Question

A disc is placed on a surface of pond which has refractive index $5/3$. A source of light is placed $4\,m$ below the surface of liquid. The minimum radius of disc needed so that light is not coming out is

• A. $\infty$
• B. $3\,m$
• C. $6\,m$
• D. $4\,m$

Answer 1

Answer: (B)

$3\,m$

Answer 2

Here $OP = 4\,m$, $\mu = \frac{5}{3}$ Let $PR = r =$ radius of disc $sin\, C = \frac{1}{\mu} = \frac{3}{5} \quad...(i)$ From figure, $sin\,C = \frac{PR}{OR}$ =$\frac{PR}{\sqrt{OP^{2} +PR^{2}}} = \frac{r}{\sqrt{4^{2}+r^{2}}} \quad...\left(ii\right)$ Equating $\left(i\right)$ and $\left(ii\right)$, we get $\frac{3}{5} = \frac{r}{\sqrt{4^{2}+r^{2}}}; \frac{9}{25} = \frac{r^{2}}{4^{2}+r^{2}}$ $25r^{2} = 9\left(4^{2}+r^{2}\right)$ $\Rightarrow 25 r^{2} = 144 +9r^{2}$ $16r^{2} = 144$ $r^{2} = \frac{144}{16} = 9$ $r= 3 \,m$