If n₁ = 1.3 and n₂ = 2.2,what is the condition for the angle $\theta$ for no light escaping?

Question

Quizplus · Accepted Answer

The condition for no light escaping is when the angle of incidence is greater than the critical angle, which is given by:
sin⁡(critical angle) = n2/n1
where n1 is the refractive index of the medium in which the incident ray is travelling (in this case air), and n2 is the refractive index of the medium that the ray is entering (in this case the glass). 
Therefore, the critical angle can be calculated as:
sin⁡(critical angle) = 1.3/2.2
critical angle = sin^-1(1.3/2.2) = 36.21 degrees
Since the angle of incidence in this case is F1F1F1S1θF1F1F10, the condition for no light escaping is F1F1F1S1θF1F1F10 > 36.21 degrees, which is choice C.

If N1 = 1 θ\thetaθ For No Light Escaping? A) θ\thetaθ

If N₁ = 1 $\theta$ For No Light Escaping?
A) $\theta$