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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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संबंधित प्रश्न
In young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.
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A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?
A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the light energy falling on it. (a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
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A beam of light consisting of two wavelengths 600 nm and 500 nm is used in Young's double slit experiment. The silt separation is 1.0 mm and the screen is kept 0.60 m away from the plane of the slits. Calculate:
- the distance of the second bright fringe from the central maximum for wavelength 500 nm, and
- the least distance from the central maximum where the bright fringes due to both wavelengths coincide.
