16.107 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 λ1=480 nm and another due
light with λ2=600 nm. What is the separation
between third order(m=3) bright fringes of
the two patterns?
• Note: same d,D but different λ => one
pattern magnified
• d/D= 5 x 10-3 => sinθ ~tan θ ~ θ
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 λ= 550 nm, what is the thickness of the
mica?
Double Slit Interference
Solution
• In general d sin θ = mλ and
y = D tanθ
• since θ <<1, ym ~ D θm ~ m λ D/d
• for λ1=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 λ2=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
Solution
• With no mica seventh bright fringe corresponds
to path difference = dsin θ = 7 λ
• with one slit covered, there is an additional path
difference due to change of λ` = λ/n in the
mica
• Let unknown thickness of mica be L
• path difference in mica compared to no mica is
[L/ λ` - L/ λ] λ = [nL -L] = [1.58L - L] =.58L
• to shift seventh bright fringe to center, this must
correspond to 7 λ = .58 L
• hence L= 7 λ /(.58) = 7(550)x10-9/.58 = 6.64µm
Double Slit Interference
d sinθ = m λ , m=0,1,2,… bright fringe
for m=1, need λ < d
d sinθ = (m+1/2) λ
, m=0,1,2,…
dark fringe
Light gun
1
16.107 L01 Feb08/02
Intensity of Interference Pattern
Intensity of Interference Pattern
• Each slit sends out an electromagnetic wave
with an electric field E(r,t)=Emsin(kr-ωt)
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
φ=k ∆L
β=φ/2=k ∆L/2
∆L= d sinθ
k =2π/λ
• net effect at point P is the superposition of two
waves
• E(r,t) = Emsin(kr1-ωt) + Emsin(kr2- ωt)
=>same wavelength and frequency but travel
different distances r1 and r2= r1+ ∆L
• E(r,t) = Emsin(kr1- ωt) + Emsin[k(r1+ ∆L)- ωt]
•
= Em[sin(kr1- ωt) + sin(kr1- ωt+ φ)]
• where φ=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(β) sin(kr1- ωt+ β) ; β= φ/2
d
β=φ/2=πd sinθ /λ
E(r,t) = [2Emcos(β)] sin(kr1-ωt+ β)
Amplitude= 2Emcos(β)
Intensity∝ (amplitude)2
Intensity of Double Slit
• Amplitude of electric field at P is
2Emcos(β)= 2Emcos(πd sinθ/λ)
• intensity of field ∝ the amplitude squared
• I = I0 cos2 (πd sinθ/λ) where I0 =4(Em)2
• I = I0 when β =mπ => d sinθ=mλ
• I = 0 when β = (m+1/2)π
=> d sinθ=(m+1/2)λ
2
• I = I0 cos (β) with β = πd sinθ/λ
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
E= E1 + E2
I= E2 = E12 + E22 + 2 E1 E2
= I1 + I2 + “interference” <== vanishes if incoherent
2
16.107 L01 Feb08/02
Solution
y=rsinθ
r
θ
x=rcosθ
y=y1sin(kx-ωt+φ1)+ y2sin(kx-ωt+φ2)+ y1sin(kx-ωt+φ3)
• Consider adding three vectors of equal length to get zero
resultant vector
• choose φ1=0 , φ2 = 2π/3, φ3 = 4π/3
• add any two => phase difference = 2π/3
• amplitude = 2y1cos(∆φ/2)=2y1cos(π/3)=y1
• hence intensity is I1 => c)
3