Advertisements
Advertisements
प्रश्न
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Advertisements
उत्तर
Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 × 10−2 m
In case of constructive interference, we have the relation for the distance between the two fringes as:
u = `nlambda D/d`
Where,
n = Order of fringes = 4
λ = Wavelength of light used
∴ `lambda = (ud)/(nD)`
= `(1.2 xx 10^(-2) xx 0.28 xx 10^(-3))/(4xx1.4)`
= 6 × 10−7
= 600 nm
Hence, the wavelength of the light is 600 nm.
APPEARS IN
संबंधित प्रश्न
In Young' s experiment the ratio of intensity at the maxima and minima . in the interference pattern is 36 : 16. What is the ratio of the widths of the two slits?
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
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.
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.
What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
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
If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
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.
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
A Young's double slit experiment is performed with white light.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to the central will be red.
(d) The fringe next to the central will be violet.
Consider the arrangement shown in the figure. The distance D is large compared to the separation d between the slits.
- Find the minimum value of d so that there is a dark fringe at O.
- Suppose d has this value. Find the distance x at which the next bright fringe is formed.
- Find the fringe-width.
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) \[\frac{2D\lambda}{d}\]
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.
Young's double slit experiment is made in a liquid. The 10th bright fringe lies in liquid where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately
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 :
Monochromatic green light of wavelength 5 × 10-7 m illuminates a pair of slits 1 mm apart. The separation of bright lines in the interference pattern formed on a screen 2 m away is ______.
Two beams of light having intensities I and 41 interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π/2 and π/3 at points A and B respectively. The difference between the resultant intensities at the two points is xl. The value of x will be ______.
- 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).
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?
