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प्रश्न
In an interference experiment, a third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point?
विकल्प
420 nm
750 nm
630 nm
500 nm
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उत्तर
420 nm
Explanation:
Position of nth bright fringe,
`y_n = (nlambdaD)/d`
Given: λ = 700 nm
∴ `y_3 = (3 xx 700 xx D)/d`
⇒ `y_3 = (2100D)/d`
The 5th bright fringe due to light of wavelength λ' is formed at y3,
∴ y5 = y3
or `(5 xx lambda^'D)/d = (2100 nmD)/d`
⇒ `lambda^' = 2100/5` = 420 nm
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संबंधित प्रश्न
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Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit.
Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing λ = 400 nm). Describe the nature of the fringe pattern observed.
In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength \[\lambda.\] Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.
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) \[\frac{2D\lambda}{d}\]
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- 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).
