Question

A small bulb is placed at the bottom of a tank containing water to a depth of $1m$. Find the critical angle for water air interface and also calculate the diameter of the circular bright patch of light formed on the surface of water [Refractive index of water$=1.33$]

Answer 1

$PR=1m$

Let C be the critical angle

$n_2\sin C=n_1\sin 90$

$\frac{n_2}{n_1}\sin =1$

$1.33\sin C=1$

$\sin C=\frac{3}{4}$

$C={\sin }^{-1}\frac{3}{4}$

$={48}^o$

radius = PQ

$\frac{PQ}{PR}=\tan \angle PRQ=\tan C$

$\frac{\sin C}{\cos C}=\frac{\sin C}{\sqrt{1-{\sin }^{2}C}}$

$\frac{\frac{3}{4}}{\sqrt{1-\frac{9}{16}}}=\frac{3}{\sqrt{7}}$

$PQ=\frac{3}{\sqrt{7}}PR=\frac{3}{\sqrt{7}}m$