Interference of Light


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experimental set up of Young s experiment

The above experiment gives a rough experimental set up of Young's experiment.

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.

superposition of wavefront and displacement

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 following figures show the fringes when red, green and blue light are used separately in Young's double slit experimental set up (in the fringes width).

red fringe in young s double slit experiment

green fringe in young s double slit experiment

blue fringe in young s double slit experiment

(The blue fringes are narrower than the red and green).

Note:

If S in replaced by a white light, we see a whole series of colors since each colour has a different wavelength at a given point. It may be a constructive interference for one colour and destructive interference for another. Therefore the former colour alone is visible.



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