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Refraction Of Light Through Glass Slab - Practice Questions & MCQ

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

Quick Facts

  • Refraction Through A Glass Slab is considered one the most difficult concept.

  • 20 Questions around this concept.

Solve by difficulty

A ray of light AO in a vacuum is incident on a glass slab at angle 60 and refracted at angle 30 along OB as shown in the figure. The optical path length of light ray from A to B is: 

A ray of light enters a rectangular slab of refractive index 3 at an angle of incidence 60. It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between incident and emergent rays is:

Concepts Covered - 2

Refraction Through A Glass Slab

Consider an object O placed at distance d in front of a glass slab of thickness "t" and refractive index μ. The observer is on
the other side of the slab. A ray of light from the object first refracts at the surface 1 and then refracts at the surface 2 before reaching the observer as shown in the above figure.

So for the refraction at the surface (1)
Apparent depth, d1=dreal nrelative =d(nincident nreftaction )=d1μ=dμ

Similarly for the refraction at the surface (2)
Apparent depth, d2=dreal nrelative =d1+t(nincident nreftaction )=d1+tμ1=d1μ+tμ

As you observe, The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges out parallel to its incident direction.
i.e. the ray undergoes no deviation (δ=0 ).
the object appears to be shifted towards the slab by the distance known as apparent shift or Normal shift.
And the apparent shift= OAI2 A
I.e Apparant shift =t{11μ}

If the slab is placed in the medium of the refractive index μsur 
then Apparant shift =t{1μourμ}

Lateral Displacement Of Emergent Ray Through A Glass Slab

In the above figure Incident, ray AO is an incident on the EF surface of the slab at an angle of incident i, and PB is the emergent ray emerging out of the HG surface of the slab.

for the surface EF

Applying Snell's law at the surface EF and HG
μasini=μsinr and μsinr=μasine
Usingr=r and μa=1, we get sini=sine or e=i
i.e the emergent ray is parallel to the incident ray.

If PQ is the perpendicular dropped from P on the incident ray produced.
Then PQ=d is known as lateral displacement which is given as

d=PQ=OPsin(ir)=OMcosrsin(ir)=tsin(ir)cosr


If i is very small, r is also very small, then d=(11μ)ti

 

 

Study it with Videos

Refraction Through A Glass Slab
Lateral Displacement Of Emergent Ray Through A Glass Slab

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