Advertisements
Advertisements
Question
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
Advertisements
Solution
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
RELATED QUESTIONS
(i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interferences at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.
(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.
In Young’s double-slit experiment, show that:
`beta = (lambda "D")/"d"` where the terms have their usual meaning.
In Young's double slit experiment shown in figure S1 and S2 are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has strong intensity?
In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case ______.
- 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?
If the monochromatic source in Young’s double slit experiment is replaced by white light, then ______.
Why do the two narrow slits in Young's experiment act as coherent sources?
What is the fringe width \(\beta\) in Young's double-slit experiment?
What is the distance \\(x_n-x_m\\) between the \\(n\\)-th bright fringe and the \\(m\\)-th dark fringe?
