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Question
A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1 · 4 m away. If the two slits are separated by 0·28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
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Solution
Given: −
λ1 = 800 nm = 800 × 10−9 m
λ2 = 600 nm = 600 × 10−9 m
D = 1.4 m
d = 0.28 mm = 0.28 × 10−3 m
Let n1th maximum corresponds to λ1 coincides with n2th maximum corresponds to λ2. Then,
`n_1(lambda_1D)/d =n_2 ((lambda_2)D)/d`
`or,n_1/n^2 =lambda^2/lambda^1 = 600/800 =3/4`
The minimum integral value of n1 is 3 and of n2 is 4. Therefore, the minimum value of y is,
`y_(min) = n_1(lambda_1D)/d=(3 xx 800 xx 10^-9 xx 1.4)/((0.28) xx 10^-3)`
`y_(min) =12mm`
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