CBSE Class 12-science Answered

Let A and B be two fine slits, a small distance 'd' apart. Let them be illuminated by a monochromatic light of wavelength l.

MN in the screen is at a distance D from the slits AB. The waves from A and B superimpose upon each other and an interference pattern is obtained on the screen. The point C is equidistant from A and B and therefore the path difference between the waves will be zero and so the point C is of maximum intensity. It is called the central maximum.

For another point P at a distance 'x' from C, the path difference at P = BP - AP.

Now AB = EF = d, AE = BF = D

\D BPF

[Pythagoras theorem]

Similarly in D APE

(on expanding Binomially)

For bright fringes (constructive wavelength) the path difference is integral multiple of wavelength i.e., path difference is nl.

(x therefore represents distance of n^th bright fringe from C)

Now

and so on.

Therefore separation between the centers of two consecutive bright fringe is the width of a dark fringe.

Similarly for dark fringes,

The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe.

The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe.

All bright and dark fringes are of equal width as b₁ = b₂.

Hope this clarifies your doubt.

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