Advertisements
Advertisements
प्रश्न
How will the interference pattern in Young's double-slit experiment be affected if the phase difference between the light waves emanating from the two slits S1 and S2 changes from 0 to π and remains constant?
Advertisements
उत्तर
Fringe width = β = `"Dλ"/"d"`
If the phase difference becomes π, then the central fringe will be a dark fringe, since
I = `4"I"_0 cos^2 θ/2`
= `4"I"_0 cos^2 π/2`
= 0
APPEARS IN
संबंधित प्रश्न
The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.
How does an unpolarized light incident on a polaroid get polarized? Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.
A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light \[\left( \lambda = 700\text{ nm in vacuum} \right).\] Find the fringe-width of the pattern formed on the screen.
In Young's double-slit experiment, the two slits are separated by a distance of 1.5 mm, and the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths of 650 nm and 520 nm is used to obtain interference fringes.
Find the distance of the third bright fringe for λ = 520 nm on the screen from the central maximum.
In Young’s double slit experiment, what is the effect on fringe pattern if the slits are brought closer to each other?
In Young’s double slit experiment, what should be the phase difference between the two overlapping waves to obtain 5th dark band/fringe on the screen?
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:
In Young's double slit experiment, the minimum amplitude is obtained when the phase difference of super-imposing waves is: (where n = 1, 2, 3, ...)
In Young's double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate
- the fringe width
- the distance of (a) third minimum and (b) fifth maximum, from the central maximum.
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
