A ball is kept at a height h above the surface of a heavy transpa… - Vidyadip

Question

A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index $\mu.$ 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 $\text{t}<\sqrt{\frac{2\text{h}}{\text{g}}}.$ Consider only the image by a single refraction.

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Answer

Let us assume that it has taken time ‘t’ from A to B.$\therefore \ \text{AB}=\frac{1}{2}\text{gt}^2$
$\therefore \ \text{BC}= \text{h}-\frac{1}{2}\text{gt}^2$
This is the distance of the object from the lens at any time ‘t’. Here, $\text{u}=-\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)$$\mu_2=\mu$ (given) and $\mu_1=\text{i}$ (air)
So, $\Rightarrow\frac{\mu}{\text{v}}=\frac{1}{-\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}=\frac{\mu-1}{\text{R}}$$\Rightarrow\frac{\mu}{\text{v}}=\frac{\mu-1}{\text{R}}=\frac{1}{\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}=\frac{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}}{\text{R}\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}$
So, v = image distance at any time ‘t $=\frac{\mu\text{R}\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}}$ So, velocity of the image $=\text{V}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\begin{bmatrix}\frac{\mu\text{R}\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}} \end{bmatrix}=\frac{\mu\text{R}^2\text{gt}}{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}}$ (can be found out).

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