Question

A spherical refractive surface of radius $10$ cm separates two media of refractive indices $\mu =1$ and $\mu =3/2$ respectively. A point object P starts from rest at $t=0$ with an acceleration $2cm{s}^{-2}$. Find the speed of image at $t=1$ s.

• A. $2cm{s}^{-}1$
• B. $6cm{s}^{-}1$
• C. $3.7cm{s}^{-}1$
• D. none of these

Answer 1

The correct option is C $3.7cm{s}^{-}1$

Velocity of P at $t=1s$

$\frac{du}{dt}=-(0+2\times 1)=-2cm/s^2$

We have, $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

$\frac{n_2}{v}=\frac{n_2-n_1}{R}+\frac{n_1}{u}$

$v=\frac{n_{2}Ru}{(n_2-n_1)u+n_{1}R}$

$\frac{dv}{dt}==\frac{n_{2}R\frac{du}{dt}((n_2-n_1)u+n_{1}R)-(n_2-n_1)\frac{du}{dt}{n}_{2}Ru}{{[(n_2-n_1)u+n_{1}R]}^2}$

$=\frac{\left(n_2{n}_1\right)R^2\frac{du}{dt}}{{[(n_2-n_1)u+n_{1}R]}^2}=\frac{1\times \frac{3}{2}\times {10}^2\times (-2)}{{[0.5\times (-2)+10]}^2}=-3.703cm/s$