An object of height 2 cm is placed in air 40 cm to the left of the pole P of a spherical refracting surface. The centre of curvature lies 20 cm to the left of P. The refractive index is 1 on the object side and 1.54 on the other side. What is the approximate height of the image?

Question

G1 Physics · Accepted Answer

Use the refraction formula at a spherical surface: n₂ ÷ v - n₁ ÷ u = (n₂-n₁) ÷ R. Using the Cartesian sign convention from the source figure: n₁ = 1, n₂ = 1.54, u = -40 cm, R = -20 cm. Therefore, 1.54 ÷ v - 1 ÷ (-40) = (1.54-1) ÷ (-20). Thus, 1.54 ÷ v + 1 ÷ 40 = -0.54 ÷ 20. Therefore, 1.54 ÷ v = -0.027-0.025 = -0.052. Hence, v = 1.54 ÷ (-0.052) ≈ -29.6 cm. For a spherical refracting surface, lateral magnification is m = n₁v ÷ (n₂u). Therefore, m = [1×(-29.6)] ÷ [1.54×(-40)] ≈ 0.48. Image height…