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CBSE Class 12-science Answered

Question 3.b
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Asked by ritikabehera01 | 03 Mar, 2019, 12:42: PM
answered-by-expert Expert Answer
Fig.(1) shows the ray incident on face AB of right angled prism will incident on diagonal face with critical angle.
 
Let i, r and C are incident, refraction and crictical angle respectively.
 
from geometry, (90-r)+(90-c)+45 = 180   or r = 45-C  ..................(1)
 
For critical angle , sinC is reciprocal of refractive index,  hence  sinC = n1 /n ......(2)
 
At face AB,  we have ,  sin i = n×sin r  ...................(3)
 
using (1) and (2), eqn.(3) can be written as  sin i = n×sin(45-C) = (n/√2)[ cosC - sinC ] = begin mathsize 12px style fraction numerator n over denominator square root of 2 end fraction open parentheses space square root of 1 minus open parentheses n subscript 1 over n close parentheses squared end root space minus space n subscript 1 over n space close parentheses space equals space fraction numerator 1 over denominator square root of 2 end fraction open parentheses square root of n squared minus n subscript 1 superscript 2 end root space minus space n subscript 1 close parentheses end style
hence incident angle i = begin mathsize 12px style sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 2 end fraction open parentheses square root of n subscript 1 superscript 2 space minus space n squared end root space space minus space n subscript 1 close parentheses close parentheses end style
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Fig(2) shows the ray incident on face AB of right angled prisim will incident normally on diagonal face and emerges without any refraction.
 
By geometry, it can be worked out easily that angle of refraction at face AB is 45
 
 
hence we have,  sin i = n×sin r  = ( 1.352 / √2 ) = 0.956 ≈ 73°
Answered by Thiyagarajan K | 03 Mar, 2019, 07:45: PM

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