Advertisements
Advertisements
प्रश्न
In Young's double-slit experiment, the separation between the two slits is d and the distance of the screen from the slits is 1000 d. If the first minima fall at a distance d from the central maximum, obtain the relation between d and λ.
उत्तर
In Young's double-slit experiment, the separation between two slits is d.
Distance of the screen from the slit D = 1000 d
By using Bragg's law,
nλ = 2D sinθ
⇒ n = 1, D = 1000 d
⇒ 1 × λ = 2 × 1000 d sinθ
⇒ λ = 2000 d sinθ
d = `lambda/2000sintheta`
Hence, the relation between d and λ is `d = lambda/2000sintheta`.
APPEARS IN
संबंधित प्रश्न
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?
Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is \[2 \cdot 0 \times {10}^{- 3}m.\]
A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the light energy falling on it. (a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
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 a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit- width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
- 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 Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
