Question

Prove angle of incidence is equal to angle of emergence in rectangular glass slab.

Answer 1

Assume ABCD to be the glass plate with sides AB parallel to CD.

Let the refractive index of air and glass be n1and n2.

Consider a ray of light EF incident on surface AB of the glass slab with angle of incidence, ∠i.
The light ray is refracted into the glass slab along the path FG and towards the normal, NN’ . ∠ r1 is the angle of refraction.
The light ray is refracted out of the glass slab at the surface CD along the path GH. The angle of incidence at this surface is ∠r2.

The light ray emerges out of the slab forming ∠e as angle of emergence.

The surfaces AB and CD are parallel and FG is the transversal,
Therefore,  ∠r​1= ∠r1

Applying Snell’s Law at the surface AB,
sin isin r1=n2n1 .......(1)
Similarly, at surface CD, according to Snell’s Law
sin r2sin e=n1n2 .......(2)
Multiplying equation 1 and 2,

sin isin r1×sin r2sin e=n2n1×n1n2sin isin r1×sin r2sin e=1

As AB is parallel to CD, the perpendiculars to the AB and CD, NN’ and MM’ are also parallel
Therefore, alternate interior angles,  r1 and r2are congruent i.e. r1 =  r2 
Hence,
sin isin r1×sin r1sin e=1sin isin e=1Therefore, sin i=sin eor, i=e

Hence, angle of incidence equals angle of emergence.


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Answer 2

Assume ABCD to be the glass plate with sides AB parallel to CD.

Let the refractive index of air and glass be n1and n2.

Consider a ray of light EF incident on surface AB of the glass slab with angle of incidence, ∠i.

The light ray is refracted into the glass slab along the path FG and towards the normal, NN’ . ∠ r1 is the angle of refraction.

The light ray is refracted out of the glass slab at the surface CD along the path GH. The angle of incidence at this surface is ∠r2.

The light ray emerges out of the slab forming ∠e as angle of emergence.

The surfaces AB and CD are parallel and FG is the transversal,

Therefore,  ∠r​1= ∠r1

Applying Snell’s Law at the surface AB,

sin isin r1=n2n1 .......(1)

Similarly, at surface CD, according to Snell’s Law

sin r2sin e=n1n2 .......(2)

Multiplying equation 1 and 2,

sin isin r1×sin r2sin e=n2n1×n1n2sin isin r1×sin r2sin e=1

As AB is parallel to CD, the perpendiculars to the AB and CD, NN’ and MM’ are also parallel

Therefore, alternate interior angles,  r1 and r2are congruent i.e. r1 =  r2 

Hence,

sin isin r1×sin r1sin e=1sin isin e=1Therefore, sin i=sin eor, i=e

Hence, angle of incidence equals angle of emergence.

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