# An observer O, facing a mirror, observes a light

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An observer O, facing a mirror, observes a light
```An observer O, facing a mirror, observes a light
ource S. Where does O perceive the mirror
mage of S Physics
to be1112
located?
1.
2.
3.
4.
5.
Spring 2009
University of Georgia
Instructor: HBSchüttler
1
Solution to PHYS 1112 In-Class Exam #1B
2
Thu. Feb. 5, 2009, 2:00pm-3:15pm
3
4
Conceptual Problems
The image of S cannot be seen by O in this
Problem 1: An observer O, facing a mirror, observes a light source S, with the observer,
configuration.
source and mirror positioned as shown here:
mirror
O
1
3
S
4
2
Where does O perceive the mirror image of S to be located ?
(A)
(B)
(C)
(D)
(E)
Position 1.
Position 2.
Position 3.
Position 4.
The image of S cannot be seen by O in the configuration shown above.
Any rays emerging from S which strike the mirror are reflected off the mirror (acc. to
Archimedes’ law) in such a way that, after reflection, they appear to be emerging from
Position 4. This is true (a) for all rays emerging from S regardless of where the rays from S
strike the mirror; and (b) regardless of how far up or down the mirror extends, i.e., even if
the mirror surface does not cover the wall area immediately opposite of S.
It is critically important that you draw your own big, clean picture of this set-up, tracing a
few (at least two) rays from S to the mirror and then, after reflection with Θ̄ = Θ, tracing
the reflected rays backwards to see where they intersect behind the mirror: at Position 4.
This is similar to the in-class quiz on buying the ”Shortest Possible Mirror on the Wall”. You
can still see your own feet (on the ground) in this mirror, even though the mirror extends
downward to only half of your height above the ground.
1
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
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 undergo total internal reflection if incident from below the water surface for
any angle of incidence greater than 13.364o ;
(B) will undergo total internal reflection if incident from above the water surface with
an angle of incidence of 8.5o ;
(C) will undergo total internal reflection if incident from below the water surface with
an angle of incidence of 8.5o ;
(D) will have an angle of refraction smaller than the angle of incidence if the beam is
incident from below the water surface;
(E) will have an angle of refraction greater than the angle of incidence if the beam is
incident from below the water surface.
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 not both satisfied by the conditions
stated in (A), (B) or (C). For this reason, (A), (B) and (C) are wrong. [In (A) and (C)
incidence is from below, i.e. from water into air, hence TIR cannot occur, regardless of
angle of incidence. In (B), incidence is from above, i.e. from air into water, as required
for TIR condition (a); but the angle of incidence, Θ = 8.5o , is less than the critical angle
(crit)
ΘAir = 13.364o , hence again TIR does not happen.]
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!)
2
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
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
answer (D) is correct and (E) is wrong.
Problem 3: If a real object is placed more than two focal lengths away from a concave
mirror(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
Being ”more than two focal lengths away” means d > 2f . From that, and from f > 0, it
follows that
1
(1/d) < 1/(2f ) = (1/f )
2
(by taking the inverse of the inequality d > 2f ). Therefore
1
1
(1/d0 ) = (1/f ) − (1/d) > (1/f ) − (1/f ) = (1/f )
2
2
and hence [by taking the inverse of the inequality (1/d0 ) > 21 (1/f ) and using d > 2f again]
it follows that
d0 < 2f < d
but also (1/d0 ) > 0 which implies d0 > 0, i.e., the image is real.
From 0 < d0 < d and m = −d0 /d it then follows that
|m| = d0 /d < 1
i.e., the image is reduced in absolute height, compared to the object; and it also follows
that
m = −d0 /d < 0
i.e., the image is inverted, relative to the object. So, the answer is (C).
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 = 10cm so that d > 2f = 8cm, i.e., the object is placed more than two focal lengths
from mirror. Then, you’ll get d0 = (1/f − 1/d)−1 = [(1/4) − (1/10)]−1 cm = +(20/3)cm and
from that m = −d0 /d = −(20/3)/10 = −2/3. Therefore:
Since d0 > 0, the image is real.
3
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Since m < 0, the image is inverted, relative to object
Since |m| < 1, the image is reduced in absolute height, relative to object.
Problem 4: Two non-parallel light rays initially converge to a single point on a flat screen
so that the normal to the screen is enclosed between the two incident rays, as shown here:
A slab of glass is now placed somewhere in front of the screen, in the path of the light
rays, so that the slab’s two planar glass surfaces are parallel to the screen. The index of
refraction (IoR) of the glass is greater than the IoR of the surrounding air. Where is the
new convergence point of the rays?
(A)
(B)
(C)
(D)
(E)
On the screen (unchanged);
Toward the glass slab, in front of the screen (i.e., between slab and screen);
Further away from the glass slab, behind (≡ to the right of) the screen;
Inside the glass slab;
Any of the above [(A), (B), (C) or (D)] are possible, depending on the angles of
incidence of the two rays.
At the 1st refraction, when entering the slab, both rays are refracted towards the normal
inside the slab, since nGlass > nAir . So, Ray 1 is bent upward and Ray 2 is bent downward,
inside the slab, as shown in the figure above.
Then, at the 2nd refraction, when leaving the slab, both rays are refracted away from the
normal (again because nGlass > nAir ) so that the 2nd refraction is exactly ”un-doing” the
change of direction the two rays suffered as a result of the 1st refraction. This happens as a
result of Snell’s law and as a result of the fact that, inside the slab, the angle of refraction
4
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
at the left surface equals angle of incidence at the right surface (... and these two angles are
equal because the left and right slab surfaces are parallel planes).
As a consequence, (a) the direction of Ray 1 exiting the slab (to the right of slab) is parallel
to the direction of Ray 1 entering the slab, but shifted upward relative to the entering Ray
1; and (b) likewise the direction of Ray 2 exiting the slab (to the right of slab) is parallel to
the direction of Ray 2 entering the slab, but shifted downward relative to the entering Ray
2.
In the absence of the slab, without refraction, both rays would converge to their point of
intersection located exactly on the screen, as indicated by dotted lines in the figure. Hence,
as a result of the upward shift of Ray 1 and of the downward shift of Ray 2 (both caused by
the refractions at the slab), this point of convergence is shifted to the right, ı.e., behind the
screen, as shown in figure.
Numerical Problems
Problem 5: A beam of monochromatic light with a frequency of 3.571 × 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 840nm, have the same frequency as in air, and be invisible
the human eye;
(B) have a wavelength of 630nm, have the same frequency as in air, and be invisible
the human eye;
(C) have a wavelength of 840nm, have a frequency of 4.761 × 1014 Hz, and be invisible
the human eye;
(D) have a wavelength of 840nm, have a frequency of 4.761 × 1014 Hz, and be visible
the human eye.
(E) have a wavelength of 630nm, have a frequency of 4.761 × 1014 Hz, and be visible
the human eye;
to
to
to
to
to
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 ).
Therefore, f = 3.571 × 1014 Hz in both air and water and λAir = (3.00 × 108 m/s)/(3.571 ×
1014 Hz) = 840nm, but λWater = (840nm) × [1.00/(4/3)] = 630nm.
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
5
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
∼ 400nm. The light ray in this problem is not visible, since its 840nm vacuum wavelength
does not fall within the 700−400nm visible range. The 630nm 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 9.34GHz. You’re
reflecting from your car. The reflected frequency is measured to be 6.98 × 10−6 % larger than
the original frequency sent out by the radar gun. Does this percentage of frequency change
depend on the frequency sent out by the radar gun ? How fast were you driving ?
(A)
(B)
(C)
(D)
(E)
driving
driving
driving
driving
driving
20.94m/s, percentage does not depend on frequency;
10.47m/s, percentage does depend on frequency;
10.47m/s, percentage does not depend on frequency;
5.24m/s, percentage does depend on frequency;
5.24m/s, percentage does not depend on frequency.
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. Solving for u, you get u = (c/2)×(p/100%)
and its value is u = [(3.00 × 108 m/s)/2] × [(6.98 × 10−6 %)/(100%)] = 10.47m/s.
Note that it is very difficult (if not impossible!) to evaluate u from f and p accurately if you
first try to calculate the reflected wave frequency f 00 = f + pf and then solve for u, using
f 00 = f (1 + 2u/c): f 00 and f are so very close in value, that their difference, pf , is likely
smaller than the rounding error of your calculator. So, you really need to do the algebra first
in this problem [to derive p = 2(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: A flat circular mirror of radius 0.190m lies flat on the floor. Centered above
the mirror at a height of 0.590m is a light source. The circular spot formed on the ceiling by
the reflection of the light has a diameter of 2.20m. How high is the ceiling, measured from
the floor ?
(A)
(B)
(C)
(D)
(E)
1.41m
2.83m
3.00m
5.65m
6.00m
6
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
This problem is a slight variation on HW Problem P01.12. Please read and work carefully
through the posted solution of P01.12, for a detailed drawing, for details of the solution, and
for the notation used here:
In this version of the problem, r =radius of mirror on the floor, h =height of light above floor
and D = 2(r + x) =diameter of bright circle on ceiling are given; and we need to solve for
H =height of ceiling above floor. So, solving for x first, x = D/2 − r = (2.20/2 − 0.190)m =
0.910m. Then, using H/x = h/r, we get H = xh/r = (0.910m) × (0.590/0.190) = 2.83m .
Problem 8: In flint glass, red light has an index of refraction (IoR) nR = 1.765 and violet
light has an IoR nV = 1.796, while both have an IoR nAir = 1.000 in air. A beam of white
light enters a triangular prism of flint glass from air at normal incidence at the front surface
and then stikes the prism’s back surface at an angle of incidence φ = 25.0o as shown here:
φ
Red
Violet
The prism is surrounded by air. What is the angle of divergence, enclosed between the red
and the violet beam after leaving the prism through the back surface ?
(A)
(B)
(C)
(D)
(E)
0.122o
0.244o
0.488o
1.140o
2.073o
This problem is similar to HW Problem P01.19(a). Please read and work through the posted
solution of that problem carefully.
At the front (lefthand) prism surface, the incident (red+violet) beam has a 0o angle of
incidence, and therefore by Snell’s law also a 0o angle of refraction. Therefore, φ = 25.0o is
the given angle of incidence of both read and violet rays at the back (righthand) surface of
the prism.
Let φ0R and φ0V denote the angle of refraction of the red and violet light rays after leaving
the prism, respectively.
7
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Then, by Snell’s law: φ0R = arcsin[nR sin(φ)] = arcsin[(1.765 sin(25.0o )] = 48.238287o ; and
likewise, φ0V = arcsin[(nV sin(φ)] = arcsin[1.796 sin(25.0o )] = 49.378090o .
Hence, the angle of divergence between red and violet is: ∆φ ≡ |φ0R − φ0V | = 49.378090o −
48.238287o ∼
= 1.140o
Problem 9: A candle placed 40.00cm to the left of a curved mirror produces an image
60.00cm to the left of the mirror. What is the spherical radius of the mirror and where is
the center C of that sphere ?
(A)
(B)
(C)
(D)
(E)
−24.00cm, C to the left of the mirror;
+24.00cm, C to the left of the mirror;
−48.00cm, C to the right of the mirror;
+48.00cm, C to the left of the mirror;
Cannot be determined from infomation given.
The lefthand side of the mirror is both incoming side and outgoing side because the object
(candle) is placed to the left of the mirror and the image-forming rays travel to the left
after reflection at the mirror. Thus d = +40.00cm > 0 (object on incoming side)and d0 =
+60.00cm > 0 (image on outgoing side).
Then R = 2f = 2(1/d + 1/d0 )−1 or R = 2 × [(1/40.00) + (1/60.00)]−1 cm = +48cm . Since
R > 0, the center of the sphere C must be on the outgoing side, i.e., to the left of the mirror.
Problem 10: If the candle in Problem 9 is 15.0cm tall its image produced by the mirror
will be
(A)
(B)
(C)
(D)
(E)
real, inverted and 10.0cm tall;
virtual, erect and 10.0cm tall;
virtual, inverted and 22.5cm tall;
virtual, erect and 22.5cm tall;
real, inverted and 22.5cm tall.
See Problem 9: d = +40.00cm > 0 and d0 = +60.00cm > 0 are given. So, m = −d0 /d =
−60.00/40.00 = −1.50 and h0 = mh = (−1.50) × (15.0cm) = −22.5cm.
Since d0 > 0, the image is real.
Since m < 0, the image is inverted.
8
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Problem 11: A convergent lens (no.1), placed to the right of a small gem stone produces an
image of the gem 22.65cm to the right of lens no.1. If a divergent lens (no.2) of focal length
f2 = −6.42cm is now placed somewhere to the right of lens no.1, the final image produced
by lens no.2 appears approximately 35.0cm to the right of lens no.2. Approximately, how
far apart are the two lenses ? Is the object of lens no. 2 virtual or real ?
(A)
(B)
(C)
(D)
(E)
lenses
lenses
lenses
lenses
lenses
are
are
are
are
are
17.2cm
17.2cm
14.8cm
14.8cm
57.7cm
apart,
apart,
apart,
apart,
apart,
lens
lens
lens
lens
lens
no.2
no.2
no.2
no.2
no.2
object
object
object
object
object
is
is
is
is
is
real
virtual
real
virtual
real
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 = +22.65cm and
d02 = +35.0cm .
Then d2 = (1/f2 − 1/d02 )−1 = [1/(−6.42) − 1/(+35.0)]−1 cm = −5.425cm So, lens no.2 has a
virtual object, since d2 < 0.
Also, the lens-to-lens spacing is L = d0 + d2 = (+22.65 − 5.425)cm ∼
= 17.2cm .
1
Problem 12: A telescope produces a final image of a skyscraper (as seen through the
eyepiece) which appears to be 3.6cm tall and located 40.0cm from the eyepiece on the
incoming side of the eyepiece lens. Assume the skyscraper is at a distance of 20km; and the
telescope has achieved a 9.0-fold angular magnification, compared to viewing the building
from the actual 20.0km distance without instrument. What is the approximate height of the
actual skyscraper ?
(A)
(B)
(C)
(D)
(E)
500m
450m
400m
300m
200m
Again, it is critically important that you draw a big, clean picture of this set-up. Given:
|h02 | = 3.6cm and d02 = −40.cm. Also, dref = 20.0km and |M | = 9.0.
9
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Then: Θe ∼
= tan(Θe ) = |h02 /d02 | = 3.6/40. = 0.090rad.
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 | =
Then, use Θref ∼
= tan(Θref ) = h1 /dref to solve for the height of the original object: h1 =
Θref dref = (0.01) × (20000.m) = 200m . So, (E) is the answer.
10
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.
11
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 = m1 × m2 ,
where m12 is total lateral magnification of image by 2 relative to object of 1.
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
12
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Algebra and Trigonometry
2
az + bz + c = 0
sin θ =
opp
,
hyp
⇒
cos θ =
z=
,
hyp
−b ±
√
b2 − 4ac
2a
tan θ =
opp
sin θ
=
cos θ
sin2 θ + cos2 θ = 1
For very small angles θ (with |θ| 90o ):
sin θ ∼
= tan θ ∼
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.
13
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