Question

A man can see through a small hole the top end of the thin rod of height $10\text{cm}$ placed as shown below. The beaker height is $30\text{cm}$ and its radius $10\text{cm}$. When beaker is filled with a liquid upto $20\text{cm}$, he can see the lower end of rod. The refractive index of the liquid is

• A. $\frac{3}{2}$
• B. $\sqrt{\frac{3}{2}}$
• C. $\sqrt{\frac{5}{2}}$
• D. $\frac{5}{2}$

Answer 1

The correct option is C $\sqrt{\frac{5}{2}}$


In figure,
$\sin r=\frac{CE}{BC}$ $($In $△$$\text{BCE})$

$\Rightarrow \sin r=\frac{10}{\sqrt{{20}^2+{10}^2}}=\frac{10}{10\sqrt{5}}=\frac{1}{\sqrt{5}}$

and from $△BFD$

$\sin i=\frac{BF}{BD}=\frac{10}{\sqrt{{10}^2+{10}^2}}=\frac{10}{10\sqrt{2}}$

$\Rightarrow \sin i=\frac{1}{\sqrt{2}}$

From Snell's law , ${\mu }_1\sin i={\mu }_2\sin r$

Here, ${\mu }_1$ is refractive index of air and ${\mu }_2$ is the refractive index of liquid.

$\Rightarrow \frac{\sin i}{\sin r}=\frac{{\mu }_2}{{\mu }_1}$

Substituting the given values we get,

${\mu }_2=\frac{\frac{10}{10\sqrt{2}}}{\frac{10}{10\sqrt{5}}}\times 1=\sqrt{\frac{5}{2}}$

Hence, option (c) is the correct answer.