A ray of sunlight enters a spherical water droplet ( n = 4 3 ) at an angle of incidence 53 ° measured with respect to the normal to the surface. It is reflected from the back surface of the droplet and re-enters into the air. The angle between the incoming and outgoing ray is [Take sin 53 ° = 0 . 8 ]

Step-by-step solution

Applying law of refraction ⇒ 1 × sin  5 3 ° = 4 3 × sin  r 4 5 = 4 3 sin  r ⇒ sin  r = 3 5 ⇒ sin  r = 0 . 6 ⇒ r = 3 7 ° δ 1 = 5 3 ° - 3 7 ° ⇒ δ 1 = 1 6 ° C.W. δ 2 = 1 8 0 - 2 × 3 7 ° ⇒ δ 2 = 1 0 6 ° C.W. δ 3 = 5 3° - 3 7 ° ⇒ δ 3 = 1 6 ° C.W. δ net = δ 1 + δ 2 + δ 3 = 1 3 8 °