Right up his head! (Optics)

Classical Mechanics Level 3

On a sunny day, consider a scuba diver immersed in a wide swimming pool filled with water and whose walls and floor were painted in black. Looking upward, the diver sees the surface of water almost completely dark, except for an approximately circular region with radius $R$ above his/her head. If $n$ is the refractive index of water (relative to that of air) and $h$ is the depth of the eyes of the diver relative to the surface of water, then find the radius $R$ of the clear circular region in terms of $h$ and $n$ .

R = h(n^2 + 1)

R = h \sqrt{n^2 - 1}

R = \frac h{n^2 + 1 }

R = \frac h{\sqrt{n^2 -1 }}

1 solution

Ayon Ghosh
Mar 2, 2017

Its quite easy really.Just apply principle of $TIR$ .
This my solution.Yours may be different.

$sin$ $i_c$ = $1/n$ ...(1)

But $sin$ $i_c$ = $R/ [root (R^2+h^2)]$ ...(2)

Equate (1) and (2),rearrange the terms,factorise and you get answer as $(3)$ .

Ya it was easy.

Md Zuhair - 4 years, 3 months ago

Right up his head! (Optics)

1 solution

0 pending reports