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Question

A wide slab consisting of two media of refractive indices n1 and n2 is placed in air as shown in the figure. A ray of light is incident from medium n1 to n2 at an angle θ, where sinθ is slightly larger than 1n1. Take refractive index of air as 1, which of the following statement(s) is (are) correct?

A
The light ray enters air if n2=n1
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B
The light ray is finally reflected back into the medium of refractive index n1 if n2<n1
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C
The light ray is finally reflected back into the medium of refractive index n1 if n2>n1
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D
The light ray is reflected back into the medium of refractive index n1 if n2=1
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Solution

The correct option is D The light ray is reflected back into the medium of refractive index n1 if n2=1
We have

sinθ>1n1 i.e. sinθ1>1n1 n1sinθ1=n2sinθ2
sinθ2=n1sinθ1n2
If n1=n2 then, θ2=θ1
n2sinθ2=(1)sinθ3 sinθ3=n2sinθ2=n1sinθ1
sinθ1=sinθ3n1>1n1 sinθ3>1
θ3>90
This means ray can't enter air.
For n1>n2;sinθ1=n2n1>1n1
sinθ2>1n2
For surface 2air interface
n2sinθ2=sinθ3 sinθ2=sinθ3n2>1n2
θ2>90, it means ray is reflected back in medium of refractive index n2

For surface 12 inteface
n2sinθ2=n1sinθ1
sinθ2c=n1n2, where θ2c is critical angle.
For ray to enter medium of refractive index n1
θ2<θ2c
sinθ2<sinθ2c
n1n2sinθ1<n1n2
sinθ1<1 θ1<90, which is true
Hence ray enters medium of refractive index n1.
For n2>n1
n2n1sinθ2>n2n1
also, sinθ2>1n2
For surface 2air interface
n2sinθ2=sinθ3
sinθ2=sinθ3n2>1n2
θ2>90, it means ray is reflected back in medium of refractive index n2.

n2sinθ2=n1sinθ1
sinθ1=n2n1sinθ2
sinθ2c=n1n2, where, θ2c is critical angle.
For ray to enter medium of refractive index n1
θ2<θ2c
sinθ2<sinθ2c
n1n2sinθ1<n1n2
sinθ1<1 θ1<90, which is true
Hence, ray enters medium of refractive index n1
Let n2=1

n1sinθ1=n2sinθ2
n2=1 n1sinθ1=sinθ2
sinθ1=sinθ2n1>1n1
sinθ2>1 θ2>90
Ray is reflected back in the medium of reflective index n1

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