Question

A light ray moving at a 54.3 deg

angle in water (n = 1.33) hits a
boundary with oil (n = 1.52). At
what angle is it refracted in the oil?
(Water n = 1.33, Air n = 1.00)
(Unit = deg)​

Answer 1

`45.25^(\circ)`

Step-by-step explanation:

Applying Snell's law

`n_1sin\theta_1 = n2*sin\theta_2\\1.33*sin(54.3) = 1.52*sin\theta_2\\sin\theta_2 = 1.33*0.81208353/1.52 = 0.71057\\\theta_2=sin^(-1) (0.71057)\\\theta_2 = 45.25^(\circ)`