Download Solution to PHYS 1112 In-Class Exam #1A

Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #1A
Thu. Feb. 5, 2009, 11:00am-12:15pm
Conceptual Problems
Problem 1: A student runs westward at 3m/s, away from a vertical plane mirror, while
the mirror, mounted on wheels, travels eastward 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)
4m/s westward
4m/s eastward
10m/s eastward
17m/s eastward
17m/s westward
Answer: (D)
This is the same problem as HW P02.04, except that the direction of motion of the runner,
of the mirror and hence the image has been reversed: both runner and image are moving away from the mirror instead of towards it. The speed of runner relative to mirror is
(3 + 7)m/s = 10m/s in westward direction. Hence, the image’s speed is also 10m/s, but
in eastward direction, relative to the mirror. Since the mirror also moves eastward at 7m/s
relative to the ground, the speed of mirror relative to ground and of image relative to mirror
add up, to (10 + 7)m/s = 17m/s eastward.
Problem 2: Sound waves (including ultrasound) have a speed of wave propagation vAir =
346m/s in air and vWater = 1497m/s in water. Also, note that sin(13.364o ) = 346/1497 .
A narrow ultrasound beam striking the flat water surface of your swimming pool
(A) will have an angle of refraction smaller than the angle of incidence if the beam is
incident from above the water surface;
(B) will have an angle of refraction greater than the angle of incidence if the beam is
incident from below the water surface;
(C) cannot undergo total internal reflection if incident from above the water surface,
regardless of the angle of incidence;
(D) will undergo total internal reflection if incident from above the water surface with
an angle of incidence of 30o
(E) will undergo total internal reflection if incident from below the water surface with
an angle of incidence of 30o .
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Answer: (D)
Refraction of sound at the air-water interface differs from refraction of light at the same
interface in that light travels slower in water than in air, whereas sound travels faster in
water than in air.
Total internal reflection (TIR) occurs if (a) the sound beam is incident from the low-speed
medium (=air, for sound) and gets refracted into the high-speed medium (=water, for sound)
and (b) the angle of incidence Θ1 exceeds the critical angle, defined by sin(Θ(crit) ) = v1 /v2 ,
with v1 < v2 . Hence, the sound beam can undergo TIR only if incident from air into
water, but never if incident from water into air! Also, for incidence from air into water,
(crit)
(crit)
sin(ΘAir ) = 346/1497, i.e., ΘAir = 13.364o from the information given.
The foregoing two TIR prerequsites (a) and (b) are satisfied by the conditions stated in (D),
but not by the conditions stated in (E). For this reason, (D) is correct and both (E) and (C)
are wrong.
Regardless of whether the sound beam is incident from air into water or from water into air,
by Snell’s law
sin(ΘWater ) = (vWater /vAir ) sin(ΘAir )
where ΘAir and ΘWater are the angles of the sound beam in air and water, respectively,
measured from the normal, and vAir = 346m/s and vWater = 1497m/s are the corresponding
speeds. Since vWater > vAir this implies that sin(ΘWater ) > sin(ΘAir ) and hence
ΘWater > ΘAir
That is, as a matter of general principle, the beam always has the greater angle to the
normal when traveling in the medium with the greater speed. (And this is always true for
any kind of wave being refracted between any two different media!)
So, if incident from air, the sound beam is refracted away from the normal upon entering
the water; and if incident from water, the sound beam is refracted towards the normal upon
entering the air (provided that refraction occurs at all, i.e., absent TIR). For these reasons
answers (A) and (B) are wrong.
Problem 3: If a real object is placed between the focal point and the lens for a convergent
lens (f > 0), then the image is
(A)
(B)
(C)
(D)
(E)
virtual, erect and enlarged in height relative to the object
virtual, erect and reduced in height relative to the object
real, inverted and reduced in height relative to the object
real, inverted and enlarged in height relative to the object
real, erect and reduced in height relative to the object
Answer: (A)
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
For a real object d > 0. Also, ”placed between focal point and lens” means:
f >d>0.
Therefore, by taking the inverse of the inequality f > d and using d > 0, it follows
0 < 1/f < 1/d
and hence
−1/d0 = 1/d − 1/f > 0
Eq.(3.1)
From this it follows that d0 < 0, i.e., the image is virtual; and also m = −d0 /d > 0, i.e. the
image is erect.
However, from Eq.(3.1), it also follows that
−1/d0 = 1/d − 1/f < 1/d
Eq.(3.2),
since 1/f > 0. Eq.(3.2) and −d0 > 0 [from Eq.(3.1)], then implies −d0 > d > 0, and hence
m = (−d0 )/d > 1 ,
i.e. the image is enlarged. So, the answer is (A).
Lastly, if you do not like general mathematical derivations, 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 so that 0 < d < f , i.e., the object is placed between focal point and
lens. 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 virtual.
Since m > 0, the image is erect, relative to object
Since |m| > 1, the image is enlarged in absolute height, relative to object.
Problem 4: A thin lens and a curved mirror both have the same focal length when surrounded by air (with index of refraction nAir = 1). If the lens and the mirror are submerged
in water (with index of refraction nWater > 1) what happens to the focal length fL of the
lens and the focal length fM of the mirror ?
(A)
(B)
(C)
(D)
(E)
fL
fL
fL
fL
fL
changes and fM stays the same;
stays the same and fM changes;
increases and fM increases;
decreases and fM decreases;
stays the same and fM stays the same.
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Answer: (A)
The lens forms images by bending light rays by means of refraction. For each light ray
involved in forming the image, refraction occurs at the front and at the rear surface of the
lens; both of these surfaces are interfaces between the lens material (e.g., glass) and the
surrounding medium. By Snell’s law, the angles of refraction at both surfaces depend on
both the index of refraction (IoR) of the lens material and on the IOR of the surrounding medium (air or water). Therefore, for a fixed object position d, the image position d0
depends on (i.e., changes with) the IoR of the surrounding medium; and therefore the focal
length fL ≡ (1/d + 1/d0 )−1 changes with the IoR of the surrounding medium: fL changes
when the lens is submerged from air into water.
The mirror forms images by bending light rays by means of reflection. For each light ray
involved in forming the image, reflection occurs at the curved surface of the mirror. By
Archimedes’ law (Θ̄ = Θ), the angle of reflection (Θ̄) for each light ray at that surface depdends only on the angle of incidence (Θ), but it does not depend on the surrounding
medium (air or water). Therefore, for a fixed object position d, the image position d0 does
not depend on (i.e., does not changes with) the surrounding medium; and therefore the focal
length fM ≡ (1/d + 1/d0 )−1 does not change when we change the surrounding medium: fM
is the same regardless of whether the mirror is submerged in air or water.
Numerical Problems
Problem 5: A beam of monochromatic light with a frequency of 3.75 × 1014 Hz in air travels
from air into water, with indices of refraction nAir = 1.00 and nWater = 4/3, in air and water,
respectively. To an under-water observer, the light beam while traveling in water will
(A) have a wavelength of 800nm, have the same frequency as in air, and be invisible to
the human eye;
(B) have a wavelength of 800nm, have a frequency of 5.00 × 1014 Hz, and be visible to the
human eye;
(C) have a wavelength of 600nm, have a frequency of 5.00 × 1014 Hz, and be invisible to
the human eye;
(D) have a wavelength of 600nm, have the same frequency as in air, and be invisible to
the human eye;
(E) have a wavelength of 600nm, have a frequency of 5.00 × 1014 Hz, and be visible to the
human eye.
Answer: (D)
The frequency f does not change when the light wave travels from air into water. However,
the speed of light does change, from vAir = c/nAir ∼
= c = 3.00 × 108 m/s to vWater = c/nWater .
Hence, the wavelength also changes from λAir = vAir /f to λWater = vWater /f = (vAir /f ) ×
(nAir /nWater ) = λAir × (nAir /nWater ).
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Therefore, f = 3.75 × 1014 Hz in both air and water and λAir = (3.00 × 108 m/s)/(3.75 ×
1014 Hz) = 800nm, but λWater = (800nm) × [1.00/(4/3)] = 600nm.
Here, λAir is (approximately) the same as the vacuum wavelength (λVacuum ≡ c/f ). Visibility
of light to the human eye is determined only by its frequency f , or equivalently, by its vacuum wavelength: light is visible to the human eye if λVacuum ≡ c/f is between ∼ 700nm and
∼ 400nm. The light ray in this problem is not visible, since its 800nm vacuum wavelength
does not fall within the 700−400nm visible range. The 600nm value of the actual wavelength
in water is irrelevant for visibility to the human eye: it’s only the frequency in water that
matters!
Problem 6: The state highway patrol radar guns send out a frequency of 9.34 GHz. You’re
approaching a radar speed trap driving 5.24 m/s and the radar gun measures the frequency
of the radar wave reflecting from your car. By what percentage is the reflected frequency
different from the original frequency ? Does this percentage change if the radar gun frequency
changes ?
(A)
(B)
(C)
(D)
(E)
1.75 × 10−6
3.49 × 10−6
3.49 × 10−6
3.52 × 10−6
3.52 × 10−6
%,
%,
%,
%,
%,
will
will
will
will
will
not change with frequency
change with frequency
not change with frequency
change with frequency
not change with frequency
Answer: (C)
Please read and work carefully through the posted solution of, e.g., HW Problem P01.09, for
details: the frequency shift between original wave sent out by radar gun (f ) and reflected
wave (f 00 ) after reflection off your car is: ∆f ≡ f 00 − f = 2f (u/c). Therefore the percentage
change in frequency is: p ≡ (∆f /f ) × 100 % = (2u/c) × 100 %. Clearly, this percentage does
not depend on f ; it depends only on your speed u; and its value is p = 2 × [(5.24m/s)/(3.00 ×
108 m/s)] × 100 % = 3.49 × 10−6 %.
Note that it is very difficult (if not impossible!) to evaluate ∆f or p accurately if you first
try to calculate the reflected wave frequency f 00 and then subtract f from f 00 : f 00 and f
are so very close in value, that their difference ∆f is likely smaller than the rounding error
of your calculator. So, you really need to do the algebra first in this problem [to derive
∆f = 2f (u/c)] before blindly plugging numbers into your calculator. That was one of the
lessons HW P01.09 was meant to remind you of.
Problem 7: Two plane mirrors intersect at an angle θ, as shown here:
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
φ
α
θ
A light ray is incident on one of the mirrors at an angle of α = 20o . The angle φ between
the incident and outgoing rays is 94.4o . What is θ ?
(A)
(B)
(C)
(D)
(E)
34.4o
44.4o
47.2o
54.4o
67.2o
Answer: (C)
Please read and work carefully through the posted solution of HW Problem P01.11, for
details. Using the relations between the different angles given there, you can easily work
your way backwards, to get from a given φ = 94.4o to the asked-for θ = 47.2o .
More generally, if you do not fall into the trap of immediately plugging numbers into your
equations, but rather work with general variables α, φ and θ, following the approach of
P01.11, you can prove the extremely simple result that:
φ=2θ,
which is independent of α ! Hence, θ = φ/2 = 94.4o /2 = 47.2o .
Problem 8: In flint glass, red light has an index of refraction (IoR) nR = 1.865 and violet
light has an IoR nV = 1.896, while both have an IoR nAir = 1.000 in air. If a beam of white
light is incident from air upon a block of flint glass at an angle of incidence of 50o the angle
of divergence, enclosed between the red and the violet beam inside the glass, is
(A)
(B)
(C)
(D)
(E)
0.267o
0.344o
0.421o
1.68o
2.27o
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Answer: (C)
This problem is identical to HW Problem P01.19(a): please read and work through the
posted solution of that problem carefully, to get all the details. So, let Θ = 50o be the given
angle of incidence of both read and violet rays entering at the surface of the slab; and let
Θ0R and Θ0V denote the angle of refraction of the red and violet light rays after entering the
slab, respectively.
Then, by Snell’s law: Θ0R = arcsin[(1/nR ) sin(Θ)] = arcsin[(1/1.865) sin(50o )] = 24.251812o ;
and likewise, Θ0V = arcsin[(1/nV ) sin(Θ)] = arcsin[(1/1.896) sin(50o )] = 23.830474o . Hence,
the angle of divergence between red and violet is: ∆Θ ≡ |Θ0R − Θ0V | = 24.251812o −
23.830474o ∼
= 0.421o
Problem 9: A postage stamp placed 4.00cm to the left of a lens produces an image 32.0cm
to the left of the lens. What is the focal length of the lens ?
(A)
(B)
(C)
(D)
(E)
−4.57cm
+4.57cm
−3.56cm
+3.56cm
Cannot be determined from infomation given.
Answer: (B)
Given are d = +4.00cm and d0 = −32.00cm. Note here that ”to the left of the lens” is the
incoming side (=side on which rays from the object are entering the lens); hence d > 0. Also
”to the right of the lens” is the outgoing side (=side from which rays forming the image are
leaving the lens). Hence, d0 < 0, since the image is not on the outgoing side.
Hence: f = (1/d + 1/d0 )−1 = [1/(4.00cm) + 1/(−32.0cm)]−1 = +4.57cm
Problem 10: If the postage stamp in Problem 9 is 1.25cm tall its image produced by the
lens will be
(A)
(B)
(C)
(D)
(E)
real, inverted and 0.156cm tall;
virtual, erect and 0.156cm tall;
virtual, inverted and 10.0cm tall;
virtual, erect and 10.0cm tall;
real, inverted and 10.0cm tall.
Answer: (D)
See Problem 9: Given are d = +4.00cm and d0 = −32.00cm. So d0 < 0 and hence the image
is virtual. Also, given is the original object height h = 1.25cm.
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
So, m = −d0 /d = −(−32.00cm)/(+4.00cm) = +8.00 > 0, hence image is erect. Lastly, the
height of the image is h0 = mh = (+8.00) × (1.25cm) = +10.00cm.
Problem 11: A convergent lens no.1, placed to the right of a small sand grain, produces an
image of the grain 24.7cm to the right of lens no.1. If a divergent lens no.2 of focal length
f2 = −7.42cm is now placed 18.6cm to the right of lens no.1, the final image produced by
lens no.2 appears approximately
(A)
(B)
(C)
(D)
(E)
3.35cm to the left of lens no.2
3.35cm to the right of lens no.2
34cm to the right of lens no.1
34cm to the left of lens no.2
34cm to the right of lens no.2
Answer: (E)
This problem is similar to (but a lot easier than!) HW Problems P02.08 and P02.09. Please
review the posted solutions of these two problems carefully before embarking on this one.
Also, it is critically important that you draw a big, clean picture of this set-up, showing the
relative positioning of lenses, original object, image by 1st lens; and also showing the arrows
for relevant distances such as d01 and d2 and L. Then you will see that: d01 = +24.7cm and
L = 18.6cm and hence d2 = L − d01 = 18.6cm − 24.7cm = −6.1cm.
Since d2 < 0, lens no. 2 has a virtual object, located 6.1cm to the right of lens no. 2. Then
d02 = (1/f2 − 1/d2 )−1 = [(1/(−7.42cm) − 1/(−6.1cm)]−1 ∼
= +34.cm . Since d02 > 0, the image
must be on the outgoing side of lens no. 2, i.e., 34.cm to the right of lens no.2
Problem 12: A microsope produces a final image of a bacterial cell (as seen through the
eyepiece) which appears to be 1.2cm in diameter and located 25.0cm from the eyepice on the
incoming side of the eyepiece lens. Assume the microscope has achieved a 5000-fold angular
magnification, compared to viewing the bacterium from a 25.0cm nearpoint distance without
instrument. What is the approximate diameter of the actual bacterial cell ?
(A)
(B)
(C)
(D)
(E)
24. µm
4.2 µm
2.4 µm
0.42 µm
0.24 µm
Answer: (C)
Again, it is critically important that you draw a big, clean picture of this set-up. Given:
|h02 | = 1.2cm and d02 = −25.cm. Also, dref = 25.cm and |M | = 5000.
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Then: Θe ∼
= tan(Θe ) = |h02 /d02 | = 1.2/25. = 0.048rad.
Use |M | = Θe /Θref to solve for the reference angle, subtended at the eye by original
object, when viewed from reference distance dref without instrument: Θref = Θe /|M | =
0.048rad/5000. = 9.6 × 10−6 rad.
Then, use Θref ∼
= tan(Θref ) = h1 /dref to solve for the size of the original object: h1 =
Θref dref = (9.6 × 10−6 ) × (25.cm) = 2.4 × 10−4 cm = 2.4 × 10−6 m.
9
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Formula Sheet
Wave Propagation and Wave Nature of Light
Periodic Wave Condition:
v = λf =
λ
τ
Index of Refraction (IoR) for electromagnetic waves, definition:
n=
c
v
Doppler Effect:
f0 = f 1 ±
u
v
Reflection and Refraction of a Single Ray
For angle of incidence (Θ1 ), angle of reflection (Θ̄1 ) and angle of refraction (Θ2 ):
Archimedes’ Law of Reflection:
Θ1 = Θ̄1
Snell’s Law of Refraction:
sin(Θ1 )/v1 = sin(Θ2 )/v2
where v1 and v2 are the speeds of wave propagation in the respective media. For electromagnetic waves, this is also written as
Snell’s Law in IoR Form:
n1 sin(Θ1 ) = n2 sin(Θ2 )
where n1 ≡ c/v1 and n2 ≡ c/v2 are the corresponding IoR.
10
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Image Formation and Optical Instruments
Image-Object-Relations for a Single Device (Lens or Mirror):
Mirror / Thin Lens Equations:
1
1
1
+ 0 =
d d
f
m≡
,
h0
d0
=−
h
d
where d = object distance, d0 = image distance, h = object height, h0 = image height, f =
focal length, m = lateral magnification.
Focal length for reflection at a curved surface (mirror):
f = R/2
Sign Conventions for Single Device (Lens or Mirror):
d > 0 (real object) if object on ”incoming” side; else d < 0 (virtual object).
d0 > 0 (real image) if image on ”outgoing” side; else d0 < 0 (virtual image).
If m > 0 then image erect (upright) rel. to object; else, if m < 0 then image inverted
(upside-down) rel. to object.
If f > 0 then F on ”incoming” and F 0 on ”outgoing” side; else, if f < 0 then F not
on ”incoming” and F 0 not on ”outgoing” side.
R > 0 if center of spherical surface on ”outgoing” side; else R < 0. (For reflection at a
curved surface only.)
Compound Instrument:
If image by device 1 serves as object for device 2 and L =separation of devices then
d01 + d2 = L,
h2 = h01 ,
m12 ≡
h02
= m1 × m2 ,
h1
where m12 is total lateral magnification of image by 2 (h02 ) relative to object of 1 (h1 ).
Angular Magnification:
Definition:
|M | = Θe /Θref
where Θe =angle subtended at eye by optical instrument’s final image; and Θref =reference
angle subtended at eye by original object viewed without optical instrument at reference
distance dref
11
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Algebra and Trigonometry
2
az + bz + c = 0
sin θ =
opp
,
hyp
⇒
cos θ =
z=
adj
,
hyp
−b ±
√
b2 − 4ac
2a
tan θ =
opp
sin θ
=
adj
cos θ
sin2 θ + cos2 θ = 1
For very small angles θ (with |θ| 90o ):
sin θ ∼
= tan θ ∼
= θ (in radians)
Numerical Data
Speed of light in vacuum:
c = 3.00 × 108 m/s
Range of vacuum wavelengths visible to the human eye: from 700nm to 400nm.
Index of refraction of
vacuum: nV = 1.0 (exact).
air: nAir ∼
= 1.0.
water: nWater ∼
= 4/3 ∼
= 1.33.
Other numerical inputs (IoR, angles, lengths, etc.) will be provided with each problem
statement.
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