JEE Main Exam Analysis 2025 – Difficulty Level, Cutoff, and Expert Insights

Total Internal Reflection - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Total Internal Reflection is considered one of the most asked concept.

  • 25 Questions around this concept.

Solve by difficulty

A green light is incident from the water to the air-water interface at the critical angle( \mu). Select the correct statement.

Concepts Covered - 1

Total Internal Reflection

Total Internal Reflection:

When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes  90^{\circ} this angle of incidence is called critical angle (C).

When Angle of incidence exceeds the critical angle than light ray comes back into the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).

Using snell's law : 

\mu _2 \sin C = \mu _1\sin r

\implies \mu _2 \sin C = \mu _1   since,  \sin r = 1.

\implies \sin C = \frac{\mu _1}{\mu _2}=\frac{\text{R.I of rarer medium }}{\text{R.I of denser medium }}

or    \boxed{\mu=\frac{1}{\sin C}}     when   \mu_1 = 1 for air and \mu_2=\mu.

 Conditions for TIR : 

 (i) The ray must travel from denser medium to rarer medium.

 (ii) The angle of incidence 'i' must be greater than critical angle 'C' i.e  i > C.

Circle of illuminance:

     

In the figure, ray 1 strikes the surface at an angle less than critical angle C and gets refracted in rarer medium. Ray 2 strikes the
surface at critical angle and grazes the interface. Ray 3 strikes the surface making an angle greater than the critical angle and gets
internally reflected. The locus of points where ray strikes at critical angle is a circle, called circle of illuminance. All light rays striking inside the circle of illuminance get refracted in the rarer medium. If an observer is in the rarer medium, the observer will see light coming out only from within the circle of illuminance. If a circular opaque plate covers the circle of illuminance, no light will get refracted in the rarer medium and then the object cannot be seen from the rarer medium. The radius of C.O.I. can be easily found.

From the figure:

\tan(\theta _c) =\frac{R}{h}     where R is radius of C.O.I 

\implies R = h \tan(\theta _c)

Also,   \sin(\theta _c)=\frac{1}{\mu } 

Therefore using trigonometry,  \tan(\theta _c)=\frac{1}{\sqrt{\mu^2-1} }

So,  the radius of the circle of illuminance ,    \boxed{R=\frac{h}{\sqrt{\mu^2-1} }}

 

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Total Internal Reflection

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