Question

A fish is a little away below the surface of a lake. If the critical angle is ${49^ \circ }$ , then the fish could see things above the water surface within an angular range of $\theta$ where
(A) $\theta = {49^ \circ }$
(B) $\theta = {90^ \circ }$
(C) $\theta = {98^ \circ }$
(D) $\theta = {24.5^ \circ }$

Answer 1

Hint : Critical angle is the angle that the light ray will make with the water surface. Angular range is twice the critical angle. So from the value of the critical angle given in the question we can calculate the value of the angular angle.

Complete step by step answer
When the angle of incidence in water reaches a certain critical value, the refracted ray lies along the boundary, having an angle of refraction of 90-degrees. This angle of incidence is known as the critical angle; it is the largest angle of incidence for which refraction can still occur. For any angle of incidence greater than the critical angle, light will undergo total internal reflection.
The actual value of the critical angle is dependent upon the combination of materials present on each side of the boundary. The critical angle can be calculated by taking the inverse-sine of the ratio of the indices of refraction.
Given in the question, a fish is a little away below the surface of a lake and the critical angle is ${49^ \circ }$ .
Thus, the critical angle can also be defined as the angle that the light ray on one side will make with the water surface.
In that case, we can say that the angular range the fish could see things above the water surface is twice the critical angle of the lake.
Angular range $\theta = 2 \times {\text{critical angle}}$
$\Rightarrow \theta = 2 \times {49^ \circ } = {98^ \circ }.$
Thus the angular range is ${98^ \circ }$ .
The correct answer is Option C.

Note
The critical angle can be calculated by taking the inverse-sine of the ratio of the indices of refraction. The ratio of $\dfrac{{{n_r}}}{{{n_i}}}$ is a value less than 1. In fact, for the equation to even give a correct answer, the ratio of $\dfrac{{{n_r}}}{{{n_i}}}$ must be less than 1. Since total internal refraction only occurs if the refractive medium is less dense than the incident medium, the value of ${n_i}$ must be greater than the value of ${n_r}$ . If at any time the values for the numerator and denominator become accidentally switched, the critical angle value cannot be calculated. Mathematically, this would involve finding the inverse-sine of a number greater than 1 - which is not possible. Physically, this would involve finding the critical angle for a situation in which the light is travelling from the less dense medium into the denser medium - which again, is not possible.