Question

What is the critical angle for total internal reflection for a material with an index of refraction of 1.33. If the object distance is 30.0 cm and the focal point is -10.0 cm from the center of the lens, what is the magnification of the lens?

Answer 1

Given n = 1.33, it means it is water. Now, let's assume it is in contact with air and makes water - Air boundary. Therefore Critical angle for Total internal reflection is Sin theta_c = 1.00/1.33 theta_c = sin^-1(1/1.33) theta = sin^-1(0.752) theta_c = 48.7 degree d_0 = 30 cm f = -10 cm As f i8s negative, it is a diverging lens equation, 1/f = 1/d_0 + 1/d_i 1/d_i = 1/f - 1/d_0 = 1/-10 - 1/30 1/d_i = -3 - 1/30 implies d_i = -3/4 = -7.5 cm M = -d_i/d_0 = -(-7.5)/30 = 7.5/30 M = 0.25