Question

The bottom of a container is of a $4.0 \,cm$ thick glass slab. The container contains two immiscible liquids $A (\mu = 1.4)$ and $B (\mu = 1.3)$ of depths $6.0\, cm$ and $8.0 \,cm$ respectively. What is apparent position of a scratch on the outer surface of the bottom of the glass slab when viewed through the container?

• A. 4.89 cm
• B. 2.79 cm
• C. 1.89 cm
• D. 5.39 cm

Answer 1

Answer: (A)

Let $m_1, m_2, m_3$ be the refractive index of glass slab, liquid $A$ and liquid $B$ respectively. The corresponding thickness are $t_1, t_2$ and $t_3$ respectively. The apparent position of the scratch on the outer surface is obtained by taking into account refraction at the three media and is given by
$= t_{1 }\left(1 -\frac{1}{\mu_{1}}\right) + t_{2}\left(1 -\frac{1}{\mu_{2}}\right) + t_{3}\left(1 -\frac{1}{\mu_{3}}\right) $
$ =4 \left(1 -\frac{1}{1.5}\right) + 6 \left(1 -\frac{1}{1.4}\right) +8 \left(1 -\frac{1}{1.3}\right) $
$ = \frac{4 \times 0.5}{1.5} + \frac{6\times 0.4}{1.4} + \frac{8 \times 0.3}{1.3} $
= 1.33 + 1.71 + 1.85 = 4.89 cm