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Show that the Fringe Pattern on the Screen is Actually a Superposition of Slit Diffraction from Each Slit.

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प्रश्न

Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.

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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).

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2014-2015 (March) Panchkula Set 3

संबंधित प्रश्न

In Young's double slit experiment, plot a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits keeping other parameters same. What information can one obtain from the slope of the curve?


If Young's double slit experiment is performed in water, _________________ .


Two transparent slabs having equal thickness but different refractive indices µ1 and µ2are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point P0 which is equidistant from the slits?


A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle \[\theta =  \sin^{- 1}   \left( \frac{\lambda}{2d} \right)\] with the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.


What should be the path difference between two waves reaching a point for obtaining constructive interference in Young’s Double Slit experiment ?


How will the interference pattern in Young's double-slit experiment be affected if the source slit is moved away from the plane of the slits?


The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of the fourth bright fringe after a glass plate of refractive index 1.5 is introduced in the path of one of the beams. The thickness of the glass plate would 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 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.


What is fringe width?


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