Transcript Separation of Bright Fringes

Double Slit Interference
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
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 10-3 => sin ~tan  ~ 
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 .072 mm
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?
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
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
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
=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
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/
Intensity of Double Slit
E= E1 + E2
I= E2 = E12 + E22 + 2 E1 E2
= I1 + I2 + “interference” <== vanishes if incoherent
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) I1 d) 2I1 e) 4I1
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)