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प्रश्न
- 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).
पर्याय
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).
Assertion (A) is true and Reason (R) is false.
Assertion (A) is false and Reason (R) is also false.
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उत्तर
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Bright and dark fringes are produced as light travels through a slit in Young's double-slit experiment. Both the black and brilliant fringes are the same width as YDSE.
Statement 2 explanation: White light is used in Young's double-slit studies, and it is directed through the slit. Therefore, only brilliant and dark fringes may be seen utilising this source.
We know that fringe width is given by,
Width = `(lambdaD)/d`
Where,
λ = Wavelength of source used.
D = Distance between screen and slit.
d = Distance between slits.
As fringe width depends on all these factors so the fringe width remains constant.
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संबंधित प्रश्न
The fringes produced in diffraction pattern are of _______.
(A) equal width with same intensity
(B) unequal width with varying intensity
(C) equal intensity\
(D) equal width with varying intensity
A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.
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.
In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
What is the effect on the interference fringes to a Young’s double slit experiment when
(i) the separation between the two slits is decreased?
(ii) the width of a source slit is increased?
(iii) the monochromatic source is replaced by a source of white light?
Justify your answer in each case.
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
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.
In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is 0.20 W m−2, what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes?
In Young's double slit experiment, the minimum amplitude is obtained when the phase difference of super-imposing waves is: (where n = 1, 2, 3, ...)
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.
In Young's double-slit experiment, the separation between the two slits is d and the distance of the screen from the slits is 1000 d. If the first minima fall at a distance d from the central maximum, obtain the relation between d and λ.
