Question

A small source of light is $4\,m$ below the surface of a liquid of refractive index $\dfrac{5}{3}$. In order to cut off all the light coming out of liquid surface, minimum diameter of thee disc placed on the surface of liquid is:
(A) $3\,m$
(B) $4\,m$
(C) $6\,m$
(D) $\infty $

Answer 1

Hint The minimum diameter of the disc placed on the surface of the liquid is determined by using the two formulas. The first formula used is the critical angle formula and the other formula is radius of the disc formula, then the diameter is determined.

Useful formula:
The critical angle is given by,
$\sin {\theta _c} = \dfrac{1}{\mu }$
Where, ${\theta _c}$ is the critical angle and $\mu $ is the refractive index of the medium.
The radius of the disc is given by,
$r = h \times \tan {\theta _c}$
Where, $r$ is the radius of the disc, $h$ is the height of the source from the surface of the liquid and ${\theta _c}$ is the critical angle.

Complete step by step answer
Given that,
The height of the source from the surface of the liquid, $h = 4\,m$,
The refractive index of the liquid is, $\mu = \dfrac{5}{3}$,
Now,
The critical angle is given by,
$\sin {\theta _c} = \dfrac{1}{\mu }\,...................\left( 1 \right)$
By substituting the refractive index of the liquid in the above equation (1), then the above equation (1) is written as,
$\sin {\theta _c} = \dfrac{1}{{\left( {\dfrac{5}{3}} \right)}}$
By rearranging the terms in the above equation, then the above equation is written as,
$\sin {\theta _c} = \dfrac{3}{5}$
By dividing the terms in the above equation, then the above equation is written as,
$\sin {\theta _c} = 0.6$
By rearranging the terms in the above equation, then the above equation is written as,
${\theta _c} = {\sin ^{ - 1}}0.6$
From the trigonometry, the values of the ${\sin ^{ - 1}}0.6 = 36.86$, then the above equation is written as,
${\theta _c} = 36.86$
Now,
The radius of the disc is given by,
$r = h \times \tan {\theta _c}\,...................\left( 2 \right)$
By substituting the height and the critical angle in the above equation (2), then the equation (2) is written as,
$r = 4 \times \tan 36.86$
From the trigonometry, the values of the $\tan 36.86 = 0.75$, then the above equation is written as,
$r = 4 \times 0.75$
By multiplying the terms in the above equation, then the above equation is written as,
$r = 3\,m$
The relation between the radius and the diameter is,
$d = 2r$
Where, $d$ is the diameter and $r$ is the radius.
By substituting the radius in the above equation, then the above equation is written as,
$d = 2 \times 3$
By multiplying the terms in the above equation, then the above equation is written as,
$d = 6\,m$

Hence, the option (C) is the correct answer.

Note The critical angle is inversely proportional to the refractive index of the medium. As the refractive index of the medium increases, then the critical angle decreases. The radius of the disc is directly proportional to the height and the critical angle.