CHM 424
EXAM 2 CRIB - COVER PAGE
FALL 2006
There are seven numbered pages with five questions. Answer the questions on the exam. Exams
done in ink are eligible for regrade, those done in pencil will not be regraded.
1 coulomb = 6.24×1018 charges
1 volt = 1 joule per 1 coulomb
h = 6.6×10-34 J s
c = 3×108 m s-1
1 in = 2.54 cm
n=
c
v
1 amp = 1 coulomb per 1 second
ε0 = 8.85×10-12 C2 J-1 m-1
k = 1.38×10-23 J K-1
1 amu = 1.66×10-27 kg
E = hν
⎛ 1
1
1 ⎞
= ( nl − 1) ⎜
−
⎟
f
⎝ Rin Rout ⎠
2
n1 − n2 )
(
R=
( n1 + n2 )2
1
1
1
=
+
f dob dim
( n1 − n2 ) ( n1 − n2* ) ( n1 − η2 )2 + η22σ22
R=
=
*
( n1 + n2 ) ( n1 + n2 ) ( n1 + η2 )2 + η22σ22
mλ = 2nt cos θ
sin θm =
hc 1 2
= mv + ω
λ 2
RTE =
RTM =
mλ
a
A = − log (T )
sin (θ i − θ t )
2
sin (θ i + θ t )
2
tan (θ i − θ t )
2
tan (θ i + θ t )
2
n
tan θ B = 2
n1
sin θc =
n1
n2
M=
dim
d ob
n1 sin θ1 = n2 sin θ2
dθ
m
=
d λ a cos θm
ρ=
Δθ =
dl
dθ
= f
dλ
dλ
0.61λf
r
2λ
Na cosθ m
R=
λ
∝ mN
Δλ
CHM 424
EXAM 2 CRIB
Name_________________________________
FALL 2006
Score_________/150
30 pts. 1. Provide the following definitions or facts at 6 pts each.
a. Label six of the regions of the electromagnetic spectrum
b. List the types of transitions that occur in each of the regions you labeled in (a)
(i) gamma: nuclear, used to identify isotopes
(ii) x-rays: inner shell electronic, used to identify atoms in solids and provide information
about adjacent bonded atoms
(iii) ultraviolet: valence shell electronic, used to identify atoms in a sample and their
concentration, molecular electronic transitions involving single bonds, used to determine
structure and concentration
(iv) visible: valence shell electronic, used to identify atoms in a sample and
concentration, molecular electronic transitions involving conjugated double bonds, used to
determine structure and concentration
(v) infrared: vibrations, used to determine functional groups existing within a molecule
(vi) microwave: rotations, used to determine bond angles and lengths of small molecules
in the gas phase
radio: magnetic resonances, used to determine bonding in molecules, electron spin
resonance, used to determine location of unpaired electrons
Exam 2 Crib, 2006, page 1
c. Light impinges on an interference filter perpendicular to its surface. As the angle of
incidence increases, what happens to the wavelength of transmitted light?
The interference filter equation is given by mλ = 2nt cosθ, where θ is the angle of incidence.
With perpendicular incidence cos0E = 1. For any greater angle of incidence the value of the
cosine will decrease. Since everything else stays constant in the equation, the wavelength of
light will decrease.
d. Consider the situation where a lens transfers the source image onto the entrance slit of a
monochromator. What will happen if the lens has an f-number greater than the monochromator
value? Less than the monochromator?
When the lens f-number is greater than the monochromator value the grating is under filled and
the monochromator resolution suffers. When the f-number is smaller the grating is overfilled
and the monochromator is filled with scattered light.
e. Name one aberration that exists in a lens that does not exist with a mirror. Name one
aberration shared by both lenses and mirrors.
Chromatic aberrations exist in a lens because the refractive index varies with wavelength. The
change in refractive index changes the angle of refraction and moves the focal point. Mirrors do
not have chromatic aberrations. Although reflectivity varies with wavelength, this change does
not affect the angle of reflection.
Spherical aberrations and off-axis coma are shared by lenses and mirrors.
Exam 2 Crib, 2006, page 2
30 pts. 2. Perform the following calculations at 10 pts each.
a. A signal with an unknown frequency is digitized at 100 Hz, with the result aliased to 10 Hz.
When the same signal is digitized at 102 Hz, the alias is 8 Hz. What is the actual frequency?
0 Hz
90 Hz
100 Hz
110 Hz
10 Hz
200 Hz
0 Hz
50 Hz 94 Hz
102 Hz
150 Hz 110 Hz
190 Hz
204 Hz
250 Hz
8 Hz
51 Hz
153 Hz
196 Hz
255 Hz
The only frequency that is the same in both diagrams is 110 Hz.
b. If an object is 2 cm from the lens and the image is 4 cm from the lens, what is the lens focal
length.
1
1
1
=
+
f dob dim
1 1 1 3
= + =
f 2 4 4
f = 1.33
c. 600 nm light has a first order diffraction from a grating of 10E. How many lines per
millimeter characterize the grating?
sin θm =
mλ
a
mλ
1× 6.00 × 10−4 mm
a=
=
= 3.46 × 10−3 mm
sin θm
sin10°
1
= 289 mm −1
a
Exam 2 Crib, 2006, page 3
30 pts. 3. Answer the following two homework questions for 15 pts each.
a. The silver iodide bond energy is approximately 255kJ/mol (AgI is one of the possible active
components in photgray sunglasses). What is the longest wavelength of light that is capable of
breaking the bond in silver iodide?
255 kJ mol-1
6.022 × 10
23
mol
-1
= 4.23 ×10−19 J
E 4.23 × 10−19
ν= =
= 6.38 ×1014 Hz
34
−
h 6.63 ×10
λ=
3 ×108
6.38 × 1014
= 470 nm
b. Calculate the reflective loss when a beam of light passes through an empty quartz cell
assuming the refractive index of quartz is 1.55.
There are four reflections each having a reflection R. The total transmission is given by
(1 - R)4 which is then converted back to reflection.
2
1.55 − 1.00 )
(
R=
(1.55 + 1.00 )2
4
=
0.552
2.552
=
0.3025
= 0.0465
6.5025
4
T = (1 − R ) = (1 − 0.0465 ) = 0.8265
R = 1 − T = 0.1735
Exam 2 Crib, 2006, page 4
30 pts. 4. The drawing at the right (not to scale) loosely depicts a microscope with an
Amici objective and a CCD camera. This particular Amici objective is a
1-cm diameter hemisphere of glass with a refractive index of 1.52. For
the following questions use thin-lens equations, even though they are not
valid.
(a) What is the focal length (Hint: you need to ask what radius
would yield a flat lens surface?) and
(b) f/# of the objective?
(c) What is the focal spot radius at λ = 500 nm (0.5 μm)? Give the
answer in micrometers.
The CCD camera has pixels that are 9×9 μm. The biconvex lens
has to be chosen so that a circular focal spot just fits onto a single pixel.
(d) What is the required focal length of this lens?
(e) Why can't a full sphere be used as the objective to get an even
smaller spot size?
CCD camera
objective
sample surface
⎛ 1
1
1 ⎞
1 ⎞
⎛1
= ( nl − 1) ⎜
−
⎟ = 0.52 ⎜ −
⎟
(a) f
⎝ ∞ −0.5 ⎠
⎝ Rin Rout ⎠
f = 0.5 0.52 = 0.96 cm
(b) f /# =
(c) ρ =
f 0.96
=
= 0.96
d
1
0.61λ f 0.61× 0.5μm × 0.96 cm
=
= 0.58μm
r
0.5cm
(d) In order to expand a 1.16 μm diameter spot into a 9×9 μm square pixel, the image has to be
magnified by a factor of ~7.8. Thus, the focal length of the biconvex lens has to be 7.8×0.96 cm
= 7.5 cm.
(e) A full sphere would have the following focal length.
1
1 ⎞ 1.04
⎛ 1
= 0.52 ⎜
+
⎟=
f
⎝ 0.5 0.5 ⎠ 0.5
f = 0.5 1.04 = 0.48cm
This focal length would be inside the lens!
Exam 2 Crib, 2006, page 5
30 pts. 5. Flash analog-to-digital converters are used to digitize very fast signals.
They are expensive because each voltage level has a comparator.
The signal reaches all comparators at the same time, thus the time
required for conversion is much less than the successive
approximation device in the class notes. The device shown in the
figure is a 3-bit converter with eight levels of voltage resolution (0
through 7). Assume the comparators are wired so that Vin $ Vref
outputs +5 V (binary 1) and Vin < Vref outputs 0 V (binary 0).
a) Use Ohm's law to determine the reference voltage for
each comparator.
The total resistance is 7 kΩ. The reference voltage for C7 is then
(6.5/7)*7 = 6.5 V, for C6 is (5.5/7)*7 = 5.5, C5 = 4.5, C4 = 3.5, C3
= 2.5, C2 = 1.5, and C1 = 0.5.
b) For each input voltage in the following table, determine
the output (0 or 1) of each comparator. Then use the table to state
clearly how the value of the voltage is determined from the
collection of comparator outputs.
Part (c) is on the next page.
Vin
Comparator Output (0 or 1)
C1
C2
C3
C4
C5
C6
C7
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
2
1
1
0
0
0
0
0
3
1
1
1
0
0
0
0
4
1
1
1
1
0
0
0
5
1
1
1
1
1
0
0
6
1
1
1
1
1
1
0
7
1
1
1
1
1
1
1
The highest comparator having an output of 1 can be used to determine the input voltage. In this
case the comparator number gives directly the voltage, e.g. C1 = C2 = C3 = 1 is for Vin = 3.
Exam 2 Crib, 2006, page 6
+3.0 V
c) The digital logic that converts the comparator
outputs from the table into a three digit binary number (000 to
111) is called an 8-line to 3-line priority encoder. One
0.5 k
specific device that does this is a National Semiconductor
74F148 integrated circuit. Its digital logic can be solved but
not on a test!
If there are only three comparators being converted
1.0 k
(as shown in the figure) the logic is much simpler because
only a two-bit binary number results.
Vin
binary out
0
00
1
01
2
10
3
11
+
C3
+
C2
+
C1
!
!
1.0 k
0.5 k
Each bit in the binary output has its own logic that depends
upon the comparator outputs (for a two-bit number, the right
bit is called bit 0 and the left bit is called bit 1). Use the
inputs from comparators C1 through C3 to create two
separate boolean logic expressions that yield the correct
outputs. The only operators you will need are NOT, AND and
OR. Make liberal use of parentheses to clarify the digital logic.
!
0
Vin
Bit 0 will be 1 when the highest numbered comparator outputting a value of 1 is odd numbered,
e.g. C3 or C1.
bit0 = C1 OR C3
But wait this can't be the answer, since when C2 = 1, C1 also equals 1. This exception has to be
handled by the logic expression.
bit0 = (C1 AND (NOT C2)) OR C3
Bit 1 will be 1 when the highest numbered comparator outputting a value of 1 is C3 or C2.
bit1 = C2 OR C3
Exam 2 Crib, 2006, page 7