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प्रश्न
In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width.
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उत्तर
For any other incoherent source of light a steady interference pattern can never be obtained, even if the sources emit waves of equal wavelengths and equal amplitudes. This is because the waves emitted by a source undergo rapid and irregular changes of phase, so that the intensity at any point is never constant. Naturally the phase difference between the waves emitted by the two sources cannot remain constant.
The two waves interfering at P have different distances S1P = x and S2P = x + Δx.
So, for the two sources S1 and S2we can respectively write,
`I_1 = I_(01) sin (kx -wt)`
`I_1 = I_(02) sin (k(x +Deltax)-wt) =I_(02)sin(kx -wt +delta)`
`delta = kDeltax =((2pi)/lambda) xx Delta x`
The resultant can be written as,
`I =I_0sin(kx -wt +epsi)
Where` I_0^2 =I_(01)^2 + I_(02)^2 +2I_(01) I_(02) cos delta`
And tan`epsi = I_(02) sin delta /(I_(01) +I_(02)cos delta)`
The condition for constructive (bright fringe) and destructive (dark fringe) interference are as follows;
δ = 2nπ for bright fringes
δ = (2n + 1) π for dark fringes
Where n is an integer.
Now to find the fringe width,
The path difference is `Deltax =S_2P-S_1P` nearly equal to d `sintheta =d tantheta =(dy)/D`
Hence we can write, `y=(nlambdaD)/d ,n` is an integer.
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