Advertisements
Advertisements
प्रश्न
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
Using analytical method for interference bands, obtain an expression for path difference between two light waves.
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.
Two coherent sources of light having intensity ratio 81 : 1 produce interference fringes. Calculate the ratio of intensities at the maxima and minima in the interference pattern.
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, show that:
`beta = (lambda "D")/"d"` where the terms have their usual meaning.
Two slits, 4mm apart, are illuminated by light of wavelength 6000 A° what will be the fringe width on a screen placed 2 m from the slits?
In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is `1/8`th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to ______.
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is ______.
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?
If the monochromatic source in Young’s double slit experiment is replaced by white light, then ______.
