Name:__KEY____________________
Phys.116 Exam III
8 April 2015
Please do not turn the page until you are told to do so. Make sure that you have all three problems on your
copy of the test. In order to get credit on a problem, you must show your work. If you only write down an answer
without the work leading up to it, you will get no credit for it, even if it is the right answer.
The use of the intellectual property of others without attributing it to them is considered a serious academic
offense. Cheating or plagiarism will result in receiving a failing grade for the assignment or the course and may
result in student conduct charges. Repeat offenses will result in dismissal from the University.
1) A pair of glasses is designed for a person with a farpoint distance of 3.0 m so that she can read street signs 20
m away.
a) (8 points) If the glasses are to be worn 1.0 cm from her eyes, what is the needed focal length?
The object distance and image distance with respect to the glasses are:
19.99m
d o =20m − 0.01m =
di =
−(3.0m − 0.01m) =
−2.99m
d i is negative, because the image through the glasses forms on the same side as the object.
Using the lens equation, we can find the focal length of the glasses.
1
1 1
1
1
=+ =
+
=
−0.2844m-1
f d o d i 19.99m −2.99m
f = −3.52m
b) (3 points) Do the glasses have converging or diverging lenses? Why?
The glasses have diverging lenses. Diverging lenses correct for nearsightedness. Furthermore, the
focal length obtained is negative, which is consistent with diverging lenses.
c) (6 points) Compare this focal length to the focal length she would need if she chooses a style of glasses that fit
on her face so that the lenses are instead 2.0 cm away from her eyes.
The object distance and image distance with respect to the glasses will now be 19.98m and -2.98m.
1 1 1
1
1
=+ =
+
=
−0.2855m-1
f d o d i 19.98m −2.98m
f = −3.50m
The focal lengths in both situations are similar.
d) (3 points) For a nearsighted person, is the position of the glasses from the eyes as important as for a farsighted
person? Explain your reasoning.
No, most likely not. A nearsighted person wears glasses in order to see distant objects clearly,
whereas a farsighted person needs glasses to see close objects clearly. The object distance and
image distance for a farsighted person are in the centimeters range, so the same order of magnitude
as the distance from the eye to the glasses.
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2) Light with λ = 550 nm passes through a single slit and then illuminates a screen 2.0 m away.
a) (6 points) If the distance on the screen from the first dark fringe to the center of the diffraction pattern is 5.5
mm, what is the width of the slit?
w sin θ = m λ
y
Using the small angle approximation, sin θ ≈ tan θ ≈ θ , and the fact that tan θ =
(see drawing below)
D
y
mλ D 1(550 × 10−9 m)(2.0m)
So, w = =
; w=
2 × 10−4 m
mλ
⇒w =
−3
5.5 × 10 m
D
y
b) (4 points) Clearly draw the diagram, showing all the relevant information on it.
m=2
θ
w
m=1
y
m=-1
D
m=-2
c) (6 points) Now the single slit is replaced with a double slit with slit separation of 3.5 mm. If the same light
passes through the slits and forms an image on the screen, will the screen be bright or dark at the location of the
first dark fringe in part a)?
y
λ; d
=
d sin θ m=
mλ
D
−3
−3
dy ( 3.5 × 10 m )( 5.5 × 10 m )
m= =
=17.5 ⇒
m =17.5 dark fringe
Dλ
( 2.0m ) ( 550 × 10−9 m )
Since for double slit interference integer values for m give bright fringes, then half integer numbers
locate dark fringes.
d) (4 points) Clearly draw the diagram, showing all the relevant information on it.
3a) (15 points) The glass covering a piece of art is coated with a thin film of TiO 2 in order to minimize the glare.
If the glass has a refractive index of 1.62 and the TiO 2 has an index of refraction of 2.62, what are the 3 minimum
film thicknesses that will cancel light of wavelength 505 nm? (for 6 of the 15 points, draw a clear diagram,
showing all the relevant information on it. You will not get full credit if your diagram is incomplete, even if your
calculations are correct!)
λ/2
destructive interference
due to reflection
n1=1
0
n2=2.62
n3=1.62
Since the problem is asking to minimize the glare, and the glare IS minimized due to reflection, the thickness of
the thin film will be given by 2 L = m
λ
n2
.
λ
505nm
=
L m= m = 96.37m nm.
2n2
2(2.62)
The three lowest thicknesses of the film will be given by=
m 1,=
m 2,and=
m 3.
m =1 → L =96.37 nm
m = 2 → L = 192.7 nm
m = 3 → L = 289.1nm
3b) (5 points) Please draw a ray diagram, using the three main rays discussed in class. Make sure you use
solid lines for the actual rays and dashed lines for their extensions, and that you mark the direction of travel of
the rays with an arrow.
F
F
v=λf
E
c=
B
1
1 2
B
ε0E2 +
2
2 µ0
𝑆𝑆 = 𝑆𝑆0 cos 2 𝜃𝜃
𝑐𝑐
𝑛𝑛 =
𝑣𝑣
𝑛𝑛1 sin 𝜃𝜃1 = 𝑛𝑛2 sin 𝜃𝜃2
1
1
1
+
=
𝑓𝑓
𝑑𝑑0 𝑑𝑑𝑖𝑖
𝑟𝑟
𝑓𝑓 =
2
𝑑𝑑𝑖𝑖
𝑚𝑚 = −
𝑑𝑑0
ℎ𝑖𝑖
|𝑚𝑚| =
ℎ0
1
P=
f
𝑀𝑀 = 𝑚𝑚1 𝑚𝑚2
𝑛𝑛2
𝑑𝑑′ = 𝑑𝑑 � �
𝑛𝑛1
𝑛𝑛2
tan 𝜃𝜃𝐵𝐵 =
𝑛𝑛1
𝜆𝜆
𝜆𝜆𝑛𝑛 =
𝑛𝑛
𝑑𝑑 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
1
𝑑𝑑 sin 𝜃𝜃 = �𝑚𝑚 + � 𝜆𝜆
2
1 𝜆𝜆
2𝐿𝐿 = �𝑚𝑚 + �
2 𝑛𝑛2
𝜆𝜆
2𝐿𝐿 = 𝑚𝑚
𝑛𝑛2
𝑊𝑊 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
𝜆𝜆
sin 𝜃𝜃 = 1.22
𝐷𝐷
𝜆𝜆
𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 ≈ 1.22
𝐷𝐷
2𝑑𝑑 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
…constants…
u
=
𝑐𝑐 = 3 𝑥𝑥 108
𝑚𝑚
𝑠𝑠