Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separ

Question

Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0 . 35 mm , then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm .

TestHub · Accepted Answer

Fringe width for both cases can be written as. $ω_{1} = \frac{λ_{1} D}{d} & ω_{2} = \frac{λ_{2} D}{d}$ .

Using values of wavelength, we get $ω_{1} = 16 mm & ω_{2} = 12 mm$ .

Let y be the common distance of the bright fringes by the both wavelength, then

$y = n_{1} ω_{1} = n_{2} ω_{2}$ .

As LCM $(ω_{1}, ω_{2}) = 48 mm$ , therefore at $y = 48 mm$ distance both bright fringes will be found.

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