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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
Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?
If the separation between the slits in a Young's double slit experiment is increased, what happens to the fringe-width? If the separation is increased too much, will the fringe pattern remain detectable?
The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I0. If one of the slits is closed, the intensity at this point will be ____________ .
Consider the arrangement shown in the figure. The distance D is large compared to the separation d between the slits.
- Find the minimum value of d so that there is a dark fringe at O.
- Suppose d has this value. Find the distance x at which the next bright fringe is formed.
- Find the fringe-width.
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?
Consider the arrangement shown in the figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When \[z = \frac{D\lambda}{2d},\] the intensity measured at P is I. Find the intensity when z is equal to
(a) \[\frac{D\lambda}{d}\]
(b) \[\frac{3D\lambda}{2d}\] and
(c) \[\frac{2D\lambda}{d}\]
In Young's double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0·48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate:
(i) Distance between the two slits.
(ii) Fringe width, i.e. fringe separation.
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.
In Young's double-slit experiment, the screen is moved away from the plane of the slits. What will be its effect on the following?
- The angular separation of the fringes.
- Fringe-width.
If the monochromatic source in Young’s double slit experiment is replaced by white light, then ______.
