Question #101982
In a double slit experiment two interference patterns are seen on the screen.One pattern is due to light of wavelength 500nm and second pattern is due to light of wavelength 400nm. The distance between the slits is 2.5mm and the distance of screen from slits is 1m . The separation on the screen between the fifth order (m=5) bright fringes of two interference patterns will be
Ans: 0.2mm
1
Expert's answer
2020-01-29T14:07:42-0500

In double slit experiment nth fringe width is given by nλDd\dfrac{n\lambda D}{d}

The separation on the screen between the fifth order bright fringes of two interference patterns

    5λ1Dd5λ2Dd\implies\dfrac{5\lambda_1 D}{d}-\dfrac{5\lambda_2 D}{d}

    5(λ1λ2)Dd\implies\dfrac{5(\lambda_1-\lambda_2 )D}{d}

    5×(500400)×109×1/0.0025\implies5\times(500-400)\times10^{-9}\times1/0.0025

    2×103m\implies2\times10^{-3} m



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