Chapter 10 Optics homework solutions - p1
Prob 10.1
-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3-2
-1.5
-1
-0.5
0
0.5
1
n = 1.00n = 2.00
-1 -0.5 0 0.5 1 1.5 2
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
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1
n = 1.00n = 2.00
1
Chapter 10 Optics homework solutions - p2
-0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4
-0.6
-0.4
-0.2
0
0.2
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n = 1.00
n = 2.00
-0.5 0 0.5 1 1.5
-0.8
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-0.2
0
0.2
0.4
0.6
0.8
n = 1.00n = 1.50
The code on the next page produced the graphs.
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Chapter 10 Optics homework solutions - p3
[1] function c10p1(n_i,n_t,theta_i_d,N)
[2] % makes a diagram showing Huygens method of computing refraction
[3] % n_i = incident index of refraction
[4] % n_t = transmitted index of refraction
[5] % theta_i = incident angle
[6] % N = number of wavefronts to draw.
[7] L = 1;
[8] theta_i = theta_i_d*pi/180;
[9] dr_i = L*tan(theta_i) / N;
[10] dr_t = n_i * dr_i / n_t;
[11] theta_t = asin(sin(theta_i)*n_i/n_t);
[12] C = cos(theta_i);
[13] S = sin(theta_i);
[14] figure(1)
[15] plot([-0.5,1.3]*L/C,[0,0],’g’)
[16] hold on
[17] nN = round(N*n_i/n_t);
[18] Nt = N+15;
[19] for n = -nN:N
[20] x = + n * dr_i * S;
[21] y = - n * dr_i * C;
[22] if n
Chapter 10 Optics homework solutions - p4
Prob 10.3Since E does not depend on the θ or φ we have that
∇2E = 1r
∂2
∂r2(rE)
=1
r
∂2
∂r2
(rE0r0
eikr
r
)= E0r0
1
r
∂2
∂r2(eikr
)= E0r0
1
r(ik)
2eikr
= −k2E0r0eikr
r
= −k2E−→ ∇2E + k2E = 0
4
Chapter 10 Optics homework solutions - p5
Prob 10.4(a) Let Eφ = E0r0
ei(kr−ωt)
r then
~E = 0 r̂ + 0 θ̂ + Eφφ̂
(i) In spherical coordinates ~A = Ar r̂ +Aθ θ̂ +Aφφ̂ the divergence is
∇ · ~A = 1r2
∂
∂r
(r2Ar
)+
1
sin θ
∂
∂θ(sin θAθ) +
1
r sin θ
∂Aφ∂φ
so
∇ · ~E = 1r2
∂
∂r
(r20)
+1
sin θ
∂
∂θ(sin θ0) +
1
r sin θ
∂Eφ∂φ
=1
r sin θ
∂
∂φE0r0
ei(kr−ωt)
r
= 0
So Gauss’ Law is satisfied.
(ii) In the same way we see that the terms in the curl of ~E are all zero except one.
∇× ~E = −θ̂1r
∂
∂r(rEφ)
= −θ̂1r
∂
∂r
(E0r0e
i(kr−ωt))
= −ik 1rE0r0e
i(kr−ωt)θ̂
= −ikEφθ̂
−→ ∂~B
∂t= −∇× ~E = ikEφθ̂ = ikE0r0
ei(kr−ωt)
rθ̂
−→ ~B = ik−iω
E0r0ei(kr−ωt)
rθ̂
−→ ~B = −1cE0r0
ei(kr−ωt)
rθ̂
= −1cEφθ̂
−→ Bθ = −1
cEφ
Thus we have found the magnetic field from the electric field by using Faraday’s law.
(iii) Now we can check the divergence of ~B.
∇ · ~B = 1r2
∂
∂r
(r20)
+1
sin θ
∂
∂θ(sin θBθ) +
1
r sin θ
∂0
∂φ
=1
sin θ
∂
∂θ(sin θBθ)
=1
sin θ
[Bθ
∂
∂θsin θ + sin θ
∂Bθ∂θ
]=
1
sin θ[Bθ cos θ + sin θ · 0]
= −1c
cot θEφ
6= 0
(iv) Since ~B is just in the θ̂ direction and only varies in r we get just one non-zero term inthe curl.
∇× ~B = φ̂1r
∂
∂r(rBθ) = −
1
cφ̂
1
r
∂
∂r(rEφ) = −
1
cφ̂ (ikEφ) = −
ik
c~E = − iω
c2~E
while
µ0�0∂ ~E
∂t= −iωµ0�0 ~E = −
iω
c2~E
Thus
∇× ~B = µ0�0∂ ~E
∂t
and Ampere’s law is satisfied.
(b) We see that we need sin θ � 1 and kr � 1 in order for the simpler field E0r0 ei(kr−ωt)
r φ̂to be a good approximation.
5
Chapter 10 Optics homework solutions - p6
We note that the field can be written in complex form with α = kr − ωt
~E = Re
{A sin θ
r
[cosα− 1
krsinα
]}φ̂
= Re
{A sin θ
r
[cosα+
i
kri sinα
]}φ̂
= Re
{A sin θ
r
[(cosα+ i sinα) +
i
kr(cosα+ i sinα)
]}φ̂
= Re
{A sin θ
r
[1 +
i
kr
]eiα}φ̂
= Re
{A sin θ
r
√1 +
1
k2r2eiβeiα
}φ̂ with tanβ =
1
kr
= Re
{A sin θ
r
√1 +
1
k2r2ei(kr−ωt+β)
}φ̂
=A sin θ
r
√1 +
1
k2r2cos(kr − ωt+ β)φ̂
So we have an oscillation with phase kr+β and amplitude of the oscillation is A sin θr
√1 + 1k2r2 .
0.05
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0.1
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0.20.2
0.2 0.25
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0.7
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0.750.75
0.8
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0.850.85
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0.950.95
1
1
-10 -8 -6 -4 -2 0 2 4 6 8 10-10
-8
-6
-4
-2
0
2
4
6
8
10
[1] function c10p4
[2] figure(1)
[3] x = -10:0.1:10;
[4] y = x’;
[5] r2 = x.^2 + y.^2;
[6] amp = sqrt(1+1./r2).*abs(x).*(1./r2);
[7] contour(x,y,amp,0:0.05:1,’ShowText’,’on’);
[8] texText(’$kx$’,’x’)
[9] texText(’$ky$’,’y’)
[10] texText(’amplitude $\vert E\vert$’,’t’)
[11] graphToPDF(’c10p4.pdf’,6,6)
[12] end
6
Chapter 10 Optics homework solutions - p7
Prob 10.5We have the approximation
Ẽ(x, y, z) =1
iλz
∫ ∫Ẽ(x′, y′, 0)e
ik2zDdx′dy′
with D ≡[(x− x′)2 + (y − y′)2
]. We note here for future reference that
∂D
∂x= 2(x− x′)
∂D
∂y= 2(y − y′)
Now we compute the following derivatives.
∂
∂zẼ(x, y, z) =
∂
∂z
1
iλz
∫ ∫Ẽ(x′, y′, 0)e
ik2zDdx′dy′
= − 1iλz2
∫ ∫Ẽ(x′, y′, 0)e
ik2zDdx′dy′ +
1
iλz
∫ ∫Ẽ(x′, y′, 0)
∂
∂zeik2zD
= −1zẼ(x, y, z) +
1
iλz
∫ ∫Ẽ(x′, y′, 0)
(−ik2z2
)De
ik2zDdx′dy′
= −1zẼ(x, y, z)− k
2λz3
∫ ∫Ẽ(x′, y′, 0)De
ik2zDdx′dy′
2ik∂
∂zẼ(x, y, z) = −2ik
zẼ(x, y, z)− ik
2
λz3
∫ ∫Ẽ(x′, y′, 0)De
ik2zDdx′dy′
and∂
∂xẼ(x, y, z) =
1
iλz
∫ ∫Ẽ(x′, y′, 0)
∂
∂xeik2zD
=1
iλz
∫ ∫Ẽ(x′, y′, 0)
ik
2z
∂D
∂xeik2zDdx′dy′
=1
iλz
∫ ∫Ẽ(x′, y′, 0)
ik
2z2(x− x′)e ik2zDdx′dy′
=ik
z
1
iλz
∫ ∫Ẽ(x′, y′, 0)(x− x′)e ik2zDdx′dy′
so∂2
∂x2Ẽ(x, y, z) =
ik
z
1
iλz
∫ ∫Ẽ(x′, y′, 0)
∂
∂x
[(x− x′)e ik2zD
]dx′dy′
=ik
z
1
iλz
∫ ∫Ẽ(x′, y′, 0)
[1 + (x− x′) ik
2z2(x− x′)
]eik2zDdx′dy′
=ik
zẼ(x, y, z) +
(ik
z
)21
iλz
∫ ∫Ẽ(x′, y′, 0)(x− x′)2e ik2zDdx′dy′
=ik
zẼ(x, y, z) +
ik2
λz3
∫ ∫Ẽ(x′, y′, 0)(x− x′)2e ik2zDdx′dy′
In the same way
∂2
∂y2Ẽ(x, y, z) =
ik
zẼ(x, y, z) +
ik2
λz3
∫ ∫Ẽ(x′, y′, 0)(y − y′)2e ik2zDdx′dy′
Adding these last two results we see that[∂2
∂x2+
∂2
∂y2
]Ẽ(x, y, z) =
2ik
zẼ(x, y, z) +
ik2
λz3
∫ ∫Ẽ(x′, y′, 0)De
ik2zDdx′dy′
But then we see that this is exactly −2ik ∂∂z Ẽ(x, y, z). Thus we find that[∂2
∂x2+
∂2
∂y2+ 2ik
∂
∂z
]Ẽ(x, y, z) = 0
7
Chapter 10 Optics homework solutions - p8
Prob 10.6First we compute the following integral∫ a
−ae−ibxdx =
[e−ibx
−ib
]a−a
=e−iba − eiba
−ib
=eiba − e−iba
ib
=2i sin(ba)
ib
= 2asin(ba)
ba
= 2a sinc(ba)
Now we compute
E(x, y, z) = E0eikz
iλzeik2z (x
2+y2)
∫ ∆x/2−∆x/2
dx′e−ikxz x′∫ ∆y/2−∆y/2
dy′e−ikyz y′
= E0eikz
iλzeik2z (x
2+y2)2∆x
2sinc
(kx
z
∆y
2
)2
∆y
2sinc
(ky
z
∆y
2
)= E0
eikz
iλzeik2z (x
2+y2)∆x∆y sinc
(π∆x
λzx
)sinc
(π∆y
λzy
)Thus
I = 12�0c|E|2 = I0
(∆x∆y
λz
)2sinc2
(π∆x
λzx
)sinc2
(π∆y
λzy
)with I0 =
12�0c|E0|
2
8
Chapter 10 Optics homework solutions - p9
Prob 10.7For definitness we will assume that in the y direction we have an aperature of length a.
E(x, y, z) = E0eikz
iλzeik2z (x
2+y2)
∫ L/2−L/2
dx′ cosπx′
Le−i
kxz x′∫ a/2−a/2
dy′e−ikyz y′
= E0eikz
iλzeik2z (x
2+y2)
∫ L/2−L/2
dx′eiπLx′+ e−i
πLx′
2e−i
kxz x′a sinc
(πaλzy)
= E0eikz
iλzeik2z (x
2+y2)a sinc(πaλzy) 1
2
[∫ L/2−L/2
dx′e−i(kxz −
πL )x
′+
∫ L/2−L/2
dx′e−i(πL+
kxz )x
′
]
= E0eikz
iλzeik2z (x
2+y2)a sinc(πaλzy) 1
2
[L sinc
(L2 (
kxz −
πL ))
+ L sinc(L2 (
kxz +
πL ))]
= E0eikz
iλzeik2z (x
2+y2)La sinc(πaλzy) 1
2[sinc (πLλz x−
π2 ) + sinc (
πLλz x+
π2 )]
Let ξ = πLλz x then we have the x dependent part is
1
2[sinc(ξ − π/2) + sinc(ξ + π/2)]
while for the uniform slit we had just sinc(ξ). Let us graph four things E+ =12 sinc(ξ + π/2)
and E− =12 sinc(ξ − π/2) and E = E− + E+ and Eu = sinc(ξ).
-4 -2 0 2 4 6 8-0.4
-0.2
0
0.2
0.4
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0.8
1
We see that the sidelobes are mostly removed because the phase of the E+ sidelobes is oppositeof the phase of the E− side lobes, that is when one is up the other is down. This becomes evenmore apparent when we look that intensity.
-4 -2 0 2 4 6 80
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9
Chapter 10 Optics homework solutions - p10
Prob 10.8Passing to cylindrical coordinates
E(x, y, z) =1
iλ
∫aperture
E(x′, y′, 0)eikR
Rdx′dy′
=1
iλ
∫aperture
E(r, φ, 0)eikR
Rρ dφ dρ
=E0iλ
∫ 2π0
dφ
∫ D/20
ρ dρeikR
R
Starting with the Fresnel approximation
E(x, y, z) =E0iλ
∫ 2π0
dφ
∫ D/20
ρ dρeikR
R
Eb(x, y, z) =E0e
ikzeik2z (x
2+y2)
iλz
∫ 2π0
dφ
∫ D/20
ρ dρ eik2z ρ
2
e−ikz (xρ cosφ+yρ sinφ)
so
Eb(0, 0, z) =E0e
ikz
iλz
∫ 2π0
dφ
∫ D/20
ρ dρ eik2z ρ
2
=E0e
ikz
iλz2π
∫ D/20
ρ dρ eik2z ρ
2
=E0e
ikz
iλzπ
∫ D2/40
du eik2zu
=E0e
ikz
iλzπ
2z
ik
[eik2z
D2
4 − 1]
= −E0eikz[eikD2
8z − 1]
= −E0eikzeikD2
16z
[eikD2
16z − e−i kD2
16z
]= −2iE0eik(1+(
D4z )
2)z sin
(kD2
16z
)−→ Ib(0, 0, z) = I04 sin2
(k2D2
16kz
)In the Fraunhofer approximation we also toss the term e
ik2z ρ
2
, so that
Ec(0, 0, z) =E0e
ikz
iλz
∫ 2π0
dφ
∫ D/20
ρ dρ
=E0e
ikz
iλz2π
1
2
(D
2
)2= E0
eikz
i
k2D2
8kz
−→ Ic(0, 0, z) = I0(k2D2
8kz
)2This can be compared with the intensity gotten from the Huygens-Fresnel equation.
Ia = I02
[1− cos
(√14 (kD)
2 + (kz)2 − kz)]
= I04 sin2
(12
√14 (kD)
2 + (kz)2 − 12kz)
101 102 1030
1
2
3
4
5
10
Chapter 10 Optics homework solutions - p11
Prob 10.9
11
Chapter 10 Optics homework solutions - p12
Prob 10.e1First we recall some results ∫ b/2
−b/2e−iβxdx = b sinc(βb/2)
∫ a/2−a/2
dx cos(αx)e−iβx =
∫ a/2−a/2
dxeiαx + e−iαx
2e−iβx
=
∫ a/2−a/2
dxe−i(β−α)x + e−i(β+α)x
2
=a sinc
((β − α)a2
)+ a sinc
((β + α)a2
)2
=a
2
[sinc
((β − α)a
2
)+ sinc
((β + α)
a
2
)]We assume a
E(x, y, 0) = E0 cos(πxa
)[1− cos
(4nπy
b
)]over the aperture −a2 < x <
a2 and −
b2 < y <
b2 . In the Fraunhofer approximation we have
E(x, y, z) =E0e
ikzeik2z (x
2+y2)
iλz
∫ a/2−a/2
dx′ cos
(πx′
a
)e−i
kxz x′
×∫ b/2−b/2
dy′[1− cos
(4nπy′
b
)]e−i
kyz y′
=E0e
ikzeik2z (x
2+y2)
iλz
a
2
[sinc
((kx
z− πa
)a
2
)+ sinc
((kx
z+π
a)a
2
)]×∫ b/2−b/2
dy′[1− cos
(4nπy′
b
)]e−
ikyz y′
=E0e
ikzeik2z (x
2+y2)
iλz
a
2
[sinc
( a2zkx− π
2
)+ sinc
( a2zkx+
π
2
)]×
[∫ b/2−b/2
dy′e−ikyz y′−∫ b/2−b/2
dy′ cos
(4nπy′
b
)e−
ikyz y′
]
=E0e
ikzeik2z (x
2+y2)
iλz
a
2
[sinc
( a2zkx− π
2
)+ sinc
( a2zkx+
π
2
)]×[b sinc
(ky
z
b
2
)− b
2
[sinc
((ky
z− 4nπ
b)b
2
)+ sinc
((ky
z+
4nπ
b)b
2
)]]=E0e
ikzeik2z (x
2+y2)
iλz
a
2
[sinc
( a2zkx− π
2
)+ sinc
( a2zkx+
π
2
)]b
2×[2 sinc
(b
2zky
)− sinc
(b
2zky − 2nπ
)− sinc
(b
2zky + 2nπ
)]
12
