Question

A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index μ. The radius of the sphere is R. At t = 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for \[t < \sqrt{\frac{2h}{g}}\] . Consider only the image by a single refraction.

Answer 1

Given,
A transparent sphere with refractive index μ and a of radius R.
A ball is kept at height h above the sphere.
As per the question, at t = 0 the ball is dropped normally on the sphere. 

Let take the time taken to travel from point A to B as t .
Thus, the distance travelled by the ball is given by: = `1/2 "gt"^2` Where g is acceleration due to gravity.
Therefore, the distance BC is given by: 
\[= h - \frac{1}{2}g t^2\] We are assuming this distance of the object from lens at any time t.
So here,
u = \[- \left( h - \frac{1}{2}g t^2 \right)\] 
Taking:
Refractive index of air, μ₁ =  1
Refractive index of sphere, μ₂ =  μ (given)
Thus, \[\frac{\mu}{v} - \frac{1}{- \left( h - \frac{1}{2}g t^2 \right)} = \frac{\mu - 1}{R}\]
\[ \Rightarrow \frac{\mu}{v} = \frac{\mu - 1}{R} - \frac{1}{\left( h - \frac{1}{2}g t^2 \right)} = \frac{\left( \mu - 1 \right)\left( h - \frac{1}{2}g t^2 \right) - R}{R\left( h - \frac{1}{2}g t^2 \right)}\]
Let v be the image distance at any time t. Then,
\[v = \frac{\mu R\left( h - \frac{1}{2}g t^2 \right)}{\left( \mu - 1 \right)\left( h - \frac{1}{2}g t^2 \right) - R}\] 
Therefore, velocity of the image ( V ) is given by, 
\[V = \frac{dv}{dt} = \frac{d}{dt}\left[ \frac{\mu R\left( h - \frac{1}{2}g t^2 \right)}{\left( \mu - 1 \right)\left( h - \frac{1}{2}g t^2 \right) - R} \right]\]
\[ = \frac{\mu R^2 gt}{\left[ \left( \mu - 1 \right)\left( h - \frac{1}{2}g t^2 \right) - R \right]^2}\]