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Question

A solid glass sphere with radius 1 m and an index of refraction 1.5 is silvered over one hemisphere. A small object is located on the axis of the sphere at a distance 2 m to the left of the vertex of the unsilvered hemisphere. The position of final image after all refractions and reflections have taken place is

A
at vertex of silvered face
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B
at vertex of unsilvered face
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C
at centre of curvature
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D
at object itself
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Solution

The correct option is A at vertex of silvered face
Given,
refractive index of air, μ1=1
refractive index of glass sphere , μ2=1.5

For the given problem, the ray of light will suffer two refractions and one reflection.

First refraction takes place when light from object enters the sphere and then get reflected from silvered face and second refraction takes place while it returns after reflection.

For first refraction and then reflection, the ray of light travels from left to right while for the last (second) refraction it travels from right to left.

Hence, the sign convention will change accordingly.


Applying the lens makers formula,

μ2vμ1u=μ2μ1R

substituting the values for the first refraction:

1.5v11(2)=1.51(+1)

1.5v1=0

v1=

This image will act as object for silvered hemisphere.

Applying mirror formula,

1v+1u=1f=2R

Substituting the values for reflection,

1v2+1=21

v2=12 m

This image will act as object for refracting unsilvered hemisphere,

Applying lens makers formula for second refraction,

μ2vμ1u=μ2μ1R

Substituting the values,

1v31.5(1.5)=11.5(1)

(Reversed sign convention)

1v3=12

v3=2 m

Hence the final image formed at the vertex of the silvered face.

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