Question

A plane mirror is placed horizontally inside water\[\left( {\mu = \dfrac{4}{3}} \right)\]. A ray falls normally on it. Then, the mirror is rotated at an angle \[\theta \]. What is the minimum value of \[\theta \] for which the ray does not come out of the water surface?

A. \[\dfrac{\pi }{4}\]
B. \[{\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)\]
C. \[\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)\]
D. \[2{\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)\]

Answer 1

Hint:Before solving the problem, let us understand the total internal reflection and critical angle. As there are two different mediums of different materials, a light wave is entering from one medium to another. If the light wave at the intersection of these mediums, instead of being refracted, gets reflected back into the first medium we say it as total internal reflection. When the angle of incidence of this light ray leads to a refraction of light whose angle is, we call it \[{90^0}\] as critical angle.

Formula Used:
The condition for the total internal reflection is given by t,
\[\sin \left( i \right) > \sin {\theta _C}\]…………… (1)
Where, \[i\] is the angle of incidence, \[{\theta _C}\] is the critical angle and \[R\] is the radius of the flat track.

Complete step by step solution:
If the mirror is rotated at an angle of \[\theta \], then the reflected ray rotates at an angle of \[2\theta \]
So, from equation (1) we have,
\[\sin \left( i \right) > \sin {\theta _C}\]
\[\Rightarrow \sin \left( r \right) > \sin {\theta _C}\]
\[\Rightarrow \sin \left( {2\theta } \right) > \sin {\theta _C}\]
But we know the formula to find the critical angle is,
\[\sin {\theta _C} = \left( {\dfrac{1}{\mu }} \right)\]

Substituting the value of the critical angle in the above equation, we get,
\[\sin \left( {2\theta } \right) > \left( {\dfrac{1}{\mu }} \right)\]
\[\Rightarrow 2\theta > {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)\]
\[\Rightarrow \theta > \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)\]
By data we have \[\left( {\mu = \dfrac{4}{3}} \right)\].
 Substituting this value in the above equation we get,
\[\theta > \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{1}{{\left( {\dfrac{4}{3}} \right)}}} \right)\]
\[ \therefore \theta > \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)\]
Therefore, the minimum value of \[\theta \]for which the ray does not come out of the water surface is \[\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)\].

Hence, Option C is the correct answer

Note: Always remember that the conditions for the total internal reflection are;
1. Light must travel from a denser medium into a less dense medium (i.e., from glass to air).
2. The angle of incidence should be greater than the critical angle.