Q41. A spherical refracting interface of radius R separates media of refractive indices n₁ = 1 and n₂ = 1.4, as shown in the figure. A point object is placed 4R in front of the interface in the n₁ medium. What is the magnitude of the lateral magnification of the image formed by the spherical interface?

Question

G1 Physics · Accepted Answer

Using the Cartesian sign convention for the figure, u = -4R, R = +R, n₁ = 1, n₂ = 1.4. For refraction at a spherical surface, n₂ ÷ v - n₁ ÷ u = (n₂-n₁) ÷ R. Thus, 1.4 ÷ v - 1 ÷ (-4R) = 0.4 ÷ R. So, 1.4 ÷ v + 1 ÷ (4R) = 0.4 ÷ R, 1.4 ÷ v = 0.15 ÷ R. Hence, v = 1.4R ÷ 0.15 = 28R ÷ 3. For a spherical refracting surface, m = n₁v ÷ (n₂u). Therefore, m = [1×(28R ÷ 3)] ÷ [1.4(-4R)] ≈ -1.67. The required magnitude is approximately 1.66. Correct option: A.