Question

A point source of light is kept below the surface of water $(n_w=4/3)$ at a depth of $\sqrt{7}m$. The radius of the circular bright patch of light noticed on the surface of water is ______ $m$.

• A. $\frac{3}{\sqrt{7}}$
• B. $3$
• C. $\frac{\sqrt{7}}{3}$
• D. $\sqrt{7}$

Answer 1

The correct option is B $3$

Let the radius of the circular bright patch of light be $r$. The ray of light grazes at the surface of water, thus angle $\theta$ must be equal to critical angle.

Given : $h=\sqrt{7}m$ $n_w=4/3$

Thus $\sin \theta =\frac{1}{n_w}$

OR $\sin \theta =\frac{3}{4}$

$\therefore$ $\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-(\frac{3}{4}{)}^2}=\frac{\sqrt{7}}{4}$

Thus $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{3}{\sqrt{7}}$

But $\tan \theta =\frac{r}{h}$

$\therefore$ $\frac{r}{\sqrt{7}}=\frac{3}{\sqrt{7}}$ $⟹r=3m$