Question

For the observer just above the medium 1 $({\mu }_1=\frac{4}{3})$, the shift in the position of coin for the arrangement shown in figure will be :

• A. $18\text{cm}$
• B. $25\text{cm}$
• C. $7\text{cm}$
• D. $10\text{cm}$

Answer 1

The correct option is C $7\text{cm}$

Each medium with the given thickness will behave as a slab.
The formula for normal shift due to slab is,
$S=t(1-\frac{1}{\mu })$
Net shift in position of coin will be,
$S_{total}=S_1+S_2$
$\Rightarrow S_{total}=t_1(1-\frac{1}{{\mu }_1})+t_2(1-\frac{1}{{\mu }_2})$
$S_{total}=16(1-\frac{3}{4})+9(1-\frac{2}{3})$
$\therefore S_{total}=\frac{16}{4}+\frac{9}{3}=7\text{cm}$

Why this question? Tip: Visualize the medium as a transparent slab and apply principle of superposition to get net shift in position of object.