Answer:
40.62°
Explanation:
Case I
Light is travelling from slab (incident medium) into the water (refractive medium) , the critical angle = 60°.
The formula for the critical angle is:
[tex]{sin\theta_{critical}}=\frac {n_r}{n_i}[/tex]
Where,
[tex]{\theta_{critical}}[/tex] is the critical angle
[tex]n_r[/tex] is the refractive index of the refractive medium.
[tex]n_i[/tex] is the refractive index of the incident medium.
So,
Given that critical angle = 60°
[tex]n_r[/tex] = 1.33
Applying in the formula as:
[tex]{sin60^0}=\frac {1.33}{n_i}[/tex]
Refractive index of the slab = 1.5357
Case II
To find the critical angle when the air is the refractive medium (n=1).
So,
Applying in the formula as:
[tex]{sin\theta_{critical}}=\frac {1}{1.5357}[/tex]
The critical angle is = sin⁻¹ 0.6511 = 40.62°
