Given
sin‌θ>‌‌‌ i.e.
‌‌sin‌θi>‌
From Snell's law,
n1‌sin‌θ1=n2‌sin‌θ2 ⇒‌‌sin‌θ2=‌ If
n1=n2 then
θ2=θ1 Again form Snell's law,
n2‌sin‌θ2=(1)‌sin‌θ3 ⇒sin‌θ3=n2‌sin‌θ2‌‌ or,
sin‌θ3=n1‌sin‌θ1 ∴‌‌sin‌θ1=‌>‌‌‌ So,
sin‌θ3>1 ∴‌‌θ3>90∘ Hence ray cannot enter air if
n2=n1 For
n2<n1, from Snell 's law,
sin‌θ1=‌‌sin‌θ2>‌ ∴‌‌sin‌θ2>‌ Now for surface 2 - air interface
n2‌sin‌θ2=(1)‌sin‌θ3 ∴‌‌sin‌θ2=‌>‌‌‌ So,
θ2>90∘ Therefore ray is reflected back in medium-2.
Again for surface
1−2 interface
n2‌sin‌θ2=n1‌sin‌θ1 sin‌θ2C=‌
θ2C is critical angle
For ray to enter medium
−1 of refractive index
n1 θ2<θ2C‌‌∴sin‌θ2<sin‌2‌θC ‌‌sin‌θ1<‌‌‌⇒sin‌θ1<1 Here,
θ1<90∘ So ray is reflected back in medium of refractive index
n1 For
n2>n1 ‌‌sin‌θ2>‌ For surface 2 - air interface
n2‌sin‌θ2=sin‌θ3 sin‌θ2=‌>‌=0 so
θ2>90∘ Hence ray is reflected back in medium-2
n2‌sin‌θ2=n1‌sin‌θ1 or,
‌‌sin‌θ1=‌‌sin‌θ2 sin‌θ2C−‌;θ2C is critical angle
For ray to enter medium-1,
θ2<θ2C sin‌θ2<sin‌θ2C ‌‌sin‌θ1<‌‌‌∴‌‌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=‌>‌ or
sin‌θ2>1‌‌⇒θ2>90∘ Therefore the ray of light is reflected back into the medium of refractive index
n1.
