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