Question

Solve this:
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 will be
(1) - 3R (2) + 3R (3) - R (4) + R

Answer 1

Answer: (2). The distance of this point from the centre will be $OI = 3R$

Explanation:

Given that,

Refractive index of glass $\mu_{g} = 1.5$

Refractive index of liquid  $\mu_{l} = 1.25$

Now, for refraction at first surface

$\dfrac{1.5}{v} - \dfrac{1.25}{\infty} = \dfrac{1.5+1.25}{R}$

$v= 6R$

The distance of AI' is 6R.

Now, the distance of BI' is

$BI' = AI'- AB$

$BI' = 6R-2R = 4R$

For refraction at second surface

$\dfrac{1.25}{v} -\dfrac{1.5}{4R}= \dfrac{1.25-1.5}{-R}$

$\dfrac{1.25}{v}= \dfrac{0.25}{R} + \dfrac{1.5}{4R}$

$v = 2R$

Now, the distance of BI is 2R

So, the distance of I from center

$OI = OB+BI$

$OI = R+2R$

$OI = 3R$

Hence, this is the required solution.