Will NEET 2025 be Postponed? Know the Complete Details Here

Total Internal Reflection MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

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

  • 23 Questions around this concept.

Solve by difficulty

Which of the following is not due to total internal reflection?

A ray of light travelling in a transpatent medium of refractive index \mu, falls on a surface separating the medium from air at an angle of incidence of 45^{\circ}. For which of the following value of \mu the ray can undergo total internal reflection?

Light travels a distance  x  in time t_1 in air and  10x in time  t_2 in another denser medium. What is the critical angle for this medium?

NEET 2024: Cutoff (OBC, SC, ST & General Category)

NEET 2024 Admission Guidance: Personalised | Study Abroad

NEET 2025: SyllabusMost Scoring concepts NEET PYQ's (2015-24)

NEET PYQ's & Solutions: Physics | ChemistryBiology

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} }}

 

Study it with Videos

Total Internal Reflection

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top