A beam of light consisting of two wavelengths, 800nm and 600nm is used to obtain the interference fringes in Young's double slit experiment on a screen placed 1.4m away. If the two slits are separated by 0.28mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelength coincide??

Here

D = 1.4 m

d = 0.28 mm = 0.28 × 10^{-3} m

λ_{1} = 800 × 10^{-9} m

λ_{2} = 600 × 10^{-9} m

let y be the common distance of the bright fringes by both wavelength.

y = n_{1}λ_{1}D/d = n_{2}λ_{2}D/d

=> n_{1}λ_{1 }= n_{2}λ_{2}

=> 800 × 10^{-9} × n_{1} = 600 × 10^{-9} × n_{2}

=> 4 n_{1} = 3 n_{2}

n_{1} ≠ 0, n_{2} ≠ 0

For y to be minimum since n are integers

n_{1 }= 3 and n_{2} = 4

y = n_{1}λ_{1}D/d

=> y = 3×800 × 10^{-9} ×1.4/0.28 × 10^{-3}

=> y = 1.2×10^{-2} m

=> y = 1.2 cm

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