When two light waves of intensity #I_1 and I_2# respectively interfere with each other the resultant intensity at a point is given by #I_R=I_1+I_2+2sqrt(I_1I_2)cosphi#

where #phi# is the phase difference between the two light waves.

In Young's Double Slit Experiment, the two slits are identical. Therefore, the intensity equation can be shown to be

# I_R=4Icos^2(ϕ/2)#

We see that the maximum intensity produced is #4I#. Putting maximum intensity as #I_@#, we have

# I_R=I_@cos^2(ϕ/2)#

#=> I_R/I_@=cos^2(ϕ/2)# .....(1)

When two waves meet on the screen, the phase difference between them depends on two factors.

- the phase difference produced due to the difference in the path taken by the two waves.
- the initial phase difference when the wave starts at the slit.

For the given point, intensity is #I# where path difference is #lambda/6#. We know that for a path length#=lambda#, phase difference #=2pi#.

#:.# Phase difference #phi=(2pi)/lambdaxxlambda/6=pi/3#

Our intensity equation (1) becomes

# I/I_@=cos^2(pi/6)#

#=> I/I_@=3/4#

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

Suppose two coherent plane waves #1 and 2# emerge from the two slits. Let the electric fields of the two is given by

#E_1=E_@ sin (omegat)#, and

#E_2=E_@ sin (omegat+phi)#

We have assumed same maximum amplitude for both. And also assumed the point where intensity is to be calculated as the origin. The phase difference between the two is #phi#.

Assuming both fields pointing in the same direction and using superpostion principle the resultant electric field is

#E=E_1+E_2=E_@ [sin omegat+sin (omegat+phi)]#

Using trigonometric identity

#sinA+sinB=2sin((A+B)/2)cos((A-B)/2)#

above expression becomes

#E=2E_@sin(omegat+phi/2)cos((phi)/2)#

Since the intensity #I# is proportional to the time average of the square of the total electric field we get

#Iprop"<"E^2">"=4E_@^2"<"sin^2(omegat+phi/2)">"cos^2((phi)/2)#

To find the #"<"sin^2theta ">"# term we need to integrate it over one cycle and divide with the cycle interval #2pi#. We get #"<"sin^2theta">"=1/2#. Inserting #2E_@^2=I_@# we obtain

#I=I_@cos^2((phi)/2)#