Solveeit Logo

Question

Question: A right-angled prism made from a material of refractive index \(\mu \) is kept in air. A ray PQ is i...

A right-angled prism made from a material of refractive index μ\mu is kept in air. A ray PQ is incident normally on the side AB of the prism. Find (in terms of μ\mu ) the maximum value of θ\theta up to which this incident ray necessarily undergoes total internal reflection at the face AC of the prism.

Explanation

Solution

In this question, we know that if the angle between two lines is the same as the angle between their perpendiculars and one angle of right angles prism is always 9090^\circ . The angle of incidence should be minimum for total internal reflection.

Complete step by step solution:
In this question we have given that a right-angled prism has the refractive index μ\mu . We need to calculate the maximum value of θ\theta up to which this incident ray necessarily undergoes total internal reflection at the face AC of the prism.

In this question let us assume that the angle of incidence is ii and the angle of incidence is incident on ACAC.
The angle between two lines is same as the angle between their perpendiculars so we can write,
i=Ai = A
Since the prism is right angled, we can write,
θ=π2A\Rightarrow \theta = \dfrac{\pi }{2} - A
Since i=Ai = A we can write,
θ=π2i......(1)\Rightarrow \theta = \dfrac{\pi }{2} - i......\left( 1 \right)
We know that refractive index is expressed as,
μ=1sini\mu = \dfrac{1}{{\sin i}}
As we know that the angle of incidence should be minimum for the total internal reflection,
ic=sin11μ......(2){i_c} = {\sin ^{ - 1}}\dfrac{1}{\mu }......\left( 2 \right)
Here, ic{i_c} is a critical angle.

Now the maximum value of θ\theta for total internal expression can be calculated by substituting the expression of ic{i_c} to the equation (1) as,
θ=π2iC\Rightarrow \theta = \dfrac{\pi }{2} - {i_C}

Now we substitute the value of critical angle as expressed in equation (2) in above equation to get the maximum value of θ\theta .
θmax=π2sin11μ\therefore {\theta _{\max }} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\dfrac{1}{\mu }

Therefore, the maximum value will be θmax=π2sin11μ{\theta _{\max }} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\dfrac{1}{\mu }.

Note: As we know that the minimum angle for total internal reflection is the critical angle. The angle of the right-angled prism is 9090^\circ . The maximum value of θ\theta is calculated corresponding to the minimum value of ii.