Question

If eye is kept at a depth $h$ inside the water of refractive index and viewed outside, then the diameter of circle, through which the outer objects becomes visible, be

• A. $\frac{h}{\sqrt{\mu ^{2} + 1}}$
• B. $\frac{h}{\sqrt{\mu ^{2} - 1}}$
• C. $\frac{2 h}{\sqrt{\mu ^{2} - 1}}$
• D. $\frac{h}{\sqrt{2 \mu ^{2} - 1}}$

Answer 1

Answer: (C)

Let r be the radius of the circle through which other objects become visible. The ray of light must be incident at critical angle C.
$sin C=\frac{1}{\mu }=\frac{r}{\sqrt{r^{2} + h^{2}}}$
$\mu^2 r^2=r^2+h^2 \Rightarrow\left(\mu^2-1\right) r^2=h^2$
$\Rightarrow \, \, \, r=\frac{h}{\sqrt{\mu ^{2} - 1}}$
$\therefore \, \, \, Diameter=2r=\frac{2 h}{\sqrt{\mu ^{2} - 1}}$