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Solution to PHYS 1112 In-Class Exam #1A

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Solution to PHYS 1112 In-Class Exam #1A
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #1A
Tue. Feb. 8, 2011, 09:30am-10:45am
Conceptual Problems
Problem 1: A student runs northward at 5m/s, away from a vertical plane mirror, while
the mirror, mounted on wheels, travels northward at 7 m/s (with both speeds given relative
to the ground). The speed at which the student’s image moves and its direction, relative to
the ground, is
(A)
(B)
(C)
(D)
(E)
3m/s
3m/s
2m/s
9m/s
9m/s
northward
southward
northward
northward
southward
Answer: (D)
Solution: With North chosen as the positive and South then being the negative velocity
direction, let velocities be defined as follows:
v = +5m/s=runner rel. to ground; u = +7m/s=mirror rel. to ground; v 0 =runner rel. to
mirror; w0 =image of runner rel. to mirror; w=image of runner rel. to ground.
To visualize this, make a drawing and indicate distances traveled by runner, mirror and
image during a short time interval, e.g., during 1 second. Keep in mind that the runner and
his/her image must be always be at equal distances from the mirror, and on opposite sides
of the mirror.
Then v 0 = v − u = −2m/s, i.e., as seen by an observer traveling with the mirror, the runner
is moving in southward direction at speed of 2m/s (away from mirror).
Then w0 = −v 0 = +2m/s, i.e., as seen by an observer traveling with the mirror, the image
is moving in northward direction at speed of 2m/s. (away from mirror).
Then w = w0 + u = +2 + 7m/s = +9m/s i.e., as seen by an observer on the ground, the
image is moving in northward direction at speed of 9m/s.
Problem 2: UGA waves (we didn’t cover those in class, but they do obey Snell’s law!) have
a speed of wave propagation vA = 2097m/s in apple juice and vB = 522m/s in butter milk.
Also, assume that sin(14.414o ) = 522/2097 . A narrow beam of UGA waves striking a
flat horizontal interface between apple juice and butter milk, with the apple juice above
and the butter milk below the interface
(A) will not undergo total internal reflection if incident from above the interface with
an angle of incidence of 29.0o ;
1
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
(B) will always have an angle of refraction not exceeding 14.414o if the beam is incident
from below the interface without total internal reflection;
(C) will undergo total internal reflection if incident from above the interface with an
angle of incidence of 8.5o ;
(D) will not undergo total internal reflection if incident from below the interface with
an angle of incidence of 29.0o ;
(E) will have an angle of refraction greater than the angle of incidence if the beam is
incident from above the interface without total internal reflection.
Answer: (A)
Solution: Regardless of the beam’s travel direction (incident from above or incident from
below), let its angles measured from the normal to the interface be defined as follows:
ΘA =angle above the interface (in medium A=apple juice);
ΘB =angle below the interface (in medium B=butter milk).
Snell’s Law:
sin(ΘB )
sin(ΘA )
=
(S1) or
vA
vB
sin(ΘB )
vB
vB
=
(S2) or sin(ΘB ) =
sin(ΘA ) (S3)
sin(ΘA )
vA
vA
Given velocities obey vB < vA , hence the ratio (vB /vA ) < 1 in Eq.(S3). So, by Eq.(S3),
sin(ΘB ) < sin(ΘA ), and, in the absence of total internal reflection,
ΘB < ΘA .
Thus, the angle ΘB below the interface is always less than the angle ΘA above the interface.
So, for incidence from above, the beam is refracted towards the normal; for incidence from
below it’s refracted away from the normal. Make a drawing of beams, normal and interface
for incidence from above; and likewise, a drawing for incidence from below; each drawing
showing this relation between the two angles.
Which angle, ΘA or ΘB , is angle of incidence; and which is angle of refraction now depends
on the beam’s travel direction, as follows:
If incident from above: ΘA = Θ =angle of incidence; ΘB = Θ0 =angle of refraction.
If incident from below: ΘB = Θ =angle of incidence; ΘA = Θ0 =angle of refraction.
Total internal reflection (TIR) occurs when the sine of the angle of refraction, sin(Θ0 ), predicted by Snell’s Law, exceeds the value of 1, while the sine of the angle of incidence, sin(Θ),
is still less than 1. In the present case, where sin(ΘB ) < sin(ΘA ), this can happen only
for incidence from below, and only when sin(ΘB ) exceeds the critical angle of incidence,
Θcrit = 14.414o . When ΘB reaches that value, Θcrit , then, sin(ΘB ) = vB /vA ; and by Snell’s
Law, Eq.(S3), sin(ΘA ) reaches 1 and ΘA reaches 90o . When ΘB exceeds the value Θcrit , then,
by Snell’s Law, sin(ΘA ) exceeds 1 and ΘA does not exist, i.e., there is no refracted beam
anymore.
Therefore, (A) is correct, since TIR cannot occur for incidence from above, for any angle of
incidence ΘA .
2
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Also, (B) is wrong, since ΘA is the angle of refraction for incidence from below. As shown
above, when ΘB approaches Θcrit = 14.414o , the angle ΘA , approaches 90o and thereby
clearly exceeds 14.414o , as well as any other value below 90o . That’s also consistent with
ΘA > ΘB , as shown above.
Also, (C) is wrong for the same reason that (A) is correct: TIR cannot occur for incidence
from above, for any angle of incidence ΘA .
Also, (D) is wrong because TIR does occur for incidence from below when the angle of
incidence, ΘB , exceeds Θcrit = 14.414o . That is the case here, for ΘB = 29.0o .
Lastly, (E) is wrong, since ΘB < ΘA (as shown above); and ΘB is the angle of refraction,
and ΘA is the angle of incidence, for incidence from above.
Problem 3: Two identically shaped solid blocks, S and T , made from the two different
transparent materials, are immersed in the same liquid L. A ray of light strikes each block
at the same angle of incidence, as shown. According to the figure below, what is the relative
magnitude of the indices of refraction of the solid blocks, nS and nT , and liquid, nL ?
L
L
S
(A)
(B)
(C)
(D)
(E)
nT
nS
nT
nL
nS
T
< nL < nS ;
< nT < nL ;
< nS < nL ;
< nS < nT ;
< nL < nT .
Answer: (C)
Solution: First off, before you read any further, make a large (re-)drawing of the figure
above wherein the normal to the top surface of each solid block, and the corresponding angle
of incidence Θ and angles of refraction, Θ0S or Θ0T , respectively, at the top surface, are clearly
shown.
If you really know your stuff, this quick ’n dirty argument will give you the answer, fast:
3
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
(1) As shown in the figure, the initial incident ray (above block S or T ) is refracted by the
block away from the normal by both blocks S and T ; hence nS < nL and nT < nL .
(2) As also shown in the figure, in block T , the incident ray is refracted (i.e., ”bent” by
the block) away from the normal more strongly than in block S. Hence, angle of incidence
and index of refraction (IoR) nL of the liquid L being the same in both cases, we must have
nT < nS .
(3) So, combining the three inequalities from (1) and (2), we must have nT < nS < nL which
is Answer (C).
If you don’t know your stuff quite yet, you are strongly advised to work through and absorb
the the following more detailed and mathematically rigorous argument:
At the top surfaces of both blocks, compare the relative sizes of: (1) angle of incidence Θ
(same in either liquid), (2) angle of refraction Θ0S inside block S, (3) angle of refraction Θ0T
inside block T .
With normals to top surfaces carefully drawn in and angles carefully labeled, you can read
off from the figure that
Θ < Θ0S < Θ0T
Hence, since the sine increases with angle (for angles between 0o and 90o ):
sin(Θ) < sin(Θ0S ) < sin(Θ0T ) .
Thus, after dividing both inequalities by sin(Θ) on both sides:
1 <
sin(Θ0S )
sin(Θ0T )
<
.
sin(Θ)
sin(Θ)
But, by Snell’s law, nT sin(Θ0T ) = nL sin(Θ) and nS sin(Θ0S ) = nL sin(Θ), or equivalently:
sin(Θ0T )
nL
=
sin(Θ)
nT
and
sin(Θ0S )
nL
=
.
sin(Θ)
nS
Thus replacing sine-ratios in the foregoing inequalities by corresponding IoR-ratios:
1 <
nL
nL
<
nS
nT
Dividing both these inequalities by nL then gives
1
1
1
<
<
nL
nS
nT
Taking the reciprocal (1/...) on both sides of both of these two inequalities thus gives Answer
(C). (Recall here that, if 1/a < 1/b for positive numbers a and b, then a > b)
Problem 4: If a virtual object (d < 0) is presented to a divergent lens (f < 0), at an
absolute distance |d| less then the absolute focal length |f |, i.e., |d| < |f |, then the image is
4
Physics 1112
Spring 2011
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
virtual, inverted and enlarged in height relative to the virtual object
virtual, inverted and reduced in height relative to the virtual object
virtual, erect and reduced in height relative to the virtual object
real, inverted and enlarged in height relative to the virtual object
real, erect and enlarged in height relative to the virtual object
Answer: (E)
Solution: If 0 > d = −|d| (virtual object!) and 0 > f = −|f | (divergent lens!) and |d| < |f |,
then:
1/|d| > 1/|f |
and thus
1/d0 = 1/f − 1/d = −1/|f | − (−1/|d|) = 1/|d| − 1/|f |) > 0 .
Hence
"
d0 = (1/f − 1/d)−1 = (1/|d| − 1/|f |)−1 =
(|f | − |d|)
|d||f |
#−1
"
#
|f |
= |d| ×
> 0;
(|f | − |d|)
and d0 > 0 means we have a real image.
Also, together with d < 0, this implies that m = −d0 /d = +|d0 |/|d| > 0; and m > 0 means
the image is erect.
Also, since 0 < |d| < |f |, we have 0 < |f | − |d| < |f |, |f |/(|f | − |d|) > 1, and thus
"
#
"
|f |
|f |
> 1 , or |d0 | = |d| ×
(|f | − |d|)
(|f | − |d|)
and therefore
"
0
|m| = |d |/|d| =
|f |
(|d| − |f |)
#
> |d| ;
#
> 1.
Here |m| > 1 means that the image is enlarged.
Lastly, if you do not like general mathematical proofs, like the one above, you could also
answer this question simply by plugging in some numbers for d and f . Take, e.g, f = −4cm
and d = −3cm, chosen so that d < 0 and f < 0 and |d| < |f |. Then, you’ll get d0 = (1/f −
1/d)−1 = [−(1/4) + (1/3)]−1 cm = +12cm and from that m = −d0 /d = −(+12)/(−3) = +4.
Therefore:
Since d0 > 0, the image is real.
Since m > 0, the image is erect, relative to object.
Since |m| > 1, the image is enlarged in absolute height, relative to object.
5
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Numerical Problems
Problem 5: The state highway patrol radar guns send out a a microwave frequency of
8.090GHz. You have just passed, and are receding from, a radar speed trap driving 29.3m/s
and the radar gun measures the frequency of the microwave reflecting from your car. How
does this reflected microwave frequency, as detected by the receiver on the radar gun, differ
from the original frequency sent out by the gun ?
(A)
(B)
(C)
(D)
(E)
The
The
The
The
The
reflected
reflected
reflected
reflected
reflected
is
is
is
is
is
790Hz higher than the original frequency.
1580Hz higher than the original frequency.
395Hz lower than the original frequency.
1580Hz lower than the original frequency.
790Hz lower than the original frequency.
Answer: (D)
Solution: See assigned HWP on radar gun. Let u =speed of car C moving away from
radar gun; f =frequency sent out by radar gun source S; f 0 =frequency of wave received by
moving car=frequency of reflected wave, as sent out by moving car; and f 00 =frequency of
reflected wave, as received by radar gun receiver R. Then
∆fC−S := f 0 − f = −f
u
,
c
∆fR−C := f 00 − f 0 = −f 0
u
u∼
= −f
c
c
where the last approximate equality holds because u/c 1, hence ∆fC−S f . So,
∆f := f 00 − f = ∆fC−S + ∆fR−C = −2f
or
∆f = −2 × (8.090 × 109 Hz)
u
c
29.3m/s = −1580Hz .
3.00 × 108 m/s
Problem 6: Visible light has a range of wavelengths from 400nm (violet) to 700nm (red)
in vacuum. A beam of electromagnetic (EM) waves with a frequency of 714.3THz in air
travels from air into water, with indices of refraction nAir = 1.00 and nWater = 1.333. To an
under-water observer, the beam while traveling in water will
(A) have a wavelength of 420.0nm, have a frequency of 535.9THz, and be visible to the
human eye;
(B) have a wavelength of 559.9nm, have the same frequency as in air, and be visible to
the human eye;
(C) have a wavelength of 315.1nm, have the same frequency as in air, and be invisible to
the human eye;
(D) have a wavelength of 315.1nm, have a frequency of 952.2THz, and be invisible to the
human eye;
6
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
(E) have a wavelength of 315.1nm, have the same frequency as in air, and be visible to
the human eye.
Answer: (E)
Solution: The frequency f = 714.3THz is unchanged as the wave travels from one medium
into another. Its vaccum wavelength is:
λVac =
3.00 × 10−8 m/s
c
=
= 420.0 × 10−9 m = 420.0nm .
f
714.3 × 1012 Hz
This is clearly within the visible vacuum wavelength range from λVac,min = 400nm to λVac,max =
700nm, i.e., λVac,min ≤ λVac ≤ λVac,max . Therefore the EM wave beam will be visible to the
human eye.
Note: As we discussed in class, visibility of an EM wave depends only on the frequency,
not on the wavlength it may have in any particular medium it travels through before it
reaches the eye. That’s why the vacuum wavelength is the relevant quantity here: the
vacuum wavelength is determined only by the frequency. The vaccum wavelength range
from λVac,min = 400nm to λVac,max = 700nm corresponds to a frequency range from
fmin =
to
c
λVac,max
=
3.00 × 108 m/s ∼
= 429 × 1012 Hz = 429THz
−9
700 × 10 m
3.00 × 108 m/s
=
= 750 × 1012 Hz = 750THz
fmax =
−9
λVac,min
400 × 10 m
The actual frequency f = 714.3THz falls within this visible range: fmin ≤ f ≤ fmax .
Therefore, we again conclude that the EM wave beam is visible. So, one can determine
visibility either by checking whether λVac falls within the range from λVac,min = 400nm to
λVac,max = 700nm; or by checking whether f falls within the range from fmin = 429THz to
fmax = 750THz.
c
The speed of wave propagation in a medium with index of refraction n is v = c/n. Hence,
the wavelength in the medium is
λ=
c/n
1 c
1
v
=
=
= λVac .
f
f
n f
n
So, in water with n = nWater = 1.333,
λ=
420.0nm
= 315.1nm .
1.333
In summary, the EM wave beam in water has the same frequency as in air, f = 714.3THz,
and a wavelength of λ = 315.1nm; and it is visible in water.
Problem 7: A very thin, flat circular mirror lies flat on the floor. Hanging from the ceiling
and centered 1.6m above the mirror is a small lamp. The circular spot formed on the flat
ceiling 3.0m above the floor, by the reflection of the light from the lamp, has a radius of
1.29m. What is the radius of the mirror ?
7
Physics 1112
Spring 2011
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
0.45m
0.90m
0.22m
0.69m
1.38m
Answer: (A)
Solution: See HWP01.12 for details and for a drawing. Let r = radius of the mirror,
R = 1.29m= radius of circular spot on the ceiling, H = 3.0m= height of the ceiling above
floor, and h = 1.6m= height of the lamp above floor. Then, by similar triangles (see drawing
in HWP01.12 solution, with x = R − r here):
R−r
r
=
h
H
or
H
R−r
R
=
= − 1 or
h
r
r
H +h
R
=
h
r
hence, after solving for r:
r=R
1.6
h
= (1.29m)
= 0.45m .
H +h
3.0 + 1.6
Problem 8: Two laser beams are passing from water through the glass wall of an aquarium
tank into air, as shown below. The glass wall is x = 34cm thick and has two parallel planar
surfaces. Both beams enter the wall at the same point, beam 1 at normal incidence, beam 2
at an angle α = 50o to the wall. The two beams’ points of exit from the wall are y = 16.6cm
apart. Air and water have indices of refraction nAir = 1.00 and nWater = 1.333, respectively.
What is the index of refraction of the glass?
x beam 2 α beam 1 y beam 1 beam 2 (A)
(B)
(C)
(D)
(E)
2.33
1.95
1.75
1.47
0.91
Answer: (B)
Solution: Note first that beam 1 is always the normal to the surface, at both surfaces. Only
the angles of incidence and refraction of beam 2 at the left surface are needed here. Namely:
8
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
The angle of incidence of beam 2, Θ, is the angle between beam 2 and normal (≡beam 1),
to the left of the left surface:
Θ = 90o − α = 90o − 50.0o = 40.0o .
The angle of refraction of beam 2, Θ0 , is the angle between beam 2 and normal (≡beam 1)
to the right of the left surface:
tan(Θ0 ) =
y
x
or
y
Θ0 = arctan
x
16.6 = arctan
34.0
= 26.023o
Since the medium to the left of the left surface is water and to the right of it it’s glass, Snell’s
law gives nWater sin(Θ) = nGlass sin(Θ0 ), hence:
nGlass = nWater
sin(Θ)
sin(40.0o )
=
1.333
= 1.95
sin(Θ0 )
sin(26.023o )
Problem 9: A postage stamp placed 32.00cm to the left of a lens produces an image 4.00cm
to the left of the lens. If the postage stamp is 2.48cm tall its image produced by the lens
will be
(A)
(B)
(C)
(D)
(E)
real, inverted and 19.5cm tall in absolute size.
virtual, erect and 19.5cm tall in absolute size;
virtual, erect and 0.31cm tall in absolute size;
virtual, inverted and 19.5cm tall in absolute size;
real, inverted and 0.31cm tall in absolute size;
Answer: (C)
Solution: Object is to the left of lens; light originates from object and light passes through
lens. Therefore, light enters lens from the left and exits lens to the right. Therefore
left=incoming side of lens and right=outgoing side of lens.
Draw this! Show: the object, the lens, arrows indicating incoming light entering and
outgoing light rays exiting, labels for in to the left and out to the right of lens.
Object is 32.00cm to left of lens =on incoming side: d = +32.00cm> 0, by sign convention.
Image is to 4.00cm to left of lens =not on outgoing side: d0 = −4.00cm< 0, by sign
convention.
Lateral magnification:
m=−
(−4.00)
d0
=−
= +0.125
d
(+32.00)
Image size:
h0 = mh = (+0.125) × (2.48cm) = 0.31cm
9
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
From m > 0, by sign convention: image is erect.
From d0 < 0, by sign convention: image is virtual.
From h0 = 0.31cm: image is 0.31cm tall.
Problem 10: What is the focal length of the lens used in Problem 9?
(A)
(B)
(C)
(D)
(E)
−28.00cm
−4.57cm
−3.56cm
+4.57cm
+3.56cm
Answer: (B)
Solution: Object is to the left of lens; light originates from object and light passes through
lens. Therefore, light enters lens from the left and exits lens to the right. Therefore
left=incoming side of lens and right=outgoing side of lens.
Draw this! Show: the object, the lens, arrows indicating incoming light entering and
outgoing light rays exiting, labels for in to the left and out to the right of lens.
Object is 32.00cm to left of lens =on incoming side: d = +32.00cm> 0, by sign convention.
Image is to 4.00cm to left of lens =not on outgoing side: d0 = −4.00cm< 0, by sign
convention.
Use (1/f ) = (1/d) + (1/d0 ), to find focal length f :
f=
1
d
+
1 −1 1
1 −1
=
+
cm = −4.571cm
d0
(+32.00) (−4.00)
Problem 11: A microsope produces a final image of a microbial cell (as seen through the
eyepiece) which appears to be 2.4cm in diameter and located 40.0cm from the eyepiece on
the incoming side of the eyepiece lens. If the cell has an actual diameter of 2.8 µm, what is
the angular magnification achieved with the microscope, compared to viewing the cell from
a 20.0cm nearpoint (reference) distance without instrument ?
(A)
(B)
(C)
(D)
(E)
∼ 830
∼ 1990
∼ 4290
∼ 6430
∼ 9710
Answer: (C)
10
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Solution: Angle Θe subtended at eye by final image, of size he = 2.40cm, as seen through
optical instrument, at distance de = 40.0cm:
Θe ∼
= tan(Θe ) = he /de = (2.40cm)/(40.0cm) = 0.060rad
Angle subtended at eye by original objects, size ho = 2.8 × 10−4 cm, as seen without optical
instrument, from reference distance dref = dnear = 20cm:
Θref ∼
= tan(Θref ) = ho /dref = (2.80 × 10−4 cm)/(20.0cm) = 1.40 × 10−5 rad
Angular magnification:
|M | = Θe /Θref = (0.060)/(1.40 × 10−5 ) ∼
= 4290 .
Problem 12: A convergent lens (Lens1), placed to the right of a small grain of sand
produces an image of the grain to the right of Lens1. If a divergent lens (Lens2) of focal
length f2 = −6.42cm is now placed somewhere to the right of Lens1, the final image produced
by Lens2 appears approximately 32.0cm to the left of Lens2. The object of Lens2 is
(A)
(B)
(C)
(D)
(E)
real and located 5.35cm to the right of Lens2
virtual and located 5.35cm to the left of Lens2
real and located 8.03cm to the right of Lens2
virtual and located 8.03cm to the right of Lens2
virtual and located 8.03cm to the left of Lens2.
Answer: (D)
Solution: Original object=Object1 is to the left of Lens1; Lens2 is to the right of Lens1.
Light originates from Object1 and light passes through Lens1, then through Lens2. Therefore, light enters Lens1 from the left and exits Lens1 to the right. Then, light enters Lens2
from the left and exits Lens2 to the right. Therefore left=incoming side of Lens2 and
right=outgoing side of Lens2.
Draw this! Show: Object1, Lens1, Lens2, arrows indicating incoming light entering and
outgoing light rays exiting each lens, labels for in to the left and out to the right of Lens2.
Since both f2 and d02 are given and d2 is asked for, only image formation at Lens2 needs to
be considered, as follows:
Image2 is 32.0cm to the left=not on outgoing side of Lens2, hence d02 < 0, by sign
convention: d02 = −32.0cm< 0.
Use (1/f2 ) = (1/d2 ) + (1/d02 ), to find object distance of Object2, d2 :
d2 =
1 −1 1
1 −1
1
− 0
=
+
cm = −8.03cm
f2 d2
(−6.42) (−32.0)
Negative object distance, d2 < 0, means: Object2 is virtual and it is not on incoming
side, by sign convention.
11
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Since the incoming side is to the left of Lens2, being not on incoming side means: to the
right of Lens2.
So, Object2 is to the right=not on incoming side of Lens2, |d2 | = 8.03cm from Lens2.
12
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