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A few exercise solutions from the Lecture Notes,
AST2210
Chapter 2
Exercise 1
We are to find the zenith distance z and the azimuth Az given the observer’s latitude
φ, and the the declination δ as well as the hour angle H of a heavenly body.
Using the standard triangle on a sphere, with sides a, b, c and angles A, B, C, inspecting which sides correspond to our problem, then using the first of the spherical
trigonometry formulae we find
cos(a)
=
cos(b) cos(c) + sin(b) sin(c) cos(A)
cos(z)
=
cos(90 − φ) cos(90 − δ) + sin(90 − φ) sin(90 − δ) cos(H)
=
sin(φ) cos(δ) + cos(φ) cos(δ) cos(H)
Which gives the zenith distance in terms of the given known values.
We now wish to find the Azimuth. Again using the standard triangle on a sphere, but
rotated clockwise relative to the one used in the first part, we use the second of the
spherical trigonometry formulae, and find
sin(a) sin(B)
=
sin(b) sin(A)
sin(90 − δ) sin(H)
=
sin(z) sin(Az)
cos(δ) sin(H)
=
⇒ sin(Az)
=
sin(z) sin(Az)
cos(δ) sin(H)
sin(z)
Where the azimuth is now expressed in terms of known values and the zenith distance
found earlier.
Exercise 2
This time we are given the observer’s latitude φ, the star’s zenith distance z and
azimuth Az and wish to find the star’s declination δ and hour angle H. We again set
up the standard triangle on a sphere, and note which values are known and unknown.
Again using the first spherical trigonometry formula, we find
cos(90 − δ)
=
cos(z) cos(90 − φ) + sin(z) sin(90 − φ) cos(Az)
sin(δ)
=
cos(z) sin(φ) + sin(z) cos(φ) cos(Az)
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Which yields the declination from known values. We now wish to find the hour angle,
H. Sticking with the same triangle as previously, we find using the third spherical
trigonometric formula that the hour angle is given by
cos(δ) sin(H)
sin(H)
=
sin(z) sin(Az)
sin(z) sin(A)
=
cos(δ)
Chapter 3
Exercise 3
We have a surface channel CCD. See Figures 3.8 and 3.9 in the Lecture Notes for
setup. NOTE: There is an error In Equation 3.7 in the Lecture Notes - the derivative
should be a double derivative, as per the Poisson Equation (this is also the case in
some other equations where the Poisson Equation is invoked in the Lecture Notes).
First some useful relations and boundary conditions. For plane parallel capacitors we
have the capacitance:
C=
m
s
For permeability m of the material m and separation of the plates s.
Thus we have the capacitances for the SiO2 (ox) layer and the silica layer (si) simply
as
Cox =
ox
d
,
Csi =
si
xd
We will also need the Poisson equation on the form
d2 V
ρ
=−
dx2
For permeability m of the material m and charge density ρ in the material.
Gauss’s theorem on differential form will also be useful
~ = ρf ree
∇·E
m
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The boundary condition at x = 0 is given by
i Es (x = 0)
= ox Eox (x = 0).
The boundary condition at x = xd is given by
V (x = xd )
=
0,
d2 V (x = xd )
dx2
=
0
We now wish to find the total voltage drop and the thickness of the depletion region
xd . We begin by integrating the Poisson equation on both sides:
Z
xd
Z
xd
Zxxd Zxxd
x
Z
2
d V
=
d2 V
=
x
xd
Z
xd
ρ 2
dx
x
xd
eNA 2
dx
si
x
−
Z xdx Z
x
Left-hand side:
Z xd
(dV (xd ) − dV (x))
= − (V (xd ) − V (x)) = V (x)
x
Where we used that dV (xd ) = 0 and also V (xd ) = 0.
Right-hand side:
Z
xd
x
Z
xd
x
eNA 2
dx
si
Z
xd
=
x
=
=
=
eNA
(xd − x)dx
si
eNA
1 eNA
xd (xd − x) −
si
2 si
eNA
1
x2d − xxd − x2d +
si
2
x2d − x2
1 2
x
2
eNA 2
xd − 2xxd + x2
2si
Completing the square above, and equating LH and RH again we find that the Voltage
given the distance x is
⇒ V (x) =
eNA
2
(x − xd )
2si
for
[0 ≤ x ≤ xd ] .
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We now wish to find the total voltage drop Vg over the depletion region (xd ), using
that the total voltage over the entire region must add up, we write
Vg = Vox + Vs
Where Vox and Vs are the voltages at the poly-layer/ interface and the voltage at x = 0
or Si O2 /p-layer interface respectively. This means we first need to find Vox and Vs . Vox
is given by the relation between electric field and voltage by
dV
dx
= −E(x)
Z 0
= −
Eox dx
⇒ Vox
−d
Here, due to Gauss’ law we can infer that Eox is constant, as we have that there are
no charge carriers in this region (oxide layer). We can therefore write
Vox
= Eox d
We do not have an expression for Eox , but using our condition i Es (x = 0) =
ox Eox (x = 0) the electric field in the oxide layer can be related to Es instead (this is
Gauss’ law for linear materials at the interface). Therefore,
Vox
=
i
Es d
ox
We still don’t have an expression for Es either, but this we can find from our expression of V (x) for [0 ≤ x ≤ xd ], as Vs = V (x = 0) and the general relation E(x) =
− dV
dx . Then,
Thus, Vox =
E(x)
=
⇒ Es
=
eNA
si xd d.
eNA
(x − xd )
for
si
eNA
xd
E(x = 0) =
si
Lastly, Vs = V (x = 0) =
Vg
[0 ≤ x ≤ xd ] .
eNA 2
2si xd ,
so we finally find
=
Vox + Vs
=
eNA
eNA 2
xd d +
x ,
si
2si d
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which is our first solution.
We now wish to find xd . This is now simply a problem of algebra, as we simply need
to reformulate the expression above and solve for xd . We first write
eNA
2si
x2d +
eNA
d xd − Vg
ox
=
0
This is a 2nd degree polynomial in xd , and is readily solved using the standard formula
(make sure of this!).
We then find that xd can be expressed such that
xd
si
= −
±
Cox
s
si
Cox
2
+
2si
Vg .
eNA
Here the capacitance Cox = ox /d.
Chapter 6
Exercise 1
Fermat’s Principle states that:
Light travels through the path in which it can reach the destination in
the least time.
A more accurate formulation would be that the path of light or light-ray between
points will always be an extremum in total travel time τ or optical path.
Armed with this information, we now wish to rederive Snell’s law.
The geometry to illustrate Snell’s law is given in Figure 0.1. Two materials with refraction indexes n1 and n2 are separated by the central vertical line. A light ray begins
at point P, is refracted at the interface, and continues to point Q. The travel length of
the ray in the two respective materials are d1 and d2 respectively. The smaller angle
of the normal vector (or simply the horizontal) with the interface border to the two
ray sections are θ1 and θ2 The length along the horizontal for the two rays are a and
b whereas the vertical distance between P and the normal at the refraction point is x
and the distance between P and Q in the vertical is l.
We now begin going about re deriving Snell’s law. The velocities of the two light rays
in the two materials is given by
v1 =
c
n1
and v2 =
c
n2
where n1 , n2 ≥ 1.
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Figur 0.1: A simple sketch of the problem’s geometry.
Using this, we can find the travel time for the travel from P to Q (with the refraction
in between). The total travel time is simply given by
τ
d1
d2
+ .
v1
v2
=
Further, Pythagoras gives us that the lengths are given by,
d1 =
p
p
x2 + a2 and d2 = (l − x)2 + b2 .
Which means,
√
τ=
x2 + a2
+
v1
p
(l − x)2 + b2
v2
According to Fermat’s theorem, light will minimise ( /or more generally travel along a
solution that is an extremum of) the total travel time. This is mathematically expressed
as the total derivative of the time with respect to the distance and setting it equal to
zero,
dτ
=
dx
√
(
x
a2 +x2 )v1
− “√
(l−x) ”
b2 +(l−x)2 v2
=0
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Now using the definition of the sine, namely that
sin(θ) = opposing side/hypotenuse,
and inspecting the figure (make sure you see that this is correct), one can write out
sin θ1 = √
(l − x)
x
and sin θ2 = p
2
a2 + x2
b + (l − x)2
Which implies
sin θ1
sin θ2
−
v1
v2
=
0
Reordering and inserting for the velocities yields,
n1 sin θ1
= n2 sin θ2 ,
which is Snell’s law. We also want to rederive the law of reflections - this is easily
found as it can be viewed as a special case of Snell’s law. In this case both materials
are the same, and the refraction is a reflection instead. This implies n1 = n2 , and we
find that,
sin θ1
=
sin θ2
⇒ θ1
= θ2 ,
yielding the well-known result that the incoming angle and outgoing angle with respect to the normal of the reflecting surface of a reflected ray are equal.