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(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 - Physics

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Numerical

(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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Solution

(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"`

Concept: Refraction of Monochromatic Light
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