Double split experiment

Asked by VISHNU pillai | 19th Feb, 2014, 10:37: AM

Expert Answer:

Since light has a very small wavelength, we need two slits, which send out two continuous coherent waves. Since the two slits are placed very close, these waves overlap as shown below.

 

S is a narrow slit illuminated by a monochromatic source of light. Two fine slits A and B are placed symmetrically parallel to S. When a screen is placed at a large distance from A and B, alternate bright and dark bands parallel to the slits appear on the screen. These bands are called interference fringes. The solid arcs represent the crests and the dotted arcs represent the troughs in the above diagram. When these arcs overlap, the (.) dots represent the maximum intensity i.e., constructive interference, whereas the (x) crosses represent the minimum intensity i.e., destructive interference. Therefore C, E and G are bright bands and D and F are dark bands.

The maximum intensity at a point on the screen is due to constructive interference i.e., the two wave trains having same amplitude, wavelength superpose in phase with each other (i.e., crest falls on crest or trough on trough).

Similarly, the minimum intensity at any point on the screen is due to destructive interference i.e., the two wave trains having the same amplitude and wavelength superpose out of phase with each other (i.e., crest falls on a trough and vice versa).

 

The intensity of all bright bands are the same. All dark bands also have same (zero) intensity. The intensity distribution Vs distance is shown as:

Answered by  | 21st Feb, 2014, 03:30: PM

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