2.71 Optics
Problem Set 5 Solutions
1.
2π
|�k1 | = |�k2 | =
λ
�k1 = 2π (sin 30 ◦ x̂ + cos 30 ◦ ẑ) = 2π
λ
λ
�
√ �
1
3
x̂ +
ẑ
2
2
�k2 = 2π (cos 45 ◦ x̂ + sin 45 ◦ sin 30 ◦ ŷ + sin 45 ◦ cos 30 ◦ ẑ)
λ �
√
√
√ �
2
2
6
2π
=
x̂ +
ŷ +
ẑ
λ
2
4
4
Assuming |E1 | = |E2 | = 1,
�
2π x
√
3
E1 (x, y, z) = eik1 ·�r = e
i λ ( √2 + 2 √z) ≡ √eiφ1
2π
2
2
6
�
E2 (x, y, z) = eik2 ·�r = e
i λ ( 2 x+ 4 y+ 4 z) ≡ e
iφ2
�
interference pattern I
I = |E1 + E2 |2 = |E1 |2 + |E2 |2 + 2|E1 ||E2 | cos(φ1 − φ2 )
= 2[1 + cos(φ1 − φ2 )]
(a) In the xy-plane, z = 0
φ1 =
φ2 =
2π
(x/2) √
λ √
2π
( 22 x + 42 y)
λ
�
2π
φ1 − φ2 =
λ
��
√ �
√ �
1
2
2
−
x−
y = Δφ
2
2
4
I = 2[1 + cos Δφ] so the profile is a sinusoidal profile. The maxima are along the
lines whose equation is:
��
√ �
√ �
2π
1
2
2
−
x−
y = 2mπ, where m ∈ Z
λ
2
2
4
√
√
1− 2
2
x−
y = mλ
2
4
(b) For the plane z = λ
⎫
�
�
√
�
√
√
√ �
√
3
2π 1
⎬
φ1 = λ 2 x + 2 λ
2
6
2π
1
−
2
2
3
−
�√
� Δφ =
√
√
x+
y+
λ
2
λ
2
4
4
φ2 = 2π
x + 2y + 6λ ⎭
λ
2
4
4
I = 2[1 + cos Δφ], so the interference pattern is still a sinusoid (i.e. a set of linear
fringes). The maxima occur when Δφ = 2πm, m ∈ Z. The equation of the fringe
lines are:
�
√
√
√
√ �
2π 1 − 2
2
2 3− 6
x+
y+
λ = 2πm, m ∈ Z
λ
2
4
4
1
�
√
√
√
√ �
1− 2
2
2 3− 6
x+
y = m−
λ
2
4
4
Note that the slopes are the same as 1a, but the maxima are shifted.
(c) In the yz-plane, x = 0
�√ �
3
φ1 = 2π
z
λ
� √2
φ2 = 2λπ 42 y +
� √
�√
√ � �
2
π
2
3
6
√ �
−
Δφ =
−
y+
z
6
⎭
λ
2
4
4
z
4
⎫
⎬
√
We
also observe a set of fringes along the lines where Δφ = 2mπ, i.e. −
√ √
2 3− 6
z = mλ, m ∈ Z
4
2
y
4
+
2.
2π
Plane wave: Epl = |Epl |ei λ z
|
Esp |
i 2π z iπ (x2 +y2 )
Spherical wave: Esp =
e λ e
λz
αz
(a) At z = 1000λ, assuming the amplitudes of the two waves are equal:
2π
z
λ
2π
π
Esp = eiφsp where φsp =
z + (x2 + y 2 )
λ
λz
π 2
2
Δφ = φsp − φpl =
(x + y )
λz
Epl = eiφpl where φpl =
2
2
We have bright fringes where Δφ = 2πm, m = 0, 1, 2 . . ., so x 2+zy = mλ.
2
2
At z = 1000λ → x2 +
√y = 2000λ m, m = 0, 1, 2, 3 . . ., which is a set of concentric
rings of radii R = λ 2000m, m = 0, 1, 2, 3 . . .
(b) At z = 2000λ, the amplitude of the spherical wave decreases by a factor of 1/2
(energy conservation).
�
� �2
� �
1
1
5
Epl = eiφpl
I =1+
+ 2(1)
cos Δφ = + cos Δφ
Esp = 21 eiφsp
2
2
4
The maxima are given by Δφ = 2πm.
x2 + y 2
= mλ ⇒ x2 + y 2 = 4000λ2 m
2z
√
Therefore the maxima are concentric circles of radii R = 20λ 10m, m = 0, 1, 2, . . .
(c) Observations:
i. The interference pattern is a set of concentric circles whose radii are given by
√
Rm = 2zλm
2
ii. The radius of the first fringe (R1 ) increases with both λ and z
iii. At a certain distance z, the spacing between the fringes decreases as we go
radially outwards.
√
√
√
ΔRm = 2zλ( m − m − 1)
(d) If we insert a lens in branch 1 of a Michelson interferometer, the lens focuses the
plane wave to a point at its back focal plane.
After reflecting off the mirror, the lens is effectively imaging a point source at a
distance (d − f ) + d = 2d − f ; thus it forms a point source image at Si , where
1
1
1
1
1
2(d − f )
= −
= −
=
Si f
f
f (2d − f )
S0
2d − f
f (2d − f )
Si =
2(d − f )
If d = f , i.e. Si = ∞, we get a plane wave back and the output is a uniform
intensity, because we would be observing the interference of two on-axis plane
waves.
If d �= f, we get circular fringes due to the interference of a plane wave and a
spherical wave.
3. The general off-axis plane wave propagates at θ with respect to the z axis.
The off-axis plane wave equation is:
2π
Epl = |Epl |ei λ (x sin θ+z cos θ)
The equation of the spherical wave is:
Esp =
|Esp | i 2π z i π (x2 +y2 )
e λ e
λz
αz
3
(a) Assuming the amplitudes are equal at z = 1000λ, I = 2|Epl |2 (1 + cos Δφ), where
Δφ = φsp − φpl :
�
π
2
2
π 2 2π
π
2π
φsp = 2π
z
+
(x
+
y
)
λ
λz
Δφ =
x −
sin θx + y 2 +
z(1 − cos θ)
φpl = 2λπ (x sin θ + z cos θ)
λz
λ
λz
λ
Bright fringes occur when Δφ = 2πm:
1 2
1
x − (sin θ)x + y 2 = mλ − z(1 − cos θ)
2z
2z
x2 − 2z sin θx + z 2 sin2 θ + y 2 = 2z[mλ − z(1 − cos θ)] + z 2 sin2 θ . . .
�
��
�
(Eq. A)
Rm 2
(x − z sin θ)2 + y 2 = Rm 2
(b) At z = 2000λ, |Esp | = 12 |Epl |
2
I = |Epl |
�
�
5
+ cos Δφ
4
The fringes are still given by equation A where z = 2000λ. This gives a bigger
shift along the x-axis and lower contrast in the fringes as well as larger spacing of
peaks.
4. We sketch the system as follows:
The mth plane wave is at an angle θm = θ0 + mΔθ
2π
Em = ei λ [cos θm z+sin θm x]
Assuming small angles (paraxial approximation), θm � 1
cos θm ≈ 1, sin θm ≈ θm = θ0 + mΔθ
2π
2π
Em ≈ ei λ (z+θm x) = ei λ (z+θ0 x+mΔθx)
4
Adding all the plane waves,
ET =
=
N
−1
�
m=0
N
−1
�
Em
2π
ei λ (z+θ0 x+mΔθx)
m=0
=e
N
−1
�
i 2λπ z i 2λπ θ0 x
e
2π
(ei λ xΔθ )m
�m=0 ��
�
Geometric series: θ0 =1, r=ei
2π xΔθ
λ
2π
i 2π
z i 2λπ θ0 x
λ
∴ ET = e
e
·
1 − ei( λ N xΔθ)
2π
2π
2π
= e
i λ z e
i λ θ0 x ·
1 − eiφ1
1 − e
iφ2
1 − ei( λ xΔθ)
φ1
φ1
φ1
2π
ei 2 e−i 2 − e
i 2
i 2π
z
i
θ
x
0
= e
λ e
λ
· φ2 ·
φ2
φ2
ei 2 e−i 2 − ei 2
� φ1
�
�
�2
φ1
� e−i 2 − ei φ21 �2
2i
sin(
)
sin2 ( φ21 )
�
�
2
|ET |2 =
� φ2
=
=
�
�
e−i 2 − ei φ22 �
2i sin( φ22 )
sin2 ( φ22 )
5. Forward propagating plane wave:
�k1 =
2π
ẑ
λ
Backward propagating plane wave: �k2 = − 2λπ ẑ
2π
E1 = e
i( λ z−ωt) ,
2π
E2 = ei(− λ z−ωt)
4π
I = 2(1 + cos Δφ), where Δφ = φ1 − φ2 =
z
λ
�
�
��
�
�
4π
2π
∴ I = 2 1 + cos
= 4 cos2
z
z
λ
λ
Note that although we did not ignore the time dependence of each wave (ωt), the
interference wave is independent of time, thus the term “standing wave.”
5
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2.71 / 2.710 Optics
Spring 2009
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