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635.


Problem 45.25 (RHK)


We have to show that the half-width of the double-slit interference fringes is given by

,

if is small enough so that . The half-width is the angle between the two points in the fringe where the intensity is one-half that at the centre of the fringe.


Solution:           Click For PDF Version

We first work out the expression for the intensity of the resultant of two coherent waves with phase difference . Let the electric field components of the two waves at a point P at time t be described by the functions

where is the angular frequency of the waves and is the phase difference between them. We note that depends upon the location of the point P, which is described by the angle in a double-slit experiment.

We have

where the phase is

,

and the amplitude is

.

As the intensity I is proportional to the square of the amplitude, we note that

and for small

.

And

Let the centre of the m th- fringe be at angle . We have

We note that

Let the intensity is one-half that at the centre of the fringe at . We therefore have

.

This implies that

This implies that

The half-width is the angle between the two points in the fringe where the intensity is one-half that at the centre of the fringe. Therefore, half-width of the fringe will be