Question

11. A coin is embedded in a spherical plastic paperweight having a radius of 3.0 cm. The index of refraction of the plastic is ni -1.50. One coin is located 2.0 cm from the edge of the sphere. Find the position of the image of the coin. ANS: q -1.7 cm 3.0 cm ny 712 2.0 cm

Answer 1

Given n_1 = 1.50 R = -3.0 cm n_2 = 1.00 Because n_1 > n_2, where n_2 = 1.00 is the index of refraction for air, the rays originating from the coin fig. are refracted away from normal at the surface and diverge outward. Therefore By fundamental paraxial Eq^n: n_2/q = n_2 - n_1/R - n_1/P (P = 2.0 cm) 1/q = 1.00 - 1.50/-3.0cm - 1.50/2.0cm = -0.50/-3.0 cm - 1.50/2.0 {R is negative here} = -1 - (-4.5)/-6 = + 3.5/-6 1/q = -0.583 therefore q = 1/-0.583 therefore q = -1.7 cm the negative sign indicates that the image in front of the surface.