A circular beam of light having diameter 'd' falls on a plane liquid surface from air. The angle of incidence is 45^ and refractive index of liquid is 'n'. The diameter of the refracted beam is ( 45^0 = 1 / 2 , 45^0 = 1 / 2 )

Step-by-step solution

Let the diameter of the incident beam be d and the diameter of the refracted beam be d' . Let the length of the illuminated portion on the liquid surface be l . From the geometry of the incident beam, d = l i . From the geometry of the refracted beam, d' = l r . Therefore, d' = d r i . Given i = 45^ , i = 1 2 and i = 1 2 . Using Snell's law, 1 i = n r . r = 45^ n = 1 n 2 . Using the identity r = 1 - ^2 r , we get: r = 1 - 1 2n^2 = 2n^2 - 1 n 2 . Substituting the values of r and i in the expression for d' : d' = d 2