Question

A bulb is placed at a depth of $ 2\sqrt{7}\,m $ in water $ \left(\mu_{w} = \frac{4}{3}\right) $ and a floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. What is the minimum diameter of the disc?

• A. 8 m
• B. 12 m
• C. 15 m
• D. 20 m

Answer 1

Answer: (B)

$\frac{\sin\,\theta_c}{\sin\,90^{\circ}} = \frac{1}{\mu}$
$ \sin\,\theta_c = \frac{1}{\mu}$
$ \frac{R}{\sqrt{R^2 + D^2}} = \frac{1}{\mu}$
or $ \frac{R^2 + D^2}{R^2} = \mu^2$
or $ 1 + \frac{D^2}{R^2} = \mu^2$
or $ \frac{D^2}{R^2} = \mu^2 -1$
or $R = \frac{D}{\sqrt{\mu^2 -1}}$
$ = \frac{2\sqrt{7}}{\sqrt{(\frac{4}{3})^2 -1}} = \frac{2\sqrt{7}}{\sqrt{\frac{16}{9} - 1}}$
$ = 6 \,cm$
Diameter of disc $= 6m \times 2 = 12 \,m$