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

Question

In Young's double slits experiment, the position of 5 th bright fringe from the central maximum is 5 cm . The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm . The separation between the slits is:, 60 μm, 48 μm, 12 μm, 36 μm

MARKS App · Accepted Answer

Given here, D = 1 m , λ = 600 × 10 - 9 m and n = 5 Distance of nth bright fringe is given by, y n = n λ D d , where, d is separation between the slits. ⇒ 5 × 600 × 10 - 9 × 1 d = 5 × 10 - 2 ⇒ d = 5 × 600 × 10 - 9 × 1 5 × 10 - 2 = 60 × 10 - 6 m ⇒ d = 60 μ m .

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