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1.) You have measured the wavelength λ of a spectral lamp using a diffraction grating
and the relation λ = d sin Θ. Your uncertainty in the grating spacing d is 0.5% and your
uncertainty in your angle measurement is 1%. Write an expression for the uncertainty in
λ. If Θ = 30 degrees, what is the uncertainty in λ, and in improving your measurement,
should you concentrate on reducing the grating or angle uncertainty? (30 pts.)
σλ
2
2
 ∂λ  2
2  ∂λ 
= σ d   + σθ   = σ d 2 d 2 sin 2 θ + σθ 2 d 2 cos2 θ
 ∂d 
 ∂θ 
2
 σ λ  2  σ d  2  σθ  2 2
2
2
  =   +   cot θ = (0.005) + 3(0.01)
λ  d θ 
1.8% uncertainty in the wavelength
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Clearly the uncertainty in angle is dominant.
2.) An overhead projector forms an image of a transparency on a screen:
If the screen is moved closer to the projector, how must the lens of the projector be
moved? (up, down, unchanged). (20 pts.) UP
3.) A parallel beam of light is sent through an aquarium. If a convex glass lens is held in
the water, it focuses the beam
1. closer to the lens than
2. at the same position as
3. farther from the lens than
if the lens were outside the water. (20 pts.) FARTHER
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4.) A fish swims below the water at point P. You are looking at the fish at point O.
Drawing some typical rays, indicate on the drawing a point Q where the fish will appear
to you to be located. (35 pts.)
O
Q
P
Image appears above the fish in the water
5.) The human eye is a compound lens system, which we will approximate with the
following schematic:
Cornea
Lens
Retina
The cornea has a 28 mm focal length, the distance between the cornea and the lens is 4
mm, and the distance from the cornea to the retina is 22 mm. If an object is placed at
infinity, what is the necessary focal length of the lens to form an image on the retina. If
the object is placed at 1m, what is the necessary lens focal length? (45 pts.)
Sobj1=infinity; => sim1 = 28 mm. Then sobj2 = -24mm and sim2 = 18 mm so
1/f = 1/18 – 1/24 -> f = 72 mm
sobj1 = 1000mm sima = (1/28-1/1000)-1= 28.8mm then sobj2 = -24.8 mm
1/f = 1/18-1/24.8 -> f = 65.6 mm
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6.) You are hired by a scuba diving company to monitor the oxygen content in scuba
tanks, fixing it at 20% O2 and 80% N2. Describe an optical method to determine the gas
content. Be quantitative – give numbers to show that the technique is viable. Draw
diagrams if needed. The index of refraction of O2 at STP (P=1 atm; T = 20 C) for red
light is 1.00027, and for N2 is 1.0003. (75 pts.)
Hint: since the differences in n are quite small, think interferometers.
Possible solution: put a gas cell in one arm of Michelson interferometer. Measure
number of fringes that go by output as gas is put into cell.
Gas cell
Pick cell length L = 10 cm
change in optical path length due to gas = (%O2 nO + %N2 nN –2) 20 cm
= ((2.7 X 10-4)%O2 + (3.0X 10-4)%N2 )20 cm
No. of fringes = ((2.7 X 10-4)%O2 + (3.0X 10-4)%N2 )20 cm/633 X 10-7 cm
= (86) %O2 + (94) %N2
so a 1% change in mixture will result in a shift of .01 (86) - .01 (94) = .08 fringes.
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7.) Blue light of wavelength λ passes through a single slit of width a forming a diffraction
pattern on a screen. If the blue light is replaced by red light of wavelength 2λ, the
original diffraction pattern is reproduced if the slit width is changed to
1.
2.
3.
4.
5.
6.
(20
a/4.
a/2.
No change is necessary.
2 a.
4 a.
There is no width that can reproduce the original pattern.
pts.)
8.) An optical isolator is used to prevent reflections off surfaces entering back into a
laser. It uses a crystal placed in a magnetic field that exhibits the Faraday effect: the
sign of the optical activity depends on whether the light moves parallel or anti-parallel
to the magnetic field, i.e. linearly polarized light rotates clockwise if the light is moving
along the B field, and rotates counterclockwise if the light is traveling in the opposite
direction. Describe a one-way device (an “optical diode”) using polarizer(s) and a
Faraday crystal that can transmit light in only one direction, suppressing light traveling
in the other direction. (60 pts.)
use a vertical polarizer followed by a Faraday crystal that gives 45 deg clockwise
rotation, followed by a polarizer oriented at 45 degrees clockwise.
Faraday crystal
Vertically polarized light traveling from the left passes the 1st polarizer, gets
rotated by 45 degrees, is now aligned with the 2nd polarizer and passes through.
Any light coming from the right is polarized at 45 deg by the polarizer, and is then
rotated to be horizontal by the Faraday crystal and is blocked by the vertical
polarizer – giving one-way directionality.
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9.) Consider the double-slit diffraction pattern such as you measured in the lab. If over
one of the slits you place a transparent material with index of refraction n and thickness
d such that (n-1)d = mλ/2 where m is an integer, describe the resulting diffraction
pattern, compared to the pattern without the material. (35 pts.)
the material causes a 180 phase shift between the light emitted from the two slits,
so that pattern looks the same, except the bright regions are dark and the dark
regions bright.
10.) A rainbow is formed when sunlight undergoes two refractions and a reflection in a
spherical water droplet as shown below.
Show that the deviation angle
Ω = 180 − 2α + 4 β exhibits a maximum value when
2
sin α =
4−n
.
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(You may need to use
d
1
Sin −1θ =
).
dθ
1+ θ 2
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€
The index of refraction of water for red light is nr=1.33, and for blue light, nb= 1.34.
Find the maximum deviation angles for red and blue light, and describe how a rainbow
is formed as a result. Where must the sun €
be located with respect to the viewer?
Is total internal reflection necessary for this phenomena? (60 pts.)
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From Snell’s law,
respect to
 sin α 
β = sin−1
 . Then to find the maximum, take derivative of Ω with
 n 
α.
dΩ
cos α
= 0€= −2 + 4
dα
n
€
1
2
1−
sin α
n2
=4
1− sin 2 α
2
2
n − sin α
−2
€
4(1− sin 2 α ) = n 2 − sin 2 α
then
€
4 − n 2 = 3sin 2 α
sin α =
4 − n2
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for n= 1.33, sin α = 0.862, α= 59.5 deg; β = 40.4, and 4β−2α = 42.5 degrees
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for n = 1.34 sin α = 0.857, α= 59.0 deg; β = 39.8, and 4β−2α = 41.2 degrees
so red light appears below the blue, when the sun is behind the viewer.
There is actually no total internal reflection (if you take a ray at the edge of the drop, the
curvature at the back of the drop means the angle is still less than the critical angle.) So
the rainbow is visible due to the fraction of light that is reflected from the back of the
drop, and does not need TIR.
blue
red
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PRACTICAL EXAM – be sure to enter lab bench numbers!
Show your work- measurements and calculations.
P1) The laser is being diffracted through some sort of mask. Measuring the diffraction
pattern, determine the mask including the dimensions of the element(s) of the mask.
Assume the wavelength of the laser is λ=633 nm. (50 pts.)
Enter lab bench no. in the box
The mask was a square array of squares with a 0.1 mm spacing :
0.025 mm
0.075 mm
P2) Find the focal length of the lens. (50 pts.)
Enter lab bench no. in the box
The lens had a –4.0 cm focal length.
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