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प्रश्न
"If the slits in Young's double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit".
Justify the above statement through a relevant mathematical expression.
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उत्तर
The total intensity at a point where the phase difference is ∅, is given by `"I" = "I"_1 + "I"_2 + 2sqrt("I"_1"I"_2) "COS"∅`.
Here I1 and I2 are the intensities of two individual sources which are equal.
When ∅ is 0, I = 4I1
When ∅ is 90°, I = 0
Thus intensity on the screen varies between 4I2 and 0.
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संबंधित प्रश्न
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
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Estimate the number of fringes obtained in Young's double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit.
Find the intensity at a point on a screen in Young's double slit experiment where the interfering waves have a path difference of (i) λ/6, and (ii) λ/2.
A Young's double slit experiment is performed with white light.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to the central will be red.
(d) The fringe next to the central will be violet.
What should be the path difference between two waves reaching a point for obtaining constructive interference in Young’s Double Slit experiment ?
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Find the distance of the third bright fringe for λ = 520 nm on the screen from the central maximum.
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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 distance of (a) third minimum and (b) fifth maximum, from the central maximum.
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