A hollow sphere of glass as shown in fig. of refractive index η has a small mark on its interior surface which is observed from a point outside the sphere on the side opposite the centre. The inner cavity is concentric with the external surface and thickness of glass is every where equal to the radius of the inner surface. Prove that the mark will appear nearer than it really is by a distance 
R, where R is the radius of the inner surface.
Text Solution
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Sol. The object (mark) is at O. First of all refraction will take place from the inner surface and the image is situated at O 1 . Now this image acts as the object for the second outer surface. The refraction takes place from this surface and image is formed at O 2 .
Consider first refraction at air-glass interface
Using 
– 
= 
,
we get 
– 
= 
⇒ v 1 = – 
Now, consider second refraction at glass-air interface
From equation, 
– 
= 
, wet get
– 
= 
or v 2 = – 
=
∴ Shift towards observer = 3R – 
= 
R
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