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solve the below question
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Asked by sarveshvibrantacademy | 23 Mar, 2019, 11:10: AM
answered-by-expert Expert Answer
Path difference produced
= 3 over 4 2 pi R minus 1 fourth 2 πR equals πR
For maxima 
increment x equals n lambda
lambda equals fraction numerator pi R over denominator n end fraction
where n=1,2,3...
For minima
 increment x equals open parentheses 2 n minus 1 close parentheses lambda over 2
So
 lambda equals fraction numerator 2 pi R over denominator 2 n minus 1 end fraction
 where n=1,2,3...
 
 
Maximum intensity is given by
open parentheses square root of I subscript 1 end root plus square root of I subscript 2 end root close parentheses squared I subscript 1 equals I subscript 2 equals I subscript 0 over 2 I subscript m a x end subscript equals 2 I subscript 0
 
Maximum value of lambda to produce maxima at D
can be obtained by putting least value of n in
lambda equals fraction numerator pi R over denominator n end fraction.
The maximum value of lambda to produce minima at D 
 can be obtained by putting least possible value of n in
lambda equals fraction numerator 2 pi R over denominator 2 n minus 1 end fraction 
Answered by Utkarsh Lokhande | 25 Mar, 2019, 16:03: PM
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