NEET Class neet Answered

Asked by manasvijha | 25 Mar, 2019, 10:04: AM

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 Shiwani Sawant | 25 Mar, 2019, 12:34: PM

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