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प्रश्न
Obtain the equation for bandwidth in Young’s double slit experiment.
उत्तर
- Condition for bright fringe (or) maxima:
The condition for the constructive interference or the point P to be have a bright fringe is,
Path difference, δ = nλ
where, n = 0,1,2,…….. ∴ `"dy"/"D"` = nl
y = `"n" (lambda"D")/"d"` (or)
`"y"_"n" = "n" (lambda"D")/"d"` - Condition for dark fringe (or) minima:
The condition for the destructive interference or the point P to be have a dark fringe is,
Path difference δ = (2n -1) `λ/2`
when, n = 1, 2, 3,………..
∴ y = `(2"n" - 1)/2 (lambda"D")/"d"` (or)
`"y"_"n" = (2"n" - 1)/2 (lambda"D")/"d"` - Equation for bandwidth:
- The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.
- The distance between (n+l)th and nth consecutive brigh fringes from O is given by,
β = `"y"_("n + 1") - "y"_"n" (("n + 1")(lambda"D")/"d") - ("n" (lambda"D")/"d")`β = `(lambda"D")/"d"`
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संबंधित प्रश्न
State any one difference between interference of light and diffraction of light
Four light waves are represented by
(i) \[y = a_1 \sin \omega t\]
(ii) \[y = a_2 \sin \left( \omega t + \epsilon \right)\]
(iii) \[y = a_1 \sin 2\omega t\]
(iv) \[y = a_2 \sin 2\left( \omega t + \epsilon \right).\]
Interference fringes may be observed due to superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
In Young’s double slit experiment, the slits are separated by 0.5 mm and 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.
Answer in brief:
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"The fringe width ____________."
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(cos 60° = 0.5, `lambda` = wavelength of light, d = slit width)
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