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Physics-

General

Easy

Question

In the figure shown a point object O is placed in air. A spherical boundary of radius of curvature 1.0 m separates two media. AB is principal axis. The refractive index above AB is 1.6 and below AB is 2.0. The separation between the images formed due to refraction at spherical surface is

12 m
20 m
14 m
10 m

The correct answer is: 12 m

For refraction by upper surface
$table row cell fraction numerator 1.6 over denominator v subscript 1 end subscript end fraction minus fraction numerator 1 over denominator negative 2 end fraction equals fraction numerator 1.6 minus 1 over denominator 1 end fraction rightwards double arrow fraction numerator 1.6 over denominator v subscript 1 end subscript end fraction equals 0.6 minus 0.5 equals 0.1 end cell row cell rightwards double arrow v subscript 1 end subscript equals 16 m end cell end table$
For refraction by lower surface
$table row cell fraction numerator 2 over denominator v subscript 2 end subscript end fraction minus fraction numerator 1 over denominator negative 2 end fraction equals fraction numerator 2 minus 1 over denominator 1 end fraction rightwards double arrow fraction numerator 2 over denominator v subscript 2 end subscript end fraction equals 1 minus 0.5 equals 0.5 end cell row cell rightwards double arrow v subscript 2 end subscript equals fraction numerator 2 over denominator 0.5 end fraction equals 4 m end cell end table$
Distance between images = (16 – 4) = 12m.

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