# A hollow sphere of glass (shown in figure) of refractive index n has a small mark on its interior surface which is observed from a point outside the sphere on the side opposite to the center…

Problem: A hollow sphere of glass (shown in figure) of refractive index n has a small mark on its interior surface which is observed from a point outside the sphere on the side opposite to the center. Prove that the mark will appear nearer than it really is by a distance R, where R is the radius of the inner surface.

Solution : Take P1 as the pole for the refraction of the ray coming from O, at the point 1. Since ray is coming from air to glass

n1 = nair = 1 & n2 = ng = n

$\displaystyle \frac{n}{v} – \frac{1}{-2R} = \frac{n-1}{-R}$

$\displaystyle \frac{n}{v} = -\frac{1}{2R} – \frac{n-1}{R}$

$\displaystyle \frac{n}{v} = -\frac{(2n-1)}{2R}$

$\displaystyle v = – \frac{2 R n}{2n-1}$

Now the image at I1 will behave as an object for the refraction at the outer surface for which the pole will be P2.
The object distance will be

$\displaystyle – (\frac{2 R n}{2n-1} + R ) = -\frac{4n-1}{2n-1} R$

Repeating the procedure for the refraction at 2 since light comes from glass to air, here put n1 = n and n2=1 ( n1= Refractive index of the incident medium ( or first medium ) & n2 = R.I. of the refracted medium (second medium))

$\displaystyle \frac{1}{v’} – \frac{n}{-\frac{4n-1}{2n-1} R}= \frac{1-n}{-2R}$

$\displaystyle \frac{1}{v’} = \frac{1}{R}[\frac{-3n+1}{2(4n-1)}]$

$\displaystyle v’ = -\frac{2(4n-1)}{3n-1}R$ ; The virtual image informed at I2

The required distance = OI1 = OI2 – I2 P2 = 3R- |v’|

Putting the numerical value of v’

we obtain $\displaystyle OI_2 = 3R – \frac{2(4n-1)}{3n-1}R$

$\displaystyle OI_2 = (\frac{n-1}{3n-1}) R$

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