What is the critical angle of a light beam passed from a medium (n=2) to a medium (n=1.2)?

θ_c = 36.87°

Explanation:

Index of refraction for index medium; n_i = 2

Index of refraction for Refractive medium; n_r = 1.2

Formula to find the critical angle is given;

n_i(sin θ_c) = n_r(sin 90)

Where θ_c is critical angle.

Thus;

2 × (sin θ_c) = 1.2 × 1

(sin θ_c) = 1.2/2

(sin θ_c) = 0.6

θ_c = sin^(-1) 0.6

θ_c = 36.87°

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more