Download PHYS 101 Light, Department of Physics EXAM 1 – Solutions Name

PHYS 101 Light, Department of Physics
Prof. Ruiz/UNCA
EXAM 1 – Solutions
NO CALCULATOR
Name _____________________
February 19, 2013
Multiple Choice. Neatly mark your answer on the supplied scan card with a No. 2 pencil.
Q1. In The Bell Jar by Sylvia Plath, the author’s description of her physics class included (A)exciting slides
(B)fascinating words like polarization and interference (C)hideous, cramped, scorpion-lettered formulas
(D)oversimplified conceptual explanations (E)magic tricks. Plath’s course was opposite to what we do. She
was turned off by the math, but she worked hard and got an A.
Q2. In the “Bell Jar” demonstration in physics, which of the following is
false. (A)The “At” on the dial stands for atmospheric pressure. (B)The
bell cannot be heard after the air is pumped out. (C)The bell cannot be
seen after the air is pumped out. (D)The pump is noisy when the air is
being pumped out. (E)The sound gets quieter as the air is pumped out.
Sound needs a medium to travel in while light can travel through
vacuum.
Q3. Which are true concerning the method of science? If all are true,
choose E. (A)Science is an art, much like the art of being a detective trying to solve a mystery. (B)You can use
observations, theoretical analysis, hunches, clues, measurements, testing samples, chance, luck, and control
variables. (C)Your work is based on reason and logic, supported by observation and data. (D)Science is
reproducible; i.e., given ample observations and understanding of laws, two scientists will reach the same
conclusion. (E)All of the above. See your e-text (Chapter A3) for discussion.
Q4. If the angle of incidence is 50° , then the angle of reflection is (A)less than 50° (B)50° (C)greater than 50°.
The law of reflection states that these angles are always equal.
Q5. Light enters glass from air at an angle of incidence equal to 50°. The angle of refraction is (A)less than 50°
(B)50° (C)greater than 50°. In going from a fast (air) to a slow (glass) medium the light bends towards the
normal. This means a smaller angle for the angle of refraction.
Q6. The index of refraction of glass is (A)1.00 (B)1.33 (C)1.50 (D)2.4 (E)3.0.
Q7. What is the speed of light (km/s) in a medium having an index of refraction of n = 2? (A)100,000
(B)150,000 (C)200,000 (D)300,000 (E)600,000 Use n = c/v. Method 1. Plugging in the numbers we have 2 =
300,000/v where the speeds are given in km/s. Therefore, v = 150,000 km/s. Method 2. n = c/v = 2.
Therefore, v = c/2 = 300,000/2 = 150,000 km/s.
Q8. For light leaving a medium to go into water, the critical
angle is greatest for the (A)air/water (B)diamond/water
(C)glass/water (D)vacuum/water (E)water/water interface. The
critical angle occurs when you go from a “slower” to a
“faster” medium. Since we are going to water, the only
possible choices are diamond (n = 2.4) and glass (n = 1.5),
i.e., those with n > 1.33 (water). Refraction is more dramatic
when there is a greater change in the index. But greater
refraction means you reach the critical angle θc more
quickly, so it will be smaller. So you need glass, the one as
close to water as possible.
Q9. A few folks at a
party are looking down
at a dime resting in the
bottom of a pan of water
20 cm deep? For these observers, the dime will appear
to be (A)not as deep as 20 cm (B)20 cm below the
water surface (C)deeper than 20 cm (D)invisible. Sketch
a ray of light leaving the coin and traveling to the
water where the angle of incidence in greater than
zero (i.e., a slanted ray). Sketch in a normal, i.e.,
vertical line where your ray meets the water-air
interface. The ray will refract, bending away from the vertical (normal). Sketch in an eye to receive the
ray. The observer thinks the ray came from a point above the coin, concluded by tracing the ray
backwards with a dotted line.
Q10. Which of the cases in the figure at the right
depicts a pencil in water correctly? (A)A (B)B (C)C
Q11. For light leaving the pencil under water and
proceeding to an observer in air at the same height as
the bottom section of the pencil, considering all
interfaces, the light passes through (A)1 (B)2 (C)3
interfaces. Water-glass and glass-air.
Q12. If you increase the wavelength, the frequency
(A)decreases (B)stays the same (C)increases. Remember that if one goes up, the other goes down as
necessary by the formula: speed = wavelength x frequency, i.e., v = λ f.
Q13. WCQS broadcasts at 88 MHz, which is an FM frequency. The M stands for (A)103 (B)104 (C)105 (D)106
(E)107. Capital “M” stands for Mega, i.e., million. One million = 1,000,000. This is a 1 followed by 6 zeros.
Q14. The wavelength for WCQS (88.1 MHz, which you can take to be 88 MHz) is 3.4 meters. Therefore, the
wavelength for WPEK 880 kHz, an AM radio station in Fairview, NC, must be (A)3.4 (B)34 (C)340 (D)3400
(E)34,000 meters. Method 1. Note that 880 kHz = 0.88 MHz. Then, since the wavelength for 88 MHz is 3.4
meters, the wavelength for 8.8 MHz has to 34 meters. When frequency goes down, wavelength goes up
(see Q12). Finally, the wavelength for 0.88 MHz has to be 340 meters. Method 2. You note that 880
kHz/88 MHz = 880,000/88,000,000 = 880/88,000 = 88/8800 = 1/100. So this means the wavelength has to
increase a hundredfold. Your instructor grew up in a section of Camden, NJ called Fairview.
Q15. In which direction will the little sphere move in the configuration
shown at the left? (A)left (B)neither left nor right (C)right As in the
homework, we assume the larger spheres hardly move due to their
large mass or consider them to be fixed. The center sphere is
neutral, i.e., contains no net charge. Therefore, it is immune to the
electric force.
Q16. In which direction will the little sphere move if its charge is +1 instead of 0? (A)left (B)neither left nor
right (C)right “Unlikes” attract so a plus sphere in the middle will move to the left.
Q17. The force between the magnets shown at the right is
(A)attractive (B)neutral (C)repulsive. The rule with magnetic
poles is that “unlikes” attract and “likes” repel. So the south
and north poles that are almost touching are attracted to each other.
Q18. Which of the following is false concerning the electromagnet at the left? (A)If you
turn the paper clip around the electromagnet will still work. (B)If you turn the battery
around the electromagnet will still work. (C)The magnetic force increases with current.
(D)If you turn the nail inside the coils around so it faces the other way the
electromagnet will not work. The nail can face either way.
Q19. Assuming an insignificant change in current, if you wrap more windings around
the nail, your magnetic effect (A)decreases (B)stays the same (C)increases. The loops
reinforce the effect since each circular current adds to the produced magnetic field.
Q20. Each pickup in an electric guitar contains a coil wrapped around a
magnet. When the steel string vibrates near the pickup, the magnetic field
inside the coil is disturbed. Surges of generated electrical current are
produced in the coil “dancing” to the tune of the vibration. This
generation of electricity in the pickup is an excellent practical application
of a law of (A) Ampère (B)Faraday (C)Fermat (D)Galileo
(E)Michelson. Ampère goes with the electromagnet, Fermat is known
for the principle of least time in optics, Galileo was not able to measure the speed of light by the echo
technique, and Michelson was the first to accurately measure the speed of light.
Q21. Consider the “wiggling” or “shaking back and forth” of three things: (A)a charged object (B)a magnet
(C)a neutral object (charges balance out, canceling so that the net charge is zero). Which generates
electromagnetic waves? (A)A only (B)B only (C)C only (D)A and B (E)A, B, and C Since light is an
electromagnetic (EM) wave, if you produce a changing electric field you generate a magnetic field and
vice versa. So the wiggling charge changes the surrounding electric field and a wiggling magnet changes
the surrounding magnetic field. In either case, EM ripples are produced.
Q22. If the shaking in Q21 is accomplished by your hand, you then generate (A)infrared waves (B)microwaves
(C)ultraviolet waves (D)radio waves (E)x-rays. What is the fastest you can shake an object – about 4 times
per second? This is at the super low-frequency end of the EM spectrum. Therefore, shaking a charge or
magnet with your hand produces an EM wave that must be in the radio region of the spectrum.
Q23. Who in this list is not known for discussions or experiments involving the speed of light? (A)Empedocles
(B)Krebs (C)Galileo (D)Michelson (E)Roemer Krebs is an artist. Remember the laser art?
Q24. Michelson was the first to accurately measure the speed of light. His apparatus included a mirror device
with a total of (A)2 (B)4 (C)5 (D)8 (E)12 reflecting sides. He used a rotating octagonal mirror.
Q25. A light wave consists of propagating electric (E) and magnetic (M) oscillating fields. Which fields are
perpendicular to the direction of propagation for the wave? (A)only E (B)only M (C)both E and M
(D)neither. Light is a transverse wave, which means all disturbances (or changes) in the fields are
transverse, i.e., perpendicular to the direction of travel (propagation).
Use the left diagram for Q26-Q28. You see 5 resonance
graphs superimposed at the left, where the lowest two are
extremely weak in response values.
Q26. The resonance frequency for the second tallest
graph, in the appropriate units given, is (A)1 (B)2 (C)4
(D)6 (E)8. The resonance occurs where that tall
mountain is. The frequency is read off the horizontal
axis to get 1. The answer is the same for all of them.
Q27. The amplitude of response is weakest for all of the
curves at frequency value (A)0.5 (B)1 (C)2 (D)4 (E)6
seen in the figure. The response amplitude is weakest in
the graph at frequency value equal to 2. This makes
sense since at the high-frequency end the response of a
system undergoing driven oscillations is the weakest –
approaching zero as you get to even higher and higher
frequencies. Take a shoe with a shoelace and try it. If
you wiggle super fast the hanging shoe attached to the
shoelace will hardly move.
Q28. If the tension in the restoring force should increase in this system, then the resonance peak would (A)shift
to the left (B)stay where it is (C)shift to the right. Hint: What happens to the natural frequency of a wiggling
board if the board gets stiffer? If you increase the tension, the system will want to wiggle faster, i.e., at a
higher frequency. So the resonance frequency increases, i.e. gets shifted to the right.
Q29. How many kilometers are there in a Megameter? (A)100 (B)1000 (C)10,000 (D)100,000 (E)1,000,000 A
kilo represents 1000 and a Mega represents 1,000,000. The answer is 1000 since 1000 x 1000 = 1,000,000.
Q30. How many MHz are there in a GHz? (A)100 (B)1000 (C)10,000 (D)100,000 (E)1,000,000 The “M”
represents Mega (million) = 1,000,000 and Giga (billion) represents 1,000,000,000. The answer is 1000
since 1000 x 1,000,000 = 1,000,000,000. A Gigabyte is 1000 times bigger than a Megabyte.
Q31. How many nm are there in one mm? (A)100 (B)1000 (C)10,000 (D)100,000 (E)1,000,000 Method 1. A
nm is a billionth (10−9 = 0.000 000 001) and a mm is a thousandth (10−3 = 0.001). You have to move the
decimal for the billionth a total of 6 times to get to the place where it needs to be for a thousandth.
Method 2. thousandth/billionth = (1/1,000) x 1,000,000,000 using KEEP-CHANGE-INVERT for dividing
by fractions. The answer is 1,000,000,000/1,000 = 1,000,000.
Questions Q32 to Q34 refer to the figure below. Q32.
Which wave at the left has the lower frequency? (A)a
(B)b The one with the lower frequency is the one with
the less amount of wiggles.
Q33. What is the wavelength of wave “b” if “a” has a
wavelength of 6 cm? (A)2 (B)3 (C)6 (D)12 (E)24 cm The
wavelength of “b” is worth two of the wavelengths of
wave “a.”
Q34. If the frequency of wave “a” is 20 Hz, the frequency for wave “b” is (A)5 (B)10 (C)20 (D)40 (E)80 Hz.
Method 1. The top wave wiggles twice as much as the lower one. Method 2. Since the wavelength for the
lower one is twice that for the upper one, the frequency of the lower one has to be one half that of the
upper one.
Q35. The mustache on this astronomer’s face is due to (A)charcoal (B)crayon (C)a
fake attachment (D)ice (E)ink. This is a heat photo so anything real cold appears
dark. She had just “painted” some cold on her face using an ice cube.
Q36. The left image was taken in the (A)infrared (B)microwave (C)ultraviolet (D)xray.
Q37. The ozone layer is known for its important absorption in the (A)infrared
(B)microwave (C)UV (D)visible.
Q38. Carbon dioxide can trap light that is in the (A)infrared (B)radio (C)UV
(D)visible part of the spectrum.
Questions 39 through 41 refer to the figure at the left. This is Gel 701
made by Lee Filters.
Q39. The color depicted by the spectral distribution in the graph is a
(A)blue at its purest (B)green (C)magenta (D)red at its purest
(E)yellow. When you have a double peak, one in the blue and one in
the red, you have magenta.
Q40. Which spectral components are least present in this nonspectral
color? (A)blue (B)green (C)red (D)violet Magenta is lacking in green.
Q41. The company “Lee Filters” recommends this filter for stage
lighting as a(n) (A)bright yellow to simulate sunlight (B)cool lavender
for romantic evening exteriors (C)evergreen atmosphere for a forest
scene (D)hot red for a fiery scene (E)light gray for a neutral background. Method 1. You know lavender is a
magenta. Method 2. You eliminate yellow, green, red, and gray.
Questions 42-43 refer to the color depicted by the
spectral power distribution at the right. Note that the
graph includes the ultraviolet, visible, and infrared.
Q42. To the human eye, the color appears (A)blue
(B)green (C)magenta (D)red (D)yellow. Be careful. It
is not magenta since the second peak is in the
infrared which we cannot see. The tiny amount of
red present is not enough to compete with all the blue. Also, ignore the ultraviolet.
Q43. The transmission for 500-nm is (A)20% (B)40% (C)60% (D)80% (E)100% Read it off the above graph.
Q44. The spectral power distribution a light emitting diode
LED is at the left. This LED is (A)blue (B)cyan (C)green
(D)orange. The peak is over about 630 nm which is in
the orange region (600-650 nm).
Q45. The dominant wavelength for the color depicted at the
left is (A)615 (B)635 (C)655 (D)665 (E)700 nm. Read it
off the graph at the left. Find the peak and drop down
to estimate the wavelength number.
Q46. In the Munsell color system, colors get less saturated
as you (A)proceed up the color tree (B)proceed down the
color tree (C)move toward the central trunk (D)move
away from the central trunk (E)keep changing direction.
Q47. In the Munsell color system, colors get less bright as
you (A)proceed up the color tree (B)proceed down the
color tree (C)move toward the central trunk (D)move away
from the central trunk (E)keep changing direction.
Q48. In a vacuum such as outer space, green light travels (A)faster (B)slower (C)the same speed when
compared to infrared light. All EM waves travel at the same speed in a vacuum, or as physicists like to say
“in vacuum.”
Q49. Red light refracts less than blue light when white light enters glass. Therefore, the speed of the red light in
glass must be (A)less (B)the same (C)greater, when compared to the speed of the blue light in glass. The index
of refraction for red is less, which means it does not slow down as much when it enters the glass.
Therefore, its speed is greater when compared to that for the blue.
Q50. Which of the following is most pure from a spectral-power-distribution point of view? (A)Carolina blue
(B)Acapulco gold (C)wedding white (D)laser red (E)winter green A laser emits a pure color, i.e., a color
with a single wavelength. We call this light monochromatic.