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

Answer
301.8k+ views
Hint: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 $90^\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 $\mu $. We need to calculate the maximum value of $\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 $i$ and the angle of incidence is incident on $AC$.
The angle between two lines is same as the angle between their perpendiculars so we can write,
$i = A$
Since the prism is right angled, we can write,
\[ \Rightarrow \theta = \dfrac{\pi }{2} - A\]
Since $i = A$ we can write,
$ \Rightarrow \theta = \dfrac{\pi }{2} - i......\left( 1 \right)$
We know that refractive index is expressed as,
$\mu = \dfrac{1}{{\sin i}}$
As we know that the angle of incidence should be minimum for the total internal reflection,
${i_c} = {\sin ^{ - 1}}\dfrac{1}{\mu }......\left( 2 \right)$
Here, ${i_c}$ is a critical angle.
Now the maximum value of $\theta $ for total internal expression can be calculated by substituting the expression of ${i_c}$ to the equation (1) as,
$ \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 $\theta $.
$\therefore {\theta _{\max }} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\dfrac{1}{\mu }$
Therefore, the maximum value will be ${\theta _{\max }} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\dfrac{1}{\mu }$.
Note:As we know that the minimum angle for total internal reflection is the critical angle. The angle of the right-angled prism is $90^\circ $. The maximum value of $\theta $ is calculated corresponding to the minimum value of $i$.
Complete step by step solution:
In this question we have given that a right-angled prism has the refractive index $\mu $. We need to calculate the maximum value of $\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 $i$ and the angle of incidence is incident on $AC$.
The angle between two lines is same as the angle between their perpendiculars so we can write,
$i = A$
Since the prism is right angled, we can write,
\[ \Rightarrow \theta = \dfrac{\pi }{2} - A\]
Since $i = A$ we can write,
$ \Rightarrow \theta = \dfrac{\pi }{2} - i......\left( 1 \right)$
We know that refractive index is expressed as,
$\mu = \dfrac{1}{{\sin i}}$
As we know that the angle of incidence should be minimum for the total internal reflection,
${i_c} = {\sin ^{ - 1}}\dfrac{1}{\mu }......\left( 2 \right)$
Here, ${i_c}$ is a critical angle.
Now the maximum value of $\theta $ for total internal expression can be calculated by substituting the expression of ${i_c}$ to the equation (1) as,
$ \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 $\theta $.
$\therefore {\theta _{\max }} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\dfrac{1}{\mu }$
Therefore, the maximum value will be ${\theta _{\max }} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\dfrac{1}{\mu }$.
Note:As we know that the minimum angle for total internal reflection is the critical angle. The angle of the right-angled prism is $90^\circ $. The maximum value of $\theta $ is calculated corresponding to the minimum value of $i$.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

How to Convert a Galvanometer into an Ammeter or Voltmeter

Hybridisation in Chemistry – Concept, Types & Applications

What Are Current and Potential Difference in Electricity?

What Are Elastic Collisions in One Dimension?

