Wave Optics

JEE Main level, single correct, medium

70 of 191

Question 70 of 191

In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5cm5cm. The distance between slits and screen is 1m1m and wavelength of used monochromatic light is 600nm600nm. The separation between the slits is:

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Answer: (A) 60μm60\mu m

Given here,
D=1mD=1m, λ=600×10−9m\lambda =600\times {10}^{-9}m and n=5n=5
Distance of nth bright fringe is given by, yn=nλDd{y}_{n}=\dfrac{n\lambda D}{d}, where, dd is separation between the slits.
⇒5×600×10−9×1d=5×10−2\Rightarrow \dfrac{5\times 600\times {10}^{-9}\times 1}{d}=5\times {10}^{-2}
⇒d=5×600×10−9×15×10−2=60×10−6m\Rightarrow d=\dfrac{5\times 600\times {10}^{-9}\times 1}{5\times {10}^{-2}}=60\times {10}^{-6}m
⇒d=60μm\Rightarrow d=60\mu m.

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