slab is dipped in the water$(\mu =4/3)$ keeping all the distances fixed, then the shift produced by the glass slab will be :

• A. $12\text{cm}$
• B. $4\text{cm}$
• C. $6\text{cm}$
• D. $8\text{cm}$

Answer 1

The correct option is B $4\text{cm}$

Here the observer is not in air, the surroundings for slab is water $({\mu }_w=\frac{4}{3})$.

Thus, shift produced for object position:
$S=t(1-\frac{{\mu }_w}{{\mu }_{slab}})$

$\because {\mu }_{slab}=\frac{3}{2},t=36\text{cm}$

$\Rightarrow S=36[1-\frac{\left(\frac{4}{3}\right)}{\left(\frac{3}{2}\right)}]$

or $S=36\left[\frac{\frac{3}{2}-\frac{4}{3}}{\left(\frac{3}{2}\right)}\right]=36[\frac{1}{6}\times \frac{2}{3}]$

$\therefore S=36\left(\frac{2}{18}\right)=4\text{cm}$

Hence, option (b) is the correct answer.

Why this question? Caution: If the observer is not in air, pay attention to use the formula $S=t(1-\frac{{\mu }_{surrounding}}{{\mu }_{slab}})$