A light beam of diameter 3 R is incident symmetrically on a glass hemisphere of radius R and of refractive index n = 3 . The radius of the beam at the base of the hemisphere is

Step-by-step solution

From the given ray diagram shown in the figure below μ 1 s i n i = μ 2 s i n r ⇒ 1 × s i n 60 ° = 3 × s i n r ⇒ r = 30 ° In ΔMNP t a n 30 ° = x R 2 ⇒ x = R 2 3 From the geometry of the figure ⇒ r = 3 2 R - x = 3 2 R - R 2 3 = 3 - 1 2 3 R = R 3