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प्रश्न
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
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उत्तर
Fringe width is the distance between two consecutive dark or bright fringes,
so we have fringe width `=(lambdaD)/d`.
If the whole apparatus is immersed in water and refractive index of water is n then,
`v/c =1/n` = Where v is velocity of light in water
`=> n =(vlambda)/(vlambda_omega) lambda =` wavelength of light in air
`=> n =(vlambda)/(lambda_omega) lambda_omega =` wavelength of light in wetar
`lambda_omega = lambda/n v ` = frequency of light in air and water
Hence
`beta_omega =(lambda_omegad)/D =(lambda) =(lamdad)/(nD)`
`beta_omega = 1/n beta`
This shows fringe width will be reduced by the factor of the refractive index of water.
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संबंधित प्रश्न
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
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 beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1 · 4 m away. If the two slits are separated by 0·28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
If Young's double slit experiment is performed in water, _________________ .
A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment. (a) Find the change in the optical path due to introduction of the plate. (b) What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is \[\lambda.\] Neglect any absorption of light in the plate.
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) \[\frac{2D\lambda}{d}\]
How is the fringe width of an interference pattern in Young's double-slit experiment affected if the two slits are brought closer to each other?
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 an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
REASON (R): Fringe width is proportional to (d/D).
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