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)#
