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Q.

In the figure is shown Young’s double slit experiment. Q is the position of the first bright fringe on the right side of O. P is the 11th fringe on the other side, as measured from Q. If the wavelength of the light used is 6000×10-10m, then S1B will be equal to

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a

6×10-6m

b

6.6×10-6m

c

3.138×10-7m

d

3.144×10-7m

answer is A.

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Detailed Solution

P is the position of 11th bright fringe from Q. From central position O, P will be the position of 10th bright fringe. Path difference between the waves reaching at P =S1B=10λ=10×6000×10−10=6×10−6m..

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