A point object is moving with a constant speed of 2 mm/s towards the pole of a convex spherical surface separating air (refractive index = 1 ) and glass (refractive index = 1.5 ). The radius of curvature of the surface is 20 cm . At the instant when the object is at a distance of 60 cm from the pole, the speed of the image is

Step-by-step solution

For refraction at a single spherical surface, the position of the image is given by: ₂ v - ₁ u = ₂ - ₁ R Given: ₁ = 1 (air), ₂ = 1.5 (glass) R = +20 cm (convex surface) u = -60 cm Substituting these values: 1.5 v - 1 -60 = 1.5 - 1 20 1.5 v + 1 60 = 0.5 20 = 1 40 1.5 v = 1 40 - 1 60 = 3 - 2 120 = 1 120 v = 1.5 120 = 180 cm To find the velocity of the image, we differentiate the refraction formula with respect to time t : - ₂ v^2 dv dt + ₁ u^2 du dt = 0 Let v_i = dv dt be the image velocity and v_o = du dt be the obj