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Q8

 

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Asked by manasvijha 25th March 2019, 10:04 AM
Answered by Expert
Answer:
Given : D = 2m, d = 2mm, λ1 = 12000 ºA, λ2=10000 ºA
 
Let n1 be the fringe of slit 1 coinciding with the fringe n2 coming from another slit. 
Thus, the equation can be written as,
 
fraction numerator n subscript 1 lambda subscript 1 D over denominator d end fraction equals fraction numerator begin display style n subscript 2 lambda subscript 2 D end style over denominator begin display style d end style end fraction

T h u s comma space n subscript 1 lambda subscript 1 space equals space n subscript 2 lambda subscript 2
T h u s comma space n subscript 1 over n subscript 2 equals lambda subscript 2 over lambda subscript 1 equals 10000 over 12000 equals 5 over 6
 
Thus, we can say that 5th fringe coincides with 6th fringe.
 
 
Now, the minimum distance is given by,
 
x subscript m i n end subscript equals fraction numerator n subscript 1 lambda subscript 1 D over denominator d end fraction
space space space space space space space equals space fraction numerator 5 space cross times space 12000 space cross times space 10 to the power of negative 10 end exponent space cross times space 2 space space over denominator 2 space cross times space 10 to the power of negative 3 end exponent end fraction
x subscript m i n end subscript space space equals space 60 space cross times space 10 to the power of negative 4 end exponent equals space 6 space cross times space 10 to the power of negative 3 end exponent space equals space 6 space m m
 
Answered by Expert 25th March 2019, 12:34 PM
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