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Question

In a Young's double-slit experiment, the slits are 0.5 mm apart and the interference pattern is observed on a screen at a distance of 100 cm from the slits. It is found that the ninth bright fringe on one side of the central bright fringe, is at a distance of 10.5 mm from the second dark fringe lying on the opposite side of central bright fringe. The wavelength of the light used is,

A
5000 ˚A
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B
50007 ˚A
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C
2500 ˚A
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D
25007 ˚A
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Solution

The correct option is A 5000 ˚AGiven: d=0.5 mmD=100 cm Distance of ninth bright fringe, from central bright fringe is, β9,b=9(λDd) Distance of second dark fringe, from central bright fringe, is, β2,d=(2−12)λDd=3λD2d As both the above fringes are on the opposite sides of central bright fringe, the distance between them will be equal to, β9,b+β2,d=10.5 mm ⇒λDd(9+32)=10.5×10−3 ∴λ=(10.5×10−3)×(221)×(0.5×10−3100×10−2) =5×10−7 m=5000 ˚A Hence, (A) is the correct answer. Why this question? Caution: while finding the distance between any two fringes (dark or bright), it is also important to know, on which side (same or opposite) of the central maximum, do they lie.

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