Version 039 – Test #4 – Antoniewicz – (57030)
This print-out should have 20 questions.
Multiple-choice questions may continue on
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before answering.
1
Far vision
Near vision
001 10.0 points
It is found that a particular metal sample
only ejects electrons when illuminated by light
of wavelength λ ≤ λ0 .
What is the work function W of the sample
in units of eV if λ0 = 535 nm?
Given that:
h = 4.136 × 10−15 eV.s
c = 3 × 108 m/s.
1. 2.31925
2. 2.91953
3. 4.0682
4. 2.69739
5. 2.585
6. 2.82
7. 3.44667
8. 3.49521
9. 4.00258
10. 2.068
Correct answer: 2.31925 eV.
Explanation:
Apply the Energy Principle. The maximum wavelength required to eject the electron means that the photon has the minimum
energy possible, which is equal to the energy
needed to break the electron free from the
metal. This is the work function of the metal.
So,
Ephoton = W
W =
W =
hc
λ0
(4.136 × 10−15 eV.s)(3 × 108 m/s)
(535 nm)
W = 2.31925 eV
002 (part 1 of 2) 5.0 points
A person is to be fitted with bifocals. She can
see clearly when the object is between 43 cm
and 1.7 m from the eye. The upper portions
of the bifocals should be designed to enable
her to see distant objects clearly.
What power should they have?
1. -0.384615
2. -0.5
3. -1.0
4. -0.588235
5. -0.416667
6. -0.454545
7. -0.555556
8. -0.47619
9. -0.625
10. -1.25
Correct answer: −0.588235 diopters.
Explanation:
Given : p = ∞ and
q = −1.7 m .
To correct the nearsightedness, the image of
distance objects (p = ∞) should be virtual
and located at the far point (q = −1.7 m).
Thus from the lens equation and from the
power definition
P =
1
1 1
1
= + =0+
f
p q
−1.7 m
= −0.588235 diopters .
Alternative solution:
Without glass for a distant object at p =
1.7 m, it gives
1
1
1
+ =
,
1.7 m q
feye
where feye is the focal length of the eye lens at
the relaxed position and q is the corresponding image position at the retina.
With eye glass correction, it gives
1
1
1
1
+ =
+
.
∞ q
feye fglass
Therefore,
fglass = −1.7 m .
Version 039 – Test #4 – Antoniewicz – (57030)
P =
1
fglass
= −0.588235 diopters .
003 (part 2 of 2) 5.0 points
The lower portions of the bifocals should enable her to see objects comfortably at 25 cm.
What power should they have?
1. 1.2973
2. 1.91667
3. 1.95918
4. 1.72727
5. 1.82609
6. 1.36842
7. 1.87234
8. 0.666667
9. 1.67442
10. 0.875
Correct answer: 1.67442 diopters.
Explanation:
Given : p = 25 cm = 0.25 m and
q = −43 cm = −0.43 m .
To correct farsightedness, objects at p =
25 cm should form a virtual image at the near
point (q = −0.43 m).
1
1 1
= +
f
p q
1
1
=
+
0.25 m −0.43 m
2
004 10.0 points
In the figure below an unpolarized sinusoidal
electromagnetic wave with wavelength λ travels along −x direction in a region where there
are two short copper wires oriented along the
z direction, a distance L = 2.5λ apart.
You place your detector at an angle θ = 90◦
from the x-axis at some arbitrary distance
away from the two wires. Choose from among
the following:
Interference at the detector will produce
Ia) a maximum
Ib) a minimum
Ic) neither a maximum nor a minimum
P =
= 1.67442 diopters .
Alternative solution:
Without glass for a nearest object,
1
1
1
+ " = " .
43 cm q
feye
The radiation reaching the detector will be
polarized along the
IIa) x-axis
IIb) y-axis
IIc) z-axis
1. Ia, IIb
2. Ia, IIc
With glass object at 25 cm should be seen
comfortably, so
3. Ib, IIb
1
1
1
1
+ " = " + "
.
25 cm q
feye fglass
4. Ia, IIa
5. Ib, IIc correct
Therefore,
P =
1
"
fglass
=
1
1
−
25 cm 43 cm
= 1.67442 diopters .
6. Ic, IIb
7. Ib, IIa
Version 039 – Test #4 – Antoniewicz – (57030)
8. Ic, IIa
9. Ic, IIc
Explanation:
Along the vertical direction, the path difference is 2.5λ. Because this is halfway between two whole number wavelengths, the result will be a destructive overlap of the two
waves. The EM waves radiated from the copper wires must be polarized along the z-axis
since electrons in the wires accelerate along
the axes of the wires.
005 (part 1 of 3) 4.0 points
Coherent light with a wavelength of λ =
550 nm illuminates two narrow vertical slits a
distance d = 0.12 mm apart. Bright lines are
seen on a screen 2 m away.
How far apart are these stripes, center-tocenter? Express your answer in units of nm.
1. 10.5
2. 5.71429
3. 11.0
4. 5.84615
5. 7.77273
6. 8.07692
7. 13.2
8. 8.55
9. 10.0
10. 9.16667
Correct answer: 9.16667 mm.
Explanation:
Let L represent the distance to the screen,
and x represent the required quantity.
Using dsinθ = nλ and setting n = 1,
dsinθ = λ
λ
x
sinθ = ≈
d
L
Lλ
x≈
d
(2 m)(5.5 × 10−7 m)
≈
(0.00012 m)
= 0.00916667 m
= 9.16667 mm
3
006 (part 2 of 3) 3.0 points
Does the separation between stripes increase
or decrease as the slits are moved further
apart?
1. Decrease. correct
2. No change.
3. Increase.
Explanation:
Lλ
From x ≈
, we see that the distance bed
tween the two stripes is inversely proportional
to the distance between the two slits, so as the
slits move farther apart, the stripes get closer
and hence the separation decreases.
007 (part 3 of 3) 3.0 points
Does the separation between stripes increase
or decrease if you use red (λ = 700nm) light?
1. Decrease.
2. No change.
3. Increase. correct
Explanation:
Lλ
, we see that the disd
tance between the stripes is directly proportional to the light’s wavelength, so as the
wavelength increases the stripes move further
apart and the separation between them increases.
Again using x ≈
008 10.0 points
Maxwell’s great contribution to electrodynamic theory was his idea that
1. the charge to mass ratio of the electron
was a constant.
2. a time-varying electric flux acts as a current for the purposes of producing a magnetic
field. correct
Version 039 – Test #4 – Antoniewicz – (57030)
4
For incident radiation normal to the roof,
3. magnetism could be explained in terms of
circulating currents in atoms.
4. work is required to move a magnetic pole
through a current loop.
5. the magnetic force on moving charges is
perpendicular to both v and B.
6. the speed of light could be determined
from simple electrostatic and magnetostatic
experiments (finding the values of µ0 and
#0 ).
Explanation:
Maxwell’s great contribution to electrodynamics is the pair of equations that tell us
that changing time varying magnetic (electric) fields produce electric (magnetic) field.
He supplied the missing part of Ampere’s Law
that showed both occur.
009 (part 1 of 2) 5.0 points
The Sun delivers about 1421 W/m2 of electromagnetic flux to the Earth’s surface.
Calculate the total power incident on a roof
of dimensions with an area 8 m by 21 m,
assuming the radiation is incident normal to
the roof.
The velocity of light is 2.99792 × 108 m/s.
1. 319200.0
2. 108192.0
3. 238728.0
4. 306472.0
5. 137445.0
6. 180576.0
7. 195534.0
8. 501480.0
9. 69750.0
10. 347508.0
Correct answer: 2.38728 × 105 W.
Explanation:
Let :
I = 1421 W/m2 ,
a = 8 m , and
b = 21 m .
P = I A = I ab
= (1421 W/m2 ) (8 m) (21 m)
= 2.38728 × 105 W .
010 (part 2 of 2) 5.0 points
1
2
Assume that the roof is absorptive and
3
3
reflective. Let I be the same as in part 1.
Determine the force on the roof due to the
solar radiation.
1. 0.00154555
2. 0.000320689
3. 0.000585765
4. 0.00106175
5. 0.000962693
6. 0.0011337
7. 0.00105116
8. 0.000614781
9. 0.000630267
10. 0.00111975
Correct answer: 0.00106175 N.
Explanation:
The pressure on a totally absorptive surface
I
2I
is , on a totally reflective surface is
.
c
c
So, combining them together
P =
2 I
1 2I
4I
+
=
.
3 c 3 c
3c
The force is given by
F = PA =
4 Iab
= 0.00106175 N
3 c
011 10.0 points
While standing on a bank you wish to spear a
fish in front of you.
Would you aim above, below, or directly at
the observed fish to make a direct hit? If,
on the other hand, you wished to zap the fish
with a laser beam of the same color as the
fish, how would you aim?
1. Above; below
Version 039 – Test #4 – Antoniewicz – (57030)
5
The value of θ can be determined from the information given about the distance between
maxima and the grating-to-screen distance:
2. Above; above
3. Below; directly at correct
x
0.462 m
=
= 0.314286
L
1.47 m
θ = 17.4472◦ and
4. None of these
tan θ =
5. Below; above
Explanation:
You should aim the spear below the apparent position of the fish, because the effect of
refraction is to make the fish appear closer to
the surface than it really is.
In in order to zap the fish with a laser, make
no corrections and simply aim directly at the
fish. You are simple reversing the refracted
light-ray path from the fish to you. A slight
correction may be necessary, depending on
the colors of the laser beam and the fish.
sin θ = 0.299827 .
The distance d between the grating “slits”
equals the reciprocal of the number of grating
lines per centimeter:
1
1
=
n
5080 lines/cm
= 0.00019685 cm = 1968.5 nm .
d=
The wavelength λ of the laser light is
λ = d sin θ = (1968.5 nm)(0.299827)
012 10.0 points
Light from an argon laser strikes a diffraction grating that has 5080 lines/cm. The
central and first-order principal maxima are
separated by 0.462 m on a wall 1.47 m from
the grating.
Find the wavelength of the laser light.
1. 598.544
2. 486.136
3. 611.609
4. 722.933
5. 629.256
6. 492.397
7. 590.21
8. 570.135
9. 538.719
10. 528.199
Correct answer: 590.21 nm.
= 590.21 nm .
013 (part 1 of 2) 5.0 points
Consider the three systems A, B, and C (in
a perfect vacuum) depicted below. Each consists of a rod suspended by the middle, so that
the rod is free to rotate in the plane of the paper. At each end of each rod is affixed a piece
of glass with one side mirrored and the other
side black, in the manner shown in each figure. The faces of the glass are perpendicular
to the page, and, for example, shading on the
left side of the schematic means that the left
vertical face of the glass is black.
Hint: The rods are far enough apart that
they do not interfere with one another’s rotation or block one another’s view of the light
source.
Explanation:
The principal maxima are defined by the
equation
d sin θ = mλ ,
m = 0, 1, 2 . . . .
For m = 1,
A
λ = d sin θ ,
where θ is the angle between the central
(m = 0) and the first order (m = 1) maxima.
B
C
If a light source is placed to the right of
these systems, which answer best describes
how they will rotate, if at all?
Version 039 – Test #4 – Antoniewicz – (57030)
6
clockwise, C will rotate clockwise.
1. A will rotate clockwise, B, C will not
rotate. correct
2. A will not rotate, B will rotate counterclockwise, C will rotate clockwise.
3. None of them will rotate.
4. A will not rotate, B will rotate clockwise,
C will rotate counterclockwise.
5. A will rotate counterclockwise, B, C will
not rotate.
Explanation:
Basic Concept: Radiative momentum
transferred to absorbing surface.
Note: In the dime-store setup, there is gas
inside the glass bulb, in which the rods are
suspended. Light is absorbed on the dark
side, heating it up. The thermo excitation of
the gas due to the hotter dark side causes the
dark side to move away from the light source.
This is the opposite direction from that which
we expect from the consideration of radiation
pressure of light.
Solution: The momentum transfer to a mirrored surface is twice that of the transfer to
a black surface for the same intensity, so system A will have a stronger force acting to the
left on the bottom end of the rod. It will thus
rotate clockwise. The other two systems will
feel an equal force on both ends, so they will
not rotate.
014 (part 2 of 2) 5.0 points
If the light source is moved to the left of the
rods, how will they rotate?
1. A will rotate counterclockwise, B, C will
not rotate.
2. A will rotate clockwise, B, C will not
rotate. correct
3. None of them will rotate.
4. A will not rotate, B will rotate counter-
5. A will not rotate, B will rotate clockwise,
C will rotate counterclockwise.
Explanation:
This time, the stronger force on system A
will be acting to the right on the top of the
rod, so it will still rotate clockwise. B, C
will still not move, by the same argument as
before.
015 (part 1 of 2) 5.0 points
A material with an index of refraction of 1.27
is used to coat glass. The index of refraction
of glass is 1.5.
What is the minimum thickness of the coating that will minimize the reflection of light
with a wavelength of 5000 Å?
1. 0.0984252
2. 0.0906934
3. 0.135833
4. 0.113889
5. 0.119628
6. 0.109917
7. 0.134152
8. 0.104808
9. 0.0822917
10. 0.121168
Correct answer: 0.0984252 µm.
Explanation:
Phase Changes for Reflecting Waves.
Since the coating has a refraction that is less
than that for glass (but greater than that for
air), we know that the reflected light from
both the glass and the coating will undergo
a 180◦ phase shift. This means that the total trip inside the coating must be exactly
one half of the wavelength of light inside the
coating. (Hence the waves reflected from the
glass and from the coating will interfere destructively.) The wavelength of light inside
λ
the coating is
where n is the index of ren
fraction of the coating. This implies that the
thickness of the coating must be one fourth
the wavelength of light inside this medium.
Hence (calling the thickness of the material t,
Version 039 – Test #4 – Antoniewicz – (57030)
the wavelength of light in air λ, and the index
of refraction of the medium n)
t=
λ
5 × 10−7 m
=
4n
(4) (1.27)
t = 9.84252 × 10−8 m
= 0.0984252 µm .
016 (part 2 of 2) 5.0 points
Now assume that the coating’s index of refraction is 1.67. Assume that the rest of the
system (from the previous question) remains
the same.
What is the minimum thickness of the coating needed to minimize the reflection of this
light now?
1. 0.21465
2. 0.116201
3. 0.200287
4. 0.166867
5. 0.149701
6. 0.171067
7. 0.203892
8. 0.20679
9. 0.144268
10. 0.125767
Correct answer: 0.149701 µm.
Explanation:
This part is really solved in the same way
as Part 1 except now the index of refraction of the material is greater than that for
glass. Hence the light reflected from the material surface undergoes a 180◦ phase transition,
but the light reflected from the glass within
the material goes through a 0◦ phase transition. This means that the light must travel
a full wavelength within this material in order to interfere destructively. Using the same
notation as before, we then say:
t=
λ
5 × 10−7 m
=
2n
2 (1.67)
t = 1.49701 × 10−7 m
= 0.149701 µm .
7
017 (part 1 of 2) 5.0 points
An object located 33.6 cm in front of a lens
forms an image on a screen 10.8 cm behind
the lens.
Find the focal length of the lens.
1. 8.57412
2. 8.17297
3. 7.80764
4. 6.5032
5. 6.7462
6. 7.71167
7. 7.39995
8. 7.05229
9. 7.94244
10. 6.94613
Correct answer: 8.17297 cm.
Explanation:
Basic Concepts:
1 1
1
h"
q
+ =
m=
=−
p q
f
h
p
Converging Lens
f >0
∞ >p> f
f >p> 0
f <q< ∞
−∞ < q < 0
Diverging Lens
∞ >p> 0
f <q< 0
0 > m > −∞
∞ >m> 1
f <0
0 <m< 1
Solution: The formula relating the focal
length to the image distance s" and the object
distance s is
1
1
1
+ " =
s s
f
so
s s"
s + s"
(33.6 cm) (10.8 cm)
=
(33.6 cm) + (10.8 cm)
= 8.17297 cm .
f=
018 (part 2 of 2) 5.0 points
What is the magnification of the object?
1. -0.311526
Version 039 – Test #4 – Antoniewicz – (57030)
2. -0.307246
3. -0.321429
4. -0.258382
5. -0.341246
6. -0.274018
7. -0.246628
8. -0.358896
9. -0.28503
10. -0.252011
∆V =
8
2(9.11 × 10−31
(6.625 × 10−34 Js)2
kg)(1.6 × 10−19 C)(2.4 × 10−1
∆V = 26.1385 V
10.0 points
3.78 µm
020
Correct answer: −0.321429.
Explanation:
Magnification is
M =−
θ
s"
(10.8 cm)
=−
= −0.321429 .
s
(33.6 cm)
019 10.0 points
Roughly, what is the minimum accelerating potential ∆V needed in order that electrons exhibit diffraction effects in a crystal, given that the wavelength observed is
λ = 2.4 × 10−10 m?
Given that:
h = 6.625 × 10−34 Js
e = 1.6 × 10−19 C
me = 9.11 × 10−31 kg.
1. 31.107
2. 41.7058
3. 307.261
4. 66.9146
5. 150.558
6. 602.231
7. 418.216
8. 52.0961
9. 28.4608
10. 26.1385
Determine the maximum angle θ for which
the light rays incident on the end of the light
pipe shown in the figure above are subject
to total internal reflection along the walls of
the pipe. The pipe of diameter 3.78 µm has
an index of refraction of 1.38 and the outside
medium is air.
1. 53.0347
2. 71.9894
3. 34.6033
4. 65.0828
5. 63.1211
6. 42.9413
7. 44.3353
8. 28.8009
9. 30.2903
10. 74.8957
Correct answer: 71.9894◦ .
Explanation:
Given : nair = 1 ,
npipe = 1.38 , and
r = 1.89 µm .
Correct answer: 26.1385 V.
Explanation:
We can make use of the formula
λ = √
h
2me e∆V
Solving for ∆V , we obtain
∆V =
h2
2me eλ2
φ
θ
φc
d
Version 039 – Test #4 – Antoniewicz – (57030)
The critical angle φc for total internal reflection is given by
sin φc =
nair
.
npipe
!
"
nair
φc = arcsin
npipe
!
"
1
= arcsin
1.38
◦
= 46.4387 .
From the right triangle the angle φ at the
left-hand interface is
φ = 90◦ − φc
= 43.5613◦ .
Applying Snell’s Law, we have
nair sin θ = npipe sin φ
!
"
npipe sin φ
θ = arcsin
nair
"
!
1.38 sin 43.5613◦
= arcsin
1
= 71.9894◦ .
9