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 become visible, will be :

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

Answer 1

The correct option is C $\frac{2h}{\sqrt{{\mu }^2-1}}$

Let $r$ be the radius of the circle through which other objects become visible. The ray of light must be incident at critical angle $C$.

$\sin C=\frac{1}{\mu }=\frac{1}{\sqrt{r^2+h^2}}$

${\mu }^2{r}^2=r^2+h^2\Rightarrow ({\mu }^2-1)r^2=h^2$

$r=\frac{h}{\sqrt{{\mu }^2-1}}$

$\therefore Diameter=2r=\frac{2h}{\sqrt{{\mu }^2-1}}$