Advertisements
Advertisements
प्रश्न
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
Advertisements
उत्तर
The intensity variation in the fringe pattern obtained on a screen in a Young’s double slit experiment corresponds to both single slit diffraction and double slit interference because the two sources are slits of finite width in the double slit experiment.
If a lens is placed in front of the double slits and when one of the slit S1 is closed and the other kept open, a single slit diffraction is formed on the screen. A similar diffraction pattern is obtained on the screen if the slit S1 is kept open and S2 is closed. Both diffraction patterns form on the same position on the screen in the focal plane of the lens. When both slits open simultaneously, the resulting total intensity pattern on the screen is actually the superposition of the single slit diffraction pattern formed by waves from various point sources of each slit and a double slit interference pattern as shown. The actual double slit intensity pattern consists of the interference pattern (solid lines) formed within the diffraction pattern (dotted lines).
APPEARS IN
संबंधित प्रश्न
In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?
In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 × 10−7 m and the distance of the screen from the slits is 1.0m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimetre width on the screen?
In a Young's double slit experiment, \[\lambda = 500\text{ nm, d = 1.0 mm and D = 1.0 m.}\] Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
An unpolarised beam of intensity 2a2 passes through a thin polaroid. Assuming zero absorption in the polaroid, the intensity of emergent plane polarised light will be
Two balls are projected at an angle θ and (90° − θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is:
Consider a two-slit interference arrangement (Figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.
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?
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.
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.
