1. Double Slit Interference

2. Double Slit Interference
d sin = m  , m=0,1,2,… bright fringe
for m=1, need  < d
d sin = (m+1/2)  , m=0,1,2,… dark fringe
Light gun

3. L01 Feb08/02
Problem
In a double slit expt the distance between slits is d=5.0 mm and the distance to the screen is D=1.0 m. There are two interference patterns on the screen: one due to light with 1=480 nm and another due light with 2=600 nm. What is the separation between third order(m=3) bright fringes of the two patterns?
Note: same d,D but different  => one pattern magnified
d/D= 5 x => sin ~tan  ~ 

4. Solution In general d sin  = m and y = D tan
since  <<1, ym ~ D m ~ m  D/d
for 1=480 nm, D=1.0m, d=5.0 mm y3 = 3(480x 10-9)(1.0)/5.0 x 10-3)= 2.88 x 10-4 m
for 2=600 nm, D=1.0m, d=5.0 mm y3 = 3(600x 10-9)(1.0)/5.0 x 10-3)= 3.60 x 10-4 m
hence difference is mm

5. Problem
A thin flake of mica (n=1.58) is used to cover one slit of a double slit arrangement. The central point on the screen is now occupied by what had been the seventh bright fringe (m=7) before the mica was used. If = 550 nm, what is the thickness of the mica?

6. Solution
With no mica seventh bright fringe corresponds to path difference = dsin  = 7 
with one slit covered, there is an additional path difference due to change of ` = /n in the mica
Let unknown thickness of mica be L
path difference in mica compared to no mica is [L/ ` - L/ ]  = [nL -L] = [1.58L - L] =.58L
to shift seventh bright fringe to center, this must correspond to 7  = .58 L
hence L= 7  /(.58) = 7(550)x10-9/.58 = 6.64m

7. Intensity of Interference Pattern
Each slit sends out an electromagnetic wave with an electric field E(r,t)=Emsin(kr-t) where r is the distance from either slit to the point P on the screen
the two waves are coherent
net field at P is the superposition of two waves which have travelled different distances and hence have a phase difference

8. Intensity of Interference Pattern
net effect at point P is the superposition of two waves
E(r,t) = Emsin(kr1-t) + Emsin(kr2- t) =>same wavelength and frequency but travel different distances r1 and r2= r1+ L
E(r,t) = Emsin(kr1- t) + Emsin[k(r1+ L)- t]
= Em[sin(kr1- t) + sin(kr1- t+ )]
where =k L is phase shift due to path difference!
sin(A)+sin(B) = 2 cos[(A-B)/2]sin[(A+B)/2]
E(r,t) = 2Emcos() sin(kr1- t+ ) ; = /2

9. E(r,t) = [2Emcos()] sin(kr1-t+ )
=k L
=/2=k L/2
L= d sin d
k =2/
=/2=d sin /
E(r,t) = [2Emcos()] sin(kr1-t+ )
Amplitude= 2Emcos()
Intensity (amplitude)2

10. Intensity of Interference Pattern
Amplitude of electric field at P is 2Emcos()= 2Emcos(d sin/)
intensity of field  the amplitude squared
I = I0 cos2 (d sin/) where I0 =4(Em)2
I = I0 when  =m => d sin=m
I = 0 when  = (m+1/2) => d sin=(m+1/2)
I = I0 cos2 () with  = d sin/

11. Intensity of Double Slit
E= E1 + E2
I= E2 = E12 + E E1 E2
= I1 + I2 + “interference” <== vanishes if incoherent

12. MC 41-12
Three coherent equal intensity light rays arrive at P on a screen to produce an interference minimum of zero intensity.
If any two of the rays are blocked, the intensity at is I1 . What is the intensity at P if only one is blocked?
a) 0 b)I1/2 c) I d) 2I1 e) 4I1

13. Solution y=rsin r  x=rcos
y=y1sin(kx-t+1)+ y2sin(kx-t+2)+ y1sin(kx-t+3)
Consider adding three vectors of equal length to get zero resultant vector
choose 1=0 , 2 = 2/3, 3 = 4/3
add any two => phase difference = 2/3
amplitude = 2y1cos(/2)=2y1cos(/3)=y1
hence intensity is I1 => c)