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प्रश्न
(a) Can the interference pattern be produced by two independent monochromatic sources of light? Explain.
(b) The intensity at the central maximum (O) in Young's double-slit experimental set-up shown in the figure is IO. If the distance OP equals one-third of the fringe width of the pattern, show that the intensity at point P, would `"I"_°/4`
(c) In Young's double-slit experiment, the slits are separated by 0⋅5 mm and the screen is placed 1⋅0 m away from the slit. It is found that the 5th bright fringe is at a distance of 4⋅13 mm from the 2nd dark fringe. Find the wavelength of light used.
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उत्तर
(a) Two independent monochromatic sources cannot produce a sustained interference pattern. This is because the phase difference of two independent sources cannot be strictly constant throughout. A constant phase difference is essential to produce a distinguishable interference pattern. [Each of the sources produces their own diffraction pattern which interacts with each other. This interaction may or may not result in the interference pattern in case of 2 different sources but produces a clear pattern in case of coherent sources if d
(b) Fringe width, `beta = lambda"D"/"d"`
Here,
`"OP" ="x" = beta/3 = (lambda"D")/(3"d")`
`"x" = (lambda"D")/(3"d")` ..................(1)
∴ Δ X (Path difference) `= "x""d"/"D"`
∵ `"x""d"/"d" = lambda/3`......(From eq. 1)
⇒ Δ X ` = "x""d"/"D"`
φ (Phase difference) `= (2pi)/(lambda)(Delta "X")`
`= (2pi)/lambda xx lambda / 3 = (2pi)/3`
⇒ φ `= (2pi)/3`
If intensity at point O is IO, then intensity at point P will be,`"I"_"P" = "I"_"O" cos^2 (φ/2)`
`"I"_"P" = "I"_"O" cos^2 (φ/2) = "I"_"O" cos^2(pi/3)`
`= "I"_"O"(1/2)^2 = "I"_"O"/4`
⇒ `"I"_"P" = "I"_"O"/4`
(c) For Young's double slit experiment, Position of 5th bright fringe `=5xx (lambda"D")/"d"`
The position of 2th dark fringe = `(2 xx 2 xx -1) xx (lambda"D")/(2"d") = 1.5 ((lambda"D")/"d")`
Where D is the distance of the screen from the slits and d is the separation between the slits.
According to the given information,
`5 xx ((lambda"D")/"d") - 1.5 xx ((lambda"d")/"d") = 4.13 xx 10^-3 "m"`
⇒ `3.5 xx ((lambda"D")/"d") = 4.13 xx 10^-3 "m"`
⇒ `lambda = (4.13 xx 10^-3 xx 0.5 xx 10^-3)/(3.5xx1) = 0.59 xx 1^-6 "m"`
⇒ `lambda = 590 "nm"`
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