JEE Advanced 2021 Paper 1

Given sin‌θ>‌

1

n1

‌‌ i.e. ‌‌sin‌θi>‌

1

n1

From Snell's law,
n1‌sin‌θ1=n2‌sin‌θ2
⇒‌‌sin‌θ2=‌

n1‌sin‌θ1

n2

If n1=n2 then θ2=θ1
Again form Snell's law,
n2‌sin‌θ2=(1)‌sin‌θ3
⇒sin‌θ3=n2‌sin‌θ2‌‌ or, sin‌θ3=n1‌sin‌θ1
∴‌‌sin‌θ1=‌

sin‌θ3

n1

>‌

1

n1

‌‌ So, sin‌θ3>1
∴‌‌θ3>90∘
Hence ray cannot enter air if n2=n1
For n2<n1, from Snell 's law,
sin‌θ1=‌

n2

n1

‌sin‌θ2>‌

1

n1

∴‌‌sin‌θ2>‌

1

n2

Now for surface 2 - air interface
n2‌sin‌θ2=(1)‌sin‌θ3
∴‌‌sin‌θ2=‌

sin‌θ3

n2

>‌

1

n2

‌‌ So, θ2>90∘
Therefore ray is reflected back in medium-2.
Again for surface 1−2 interface
n2‌sin‌θ2=n1‌sin‌θ1
sin‌θ2C=‌

n1

n2

θ2C is critical angle
For ray to enter medium −1 of refractive index n1
θ2<θ2C‌‌∴sin‌θ2<sin‌2‌θC
‌

n1

n2

‌sin‌θ1<‌

n1

n2

‌‌⇒sin‌θ1<1
Here, θ1<90∘
So ray is reflected back in medium of refractive index n1
For n2>n1
‌

n2

n1

‌sin‌θ2>‌

n2

n1

For surface 2 - air interface
n2‌sin‌θ2=sin‌θ3
sin‌θ2=‌

sin‌θ3

n2

>‌

1

n2

=0 so θ2>90∘
Hence ray is reflected back in medium-2
n2‌sin‌θ2=n1‌sin‌θ1
or, ‌‌sin‌θ1=‌

n2

n1

‌sin‌θ2
sin‌θ2C−‌

n1

n2

;θ2C is critical angleFor ray to enter medium-1, θ2<θ2C sin‌θ2<sin‌θ2C
‌

n1

n2

‌sin‌θ1<‌

n1

n2

‌‌∴‌‌sin‌θ1<1
So, θ1<90∘
Hence ray is finally reflected back into the medium of refractive index n1.
If n2=1 then from Snell's law,
n1‌sin‌θ1=n2‌sin‌θ2
or n1‌sin‌θ1=sin‌θ2
∴‌‌sin‌θ1=‌

sin‌θ2

n1

>‌

1

n1

or sin‌θ2>1‌‌⇒θ2>90∘
Therefore the ray of light is reflected back into the medium of refractive index n1.