Home
Class 12
PHYSICS
In Young's experiment for interference o...

In Young's experiment for interference of light the slits 0.2 cm apart are illuminated by yellow light `(lamda=600nm)` What would be the fringe width on a screen placed 1 m from the plane of slits? What will be the fringe width if the system is immersed in water (refractive index `=4//3`)

Text Solution

Verified by Experts

The correct Answer is:
0.3 mm, 0.225 mm
`d=0.2cm,lamda=5896Å,D=1m`
Fringe width `beta=(lamdaD)/(d)=(5896xx10^(-10)xx1)/(0.2xx10^(-2))=0.3` mm
If system is immersed in water `(mu=1.33)`, then the fringe width becomes
`beta'=(beta)/(mu)=(0.3)/(1.3)mm=0.225mm`
Promotional Banner

Topper's Solved these Questions

  • WAVE OPTICS

    ALLEN |Exercise Exercise 4 B (Conceptual subjective Exercise)|8 Videos

  • WAVE OPTICS

    ALLEN |Exercise Exercise 5 A (Previous Years Questions)|17 Videos

  • WAVE OPTICS

    ALLEN |Exercise Exercise 3 (Miscellaneous type Questions)|27 Videos

  • UNIT & DIMENSIONS, BASIC MATHS AND VECTOR

    ALLEN |Exercise Exercise (J-A)|7 Videos

Similar Questions

Explore conceptually related problems

In young's double slit experiment the slits are illumated by light of wavelength 5890^(@) A and the distance between the fringes obtained on the screen is 0.2^(@) . If the whole apparatus is immersed in water then the angular fringe width will be, if the refractive index of water is 4//3

In a double-slit experiment the angular width of a fringe is found to be 0.2^(@) on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed i m away was found to be 0.2^(@) . What will be the angular width of the first minima if the entire experimental apparatus is immersed in water (u_("water")=4//3)

In Young's double slit experiment, the slit are 2 mm apart and are illuminated by photons of two wavelengths lambda = 12000 Å and lambda_(2)=10000 Å . At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Two slits at a distance of 1mm are illuminated by a light of wavelength 6.5xx10^-7m . The interference fringes are observed on a screen placed at a distance of 1m . The distance between third dark fringe and fifth bright fringe will be

A Young's double slit set up for interference is shifted from air to within water, then the fringe width .........

What will be the diffraction angle of first order maxima in diffraction obtained due to light of wavelength 55nm and width of slit 0.55 mm ?

The angular width of the central maximum in a single slit diffraction pattern is 60^(@) . The width of the slit is 1 mu m The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it. Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance ? (i.e. distance between the centres of each slit)

Light of wavelength 6000Å is incident on a slit of width 0.30 mm. The screen is placed 2 m from the slit. Find (a) the position of the first dark fringe and (b). The width of the central bright fringe.