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 n1 at an angle θ, where sin‌θ is slightly larger than 1/n1. Take refractive index of air as 1 . Which of the following statement(s) is(are) correct?
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 angle
For ray to enter medium-1, θ2<θ2Csin‌θ2<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.