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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 Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
Using analytical method for interference bands, obtain an expression for path difference between two light waves.
In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light \[\left( \lambda = 700\text{ nm in vacuum} \right).\] Find the fringe-width of the pattern formed on the screen.
In Young's double slit experiment the slits are 0.589 mm apart and the interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 7.5 mm from the dark fringe which is second from the center of the fringe pattern. Find the wavelength of the light used.
Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit.
Using Young’s double slit experiment, a monochromatic light of wavelength 5000Å produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000Å is used and the separation between the slits is doubled, then the new fringe width will be ______.
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ______ nm.
- 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).
