In a double slit experiment, the distance between the slits is d. The screen is at a distance D from the slits. If a bright fringe is formed exactly in front of one of the slits its order is,
A
λ2dD
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B
dλ
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C
d22Dλ
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D
D22λD
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Solution
The correct option is Cd22Dλ
When bright fringe is formed in front of one of the slits its distance(y) from the center is y=d2
The path difference at this point will be, Δx=ydD
∴Δx=d(d2)D=d22D
Let the nth bright fringe is formed at this point. So the path difference sould be equal to nλ.
⇒nλ=d22D
∴n=d22Dλ
Hence, (D) is the correct answer.
Why this question :Tip : In YDSE, for nth bright fringe, path difference is equal to n times λ.