A clear transparent glass sphere ( $μ$ = 1.5) of radius R is immersed in a liquid of refractive index 1.25. A parallel beam of light incident on it will converge to a point. The distance of this point from the centre of the sphere will be

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-3R

+3R

-R

+5R

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{μ_{1}}{- u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R} \\ \frac{1 .25}{- (- \infty)} + \frac{1 .5}{v} = \frac{1 .5 - 1 .25}{R} or \frac{1 .5}{v} = \frac{0 .25}{R} \end{array}$

$v = \frac{1 .5 R}{0 .25} = 6 R$

$\begin{array}{l} \frac{μ_{2}}{- u} + \frac{μ_{1}}{v} = \frac{μ_{1} - μ_{2}}{R} \\ \frac{1 .5}{- 4 R} + \frac{12 .5}{v} = \frac{1 .25 - 1 .5}{- R} \\ \frac{1 .25}{v} = \frac{1}{4 R} + \frac{1 .5}{4 R} = \frac{2 .5}{4 R} \\ v = \frac{1 .25 \times 4 R}{2 .5} = \frac{5 R}{2 .5} \\ v = 2 R \\ D i s \tan c e f r o m c e n t r e o f s p h e r e i s 3 R \end{array}$