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प्रश्न
In Young’s experiment interference bands were produced on a screen placed at 150 cm from two slits, 0.15 mm apart and illuminated by the light of wavelength 6500 Å. Calculate the fringe width.
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उत्तर
Given:
D = 150 cm = 1.5 m,
d = 0.15 mm = 1.5 x 10-4 m,
λ = 6500 Å = 6.5 x 10-7 m
To find:
Fringe width ( X )
Formula:
X = `"λD"/"d"`
Calculation:
From formula,
X = `[6.5 xx 10^-7 xx 1.5]/[1.5 xx 10^4]`
X = 6.5 x 10-3 m
X = 6.5 mm
The fringe width is 6.5 mm
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संबंधित प्रश्न
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(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.
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- Find the minimum value of d so that there is a dark fringe at O.
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(a) \[\frac{D\lambda}{d}\]
(b) \[\frac{3D\lambda}{2d}\] and
(c) \[\frac{2D\lambda}{d}\]
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In Young's double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate
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The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of the fourth bright fringe after a glass plate of refractive index 1.5 is introduced in the path of one of the beams. The thickness of the glass plate would be ______.
- Assertion (A): In Young's double slit experiment all fringes are of equal width.
- Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).
In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
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