A ball is held at a height h above the surface of a heavy transparent sphere having a refractive index . The radius of the sphere is R . At t = 0 , the ball is released to fall normally onto the sphere. Determine the speed of the image formed as a function of time for t < 2h g . Consider only the image produced by a single refraction.

Step-by-step solution

The distance of the ball from the pole of the spherical surface at time t is given by y = h - 1 2 gt^2 . Using the sign convention, the object distance is u = -y = - (h - 1 2 gt^2 ) . The velocity of the object is v_o = du dt = gt . The formula for refraction at a single spherical surface is: ₂ v - ₁ u = ₂ - ₁ R Substituting ₁ = 1 and ₂ = , we get: v - 1 u = - 1 R Differentiating this equation with respect to time t yields: - v^2 dv dt + 1 u^2 du dt = 0 dv dt = v^2 u^2 du dt From the refraction formula, we can find