Interference of light wave 2

Double-Slit Experiment
Suppose in the double-slit arrangement, 0.150mm, d=120cm, L=833nm,?= and . 2.00cmy=
(a) What is the path difference ? for the rays from the two slits arriving at point P?
(b) Express this path difference in terms of ?.
(c) Does point P correspond to a maximum, a minimum, or an intermediate condition?
Solutions:
14-7
(a) The path difference is given by sind??=. When Ly, ??? is small and we can make the approximationsintan/yL???=. Thus,
()242.0010m1.5010m2.5010m1.20mydL???×???=×=×????
?6
(b) From the answer in part (a), we have
672.5010m3.008.3310m????×=×
?
or 3.00??=.
(c) Since the path difference is an integer multiple of the wavelength, the intensity at point P is a maximum.
14.3 Intensity Distribution
Consider the double-slit experiment shown in Figure 14.3.1.
Figure 14.3.1 Double-slit interference
The total instantaneous electric field E
??at the point P on the screen is equal to the vector sum of the two sources:. On the other hand, the Poynting flux S is proportional to the square of the total field: 1=+EEE??????
2
2
222212121()2SEEE?=+=++?EEEE?????? ??
(14.3.1)
Taking the time average of S, the intensity Iof the light at P may be obtained as:
221212ISEE=?++?EE2
???? (14.3.2)
14-8
The cross term 122?EE???? represents the correlation between the two light waves. For incoherent light sources, since there is no definite phase relation between and , the cross term vanishes, and the intensity due to the incoherent source is simply the sum of the two individual intensities: 1E??2E??
inc12III=+ (14.3.3)
For coherent sources, the cross term is non-zero. In fact, for constructive interference, , and the resulting intensity is 1=EE????
2
14II= (14.3.4)
which is four times greater than the intensity due to a single source. On the other hand, when destructive interference takes place, 1 2 =?EE????, and 12 1I???EE????, and the total intensity becomes
1112IIII 0 =?+= (14.3.5)
as expected.
Suppose that the waves emerged from the slits are coherent sinusoidal plane waves. Let the electric field components of the wave from slits 1 and 2 at be given by P
10sinEE t ?= (14.3.6)
and
20sin()EEt??=+ (14.3.7)
respectively, where the waves from both slits are assumed have the same amplitude . For simplicity, we have chosen the point P to be the origin, so that the kxdependence in the wave function is eliminated. Since the wave from slit 2 has traveled an extra distance 0E?to, has an extra phase shift P2E? relative to from slit 1. 1E
For constructive interference, a path difference of ??=would correspond to a phase shift