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Question

Find out the no of reflection after which light ray will exit from (Given sin40°=0.64)

Solutions forall Questions of JEE Main 2019 April Physics

  1. 130000

  2. 57803

  3. 140000

  4. 150000

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Solution

The correct option is B

57803


Step 1. Given data:

sin40°=0.64

n1=refractiveindexofair=1n2=refractiveindexof2ndmedia=1.31

Step 2: Formula

n1sini=n2sinrwheren1,n2arerefractiveindexoftwomedia

Step 2. Finding out the number of reflections:

1sin40°=1.31sinr

0.64=1.31sinr

sinr=0.641.31=0.490.5

r=30°

In ABC, we have

x=20μmtanr=203μm

Therefore, n(number of reflections) is given by

n=2m203μm=1053=57803

Hence, the correct option is (B).


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