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Question

# The focal length of a thin lens made from the material of refractive index 1.5 is 20cm When it is placed in a liquid of refractive index 4/3, its focal length will be ______ cm.

A
80
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B
45
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C
60
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D
78
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Solution

## The correct option is D 801f=(n−1)(1R1−1R2)in other words 1fa=(arg−1)(1R1−1R2)...(1)ang = reflective linden of 4 lassw.r.t airfa = focal length in air& 1fl=(lng−1)(1R1−1R2)...(2)dividing (1) by (2)we getflfa=ang−1lng−1=(ang−1)(angang−1) Note itfl=(ang−1)(anganl−1)faor fl=(1.5−1)(1.54/3−1)×20=80cm1.33 = 4/3

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