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. $\frac{2h}{\sqrt{{\mu }^2-1}}$
• 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.
$sinC=\frac{1}{\mu }=\frac{r}{\sqrt{r^2+h^2}}$
${\mu }^2{r}^2=r^2+h^2\Rightarrow \left({\mu }^2-1\right)r^2=h^2$
$\Rightarrow r=\frac{h}{\sqrt{{\mu }^2-1}}$
$\therefore Diameter=2r=\frac{2h}{\sqrt{{\mu }^2-1}}$