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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
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
Show that the angular width of the first diffraction fringe is half that of the central fringe.
In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.
In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width.
The line-width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the line-width of a bright fringe in a Young's double slit experiment in terms of \[\lambda,\] d and D where the symbols have their usual meanings.
When a beam of light is used to determine the position of an object, the maximum accuracy is achieved, if the light is ______.
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.
The Young's double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then:
- 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).
What is the angular fringe width \(\alpha\)?
