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

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

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G1 Physics · Accepted Answer

For the nth bright fringe, yₙ = nλD/d. Thus d = nλD/yₙ = [5 × 600 × 10⁻⁹ × 1]/(5 × 10⁻²) = 60 × 10⁻⁶ m = 60 μm.