Applications of the Photon Model - Solutions

Hi everyone,

I want to let you know about a typo in #7 on the sheet. The shift in wavelength should be given by 2h/m_ec.

1997B Part II.
a)
b) The energy must be -1.9 eV, based on the energy emitted by the 600 nm photon.
c) 1240 nm - outside of the visible spectrum


200...Something, #7:
a) There are actually two (correct) ways to draw the diagram:

b) 1240 nm, again, outside of the visible spectrum.


7.
a) 9.9 x 10^-15 J
b) 3.3 x 10^-23 kg*m/s
c) The photon wavelength increases because some of the energy of the incident photon was lost to KE of the electron.
d) 6.0 x 10^-23 kg*m/s

1993B6
a)1.69 x 10¹⁹ Hz
b)3.73 x 10^-23 kg*m/s
c)3.31 x 10^-16 J
d)2.46 x 10^-23 kg*m/s

Quantum Theory, Photoelectric Effect and X-Rays

Photoelectric Effect
http://www.lon-capa.org/~mmp/kap28/PhotoEffect/photo.htm

More visual Photoelectric effect:
http://phet.colorado.edu/simulations/photoelectric/photoelectric.jnlp

Solutions:
2000B5
a)
i) 4.5 eV or 7.2 x 10^-19 J
ii) 1.26 x 10⁶ m/s
b) 183 nm
c) 5.56 x 10¹⁴ Hz

1980B3
a) The graph should be straight forward, but you SHOULD know that all four points cannot be on the same line!
b) 0.75 x 10¹⁴ Hz
c) 3.1 eV
d) Energy of electrons is 5 eV according to data, so the stopping potential will be 5 volts.
e) Charges move in circular paths when they are in a magnetic field perpendicular to their velocity. This is a possible reason for this behavior.

Page 855:
16. 4.14 x 10¹⁴ Hz

17. 401 nm

19. The shortest visible wavelength photons have energies of 3.1 eV (400 nm wavelength). Copper and Iron have work functions that are greater than this, so they will not eject photo electrons in visible light.
20.

a) 3.49 x 10^-19 J or 2.179 eV.
b) 0.930 V

Page 885:
P32. 4.14 x 10^-11 m (Don't worry about the longest wavelength.)
P33. 41,000 V

Diffraction, Single and Double Slit Solutions

From Handout:
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 10¹⁴ Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.

1991B6.
a) 3.9 x 10^-5 m
b) 9.6 x 10^-3 m

Problems:
7. 8.5 x 10^-5m
10. 613 nm
20. 1.57 mm 1.591 meters
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.

Thin Film Interference Solutions

From handout:

4. thickness = 1.05 x 10^-7 m

5. thickness = 96.1 nanometers

1984B5
a) 5 x 10¹⁴ Hz
b) 4.8 x 10^-7 m
c) 1.2 x 10^-7 m
d) 2.4 x 10^-7 m

38. wavelength = 643 nm, which is colored red.
40. 169 nm
43. 27*lambda/n = 2t, so t = 9045 nanometers, or 9.045 x 10^-6 m
45. 2t = n*(640 nm/1.36), and 2t = (n + 1/2)*512 nm/1.36. Solving the system, t = 471 nanometers.

Diffraction, Single and Double Slits

Huygen's Principle
http://projects.cbe.ab.ca/sss/science/physics/map_north/applets/advancedripple/advancedripple.html

Single Slit Simulation:
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/slitdiffr/index.html

Double Slit:
http://www.walter-fendt.de/ph14e/doubleslit.htm

Ripple Tank
http://www.falstad.com/ripple/

Refraction Simulations

Total Internal Reflection Simulation:
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/totintrefl/index.html

Prism/Dispersion
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/dispprizm/index.html

Glass Slab Refraction
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/dispslab/index.html

Lenses - Solutions

Multiple Choice problems from the handout:
1. B
2. E
3. C

Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.

Problems:
48b. 390 mm

50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.

52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4

60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm

83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class Monday!



AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens.













1986B5
a) 7.4%
b) 1.250 x 10⁶ W
c) We'll get to this problem in a couple weeks....
d)

Mirror Equations and Refraction Solutions

A reminder: Please DO try the simulations below. They will go a long way to giving you an intuitive understanding of how the light rays move in both the cases of refraction and reflection. If you want a preview of tomorrow's lesson, try the lenses in the bottom simulation.

Question 15 - The fish is, in reality, lower in the aquarium than is seen by the eye in the picture. Remember that the light ray from the fish is bent away from the normal when it leaves the water. This means that the actual ray from the fish to the water's surface is tilted downwards to reduce the angle from the normal to the surface.

Problems:
12. image is virtual, upright, and reduced in size. It is located at 2.09 cm behind the surface of the ball. (d_i = -2.09 cm).

13. magnification is -di/d0, so you can find di using the magnification of 3 (positive because it is upright.) Solving for the focal length, and multiplying by 2, the radius R = 3.9 meters.

14. concave mirror, R = 5.66 cm

27. n = c/v = 1.310

34. The light emerges from the first interface at an angle of 27.7 degrees. By geometry, we can see that this ray hits the other surface of the prism at 33 degrees. By Snell's Law, the ray emerges at 54.7 degrees.

35. Remember to find the angle from the normal! Theta 1 = 64.3 degrees, n1 = 1, n2 = 1.33, so theta 2 = 42.6 degrees. 2.1 meters * tan (42.6 degrees) = 1.93 meters. The total distance from the wall is therefore 2.7 m + 1.93 meters = 4.63 meters.

Mirror Equations & Refraction

Index of Refraction simulator: http://www.ps.missouri.edu/rickspage/refract/refraction.html

Site for Simulating Mirrors (lenses too, coming soon to a 6/7 period near you!)
http://www.surendranath.org/Applets/Optics/ReflRefrCurv/CurvSurfApplet.html

Plane and Curved Mirrors

Q19. Since the image is clearly magnified, this must be a concave mirror. Only concave mirrors can magnify objects. This occurs when an object is placed between the mirror and its focus.

P4. The easiest way to solve this problem is to extend the top ray through the mirror. You then obtain a pair of similar triangles. Solving the proportion, x = 75.9 cm

P11. A picture is worth a thousand words:

P16. Leave out part b, we will discuss the equations of ray tracing tomorrow in class. Make a scale sketch of the situation on graph paper (notice it is a convex mirror!) - you should obtain that the distance of the image is between 6 and 7 centimeters on the far side of the mirror, and the image should be upright.

Review Packet Answers

1.
a) amplitude = 1.5 inches
b) T = 0.6 s, f = 1/T, so f = 5/3 Hz
c) v = f*lambda = (5/3 Hz)*1.4 m = 2.33 m/s
d) using a constant accel. equation, displacement is 1.33 m.
e) We must use the Doppler effect equation: F_obs = (5/3 Hz) * (2.33 m/s + 2 m/s)/(2.33 m/s) = 3.10 Hz

2.
a) either 1.0 meters or 1.07 meters, depending on whether L1 or L2 was used.
b) lambda = v / f so f = 343 Hz.
c) The frequency of the waves will be the same in the water since it doesn't change when crossing from one medium to another. Using the wave equation, the wavelength is 4.34 meters.
d) L3 = 5*lambda/4 = 1.25 meters.
e) The speed of sound increases when temperature increases. If the temperature is higher, then the speed will be higher, which means the actual frequency will be higher. This means the number calculated in (b) is too low.

1982B2.
a) hook has T2 upwards, T1 and m2g downwards. The load has T1 up, m1g downwards.
b) T1 = 6000 N, T2 = 6600 N.

2000B2.
a) T and f directed up the incline, Fn perpendicular to the incline surface, and m1g directed downwards
b) mu = f/(m1g cos(theta))
c) M = (m1 + m2) - 3f/g
d) a = g sin(theta) - f/m1

1995B4.
a) 400 W
b) 40%
c) 1000 W
d) 600 W

2002B5
a) By the right hand rule, F_mag is directed to the left. This means the electric force must go to the right. Since the proton has a positive charge, the electric field must also go towards the right.
b) qE = qvB, so v = E/B.
c) proton moves in a circular path towards the left.
d) a = eE/m

small #1:
a) lambda = 2L
b) v = 2f0L
c) 2f0
d) f = L/2

Last problem with motor:

a) 3 milliwatts (3 mW)
b) 0.18 J
c) work = 0.118 J, so efficiency = actual output/energy in = 65.6 %
d) 5000 ohm and 1000 ohm resistors connected in series with the motor.

Interference Solutions!

Question 17.
You can use either the equation we derived at the end of class, or reasoning about the distance traveled by each wave. If the sources are close together, then the distance traveled by each wave will be similar, and the waves will have to travel a larger difference in distance in order to reach the path difference needed for constructive or destructive interference. This means the locations will be further apart when the sources are close together.

Problems:
44. The unknown whistle has to be HIGHER than the 23.5 kHz if it is to be inaudible to humans. Humans can hear up to around 20 kHz. This means the whistle frequency must be 23.5 Khz + 5 kHz = 28.5 kHz.

48.
a) 343 Hz
b) 1029 Hz, 1715 Hz

49a. Find the frequencies of each wave using the wave equation. beat frequency = 5.649 Hz

53. 77.27 Hz is the observed frequency, so the beat frequency would be 2.27 Hz.

54. The two frequencies would be 120 Hz and 126 Hz, the 126 Hz observed from the moving car. Thus the frequency of both horns must be 120 Hz.

Interference Web Sites

http://www.mta.ca/faculty/science/physics/suren/Beats/Beats.html
See two sine waves added together.

http://micro.magnet.fsu.edu/primer/java/interference/waveinteractions2/index.html
Phase Difference Demo

http://library.thinkquest.org/19537/java/Beats.html
Simulation of beats that you can hear!

http://www.colorado.edu/physics/2000/applets/fourier.html
Review of Constructive/Destructive Interference

http://www.glenbrook.k12.il.us/gbssci/phys/Class/light/u12l3b.html
Good images for demonstrating path difference

http://www.walter-fendt.de/ph11e/interference.htm
Interference of Two Circular Waves

http://projects.cbe.ab.ca/sss/science/physics/map_north/applets/advancedripple/advancedripple.html
Another Circular Waves Demo

Doppler Effect

Thanks again for making today so successful!

Question 17: The relative speed between the whistle and the listener is highest at point C. By today's discussion of the Doppler effect, this must also be the location of the highest frequency.

51.
a) Vobs = 370 m/s, Vsource = 340 m/s, fobs = 1959 Hz
b) Vobs = 310 m/s, Vsource = 340 m/s, fobs = 1631 Hz

52. The frequency of the wave as heard by the moving object is 46323 Hz. This frequency now acts as the source, and the bat is the observer. The new frequency heard by the bat is 43150 Hz.

78. You should be able to set up two equations with Fsource and v(the speed of the train) as the unknowns. Let Vs stand for the speed of sound.
First equation: Vs/(Vs - v) = 522/Fsource
Second equation: Vs/(Vs + v) = 486/Fsource.

Dividing the equations, Fsource drops out, and you can solve numerically to get v = 12.14 m/s.

Shockwave Video #1:


Shockwave Video #2:


Sonic Boom:

Standing Waves with Resonance

First, some interesting sites related to today's discussion:

http://library.thinkquest.org/19537/cgi-bin/showharm.cgi
This site contains a program that allows you to put together harmonics and see what they sound like together.

http://www.pbs.org/wgbh/nova/bridge/meetsusp.html#clips
This site describes one of the most catastrophic applications of resonance, the Tacoma Narrows Bridge collapse.

Now, some solutions:

33. Since the nodes are now at half the distance, and the wave velocity is the same, the frequency must double to 8 Hz. This is beyond the highest frequency of earthquakes.

36. The tube must be closed at one end - look at the ratio of the frequencies to each other, as they are odd fractions. The fundamental frequency is 88 Hz.

37. f = 5v/2L, so v = 247.5 m/s.

38. Open at both ends, 7v/2L = 33o Hz, L = 3.606 m

75. n*lambda/4 = L1, (n+2)*lambda/4 = L2. Subtract the two equations, and you get that lambda = 2(L2 - L1), leaving that f = 630 Hz if the speed of sound is 340 m/s.

Standing Waves

New! - Solutions

Q25. The nodes are locations of zero amplitude, so these positions could be touched without disturbing the wave.

Problems:
18. amplitude = 0.3500 meters, frequency = .8754 Hz, period = 1.142 seconds, total energy = .7411 J, KE = .6807 J and PE = .0604 J

51. 440 Hz, 880 Hz, 1320 Hz, 1760 Hz (f_n = nv/2L)
53. 70 Hz, 140 Hz, 210 Hz, 350 Hz
55. nodes are always a half wavelength apart from each other - wavelength is 19.37 cm, so nodes are half of this, 9.685 cm.
56. 87.5 Hz, n = 3 and 4 for the given frequencies.
60. m = 4f²L²*mu/(n²g), so m = 1.421 kg, .355 kg, .0568 kg for (a), (b), and (c).

Web sites from today:

http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference

http://home.austin.rr.com/jmjensen/JeffString.html

Reflected Waves


http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=19 - Waves traveling in opposite directions making standing waves



Standing Wave - nodes and antinodes

http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html - Vibrating string, standing waves fixed at both ends.

http://www.walter-fendt.de/ph11e/stlwaves.htm - open/closed Tube with standing waves

Wave Motion Solutions

Q10 . The water sloshing back and forth has a specific frequency associated with it because of the size of the pan and the speed of the waves on the surface of the water. The only way to get the water to slosh is if the swinging occurs at this special frequency.

Q18 (Not 8, I'm sorry!). Hitting across the end creates a transverse wave, hitting perpendicular to the end creates a longitudinal wave.

Problems.
34. 2.83 m/s
35. 1.259 m
39. 0.332 s
41. The speed of sound in water is 1440 m/s. The sound must travel down to the ocean floor and back. The distance is therefore 2160 meters.

EMI Problem Answers

1986B4.
a) 3 V
b) clockwise
c) 0.6 N
d) 1.8 W

1982B5.
a) 0.06 Webers
b) 0.06 V
c) -0.3 A from t = 0 to t = 2 seconds, 0 A from t = 2 to t = 4 seconds, and +0.15 A from t = 4 to t = 6 seconds.

3.
a) final speed is (2gy₀)^(1/2)
b) I = Bh/R*(2gy₀)^(1/2)
c) flux is increasing into the page, so induced current increases the flux out of the page. Current is CCW.
d) For flux:
Flux increases linearly to whB at w; Stays constant until 3w, and then decreases to zero at 4w.
For current: The magnitude of I is the value given in part (b).
Positive I from 0 to w, 0 from w to 3w, -I from 3w to 4w, and then zero from 4w to 5w.

4.
a) m = ILB/g = B^2L^2v/gR = 0.143 kg
b) 1.51 J
c) 1.51 J

5.
a) 0.3 Volts
b) 0.06 A, counter clockwise
c) 0.018 W
d) F = 0.036 N
e) 20 more turns means potential difference increases by a factor of 20. This also means twice the resistance of the wire. Since current I = deltaV/R, the factors of 20 will divide out, leaving the current the SAME.

Lenz's Law Solutions

I apologize for leaving it out, but please use the image below with Problem 2 from the handout from today's lesson:





Questions:
15. The change in flux causes an induced current within the aluminum. The magnetic field created by the induced current is attracted to the magnetic field of the bar magnet, opposing the force tending to pull the sheet out of the magnet. This induced magnetic field forms because of the induced current, not because the magnetic properties of aluminum.

16. We discussed this in class!

Problems:
3. Counter clock-wise

4. Field lines going out from magnet to the right. Moving it through loop increases the flux going into loop towards the right. By Lenz' law, the induced current will increase the magnetic flux going into the loop towards the left. This will create a current that flows clockwise through the loop, and from right to left through the resistor.

7.
a)The current through the outer loop decreases since the resistance increases. This means that the magnetic field (and therefore the flux) through the inner loop out of the page is decreasing. By Lenz's law, the induced current will attempt to increase the flux out of the page to oppose the change. This will come from a counter-clockwise current.
b) If the loop was moved outside to the left, the flux would be decreasing into the page through the loop. The induced current would try to increase flux into the page through the loop, which would result from a clockwise current.

8.
a) counter-clockwise
b) clockwise
c) zero (no change in flux)
d) counter-clockwise

Faraday's Law & Electromagnetic Induction

Question 2:
Magnetic field describes how the influence of magnetic force is transmitted through space. Magnetic force describes how magnetic flux passes through a specific region of space, specifically a loop or area through which magnetic field lines travel. It is possible to have magnetic field and not have magnetic flux (if area = 0 or if the angle between B and the normal vector is 90 degrees), but it is not possible to have magnetic flux without magnetic field.

Problems:
2. 0.147 Volts

10.
a) 0.169 Volts

12.
b) 4.29 x 10^-2 Volts
c) 1.72 mA

13.
a) .841 m/s
b) .757 N/C

89. 7.27 x 10^-3 J

Here are a couple of sites with fun EMI tools to play with:

http://www.ngsir.netfirms.com/englishhtm/Induction.htm
http://higheredbcs.wiley.com/legacy/college/halliday/0471320005/simulations6e/index.htm?newwindow=true

Magnetic Field from a Wire/Magnetic Force

1.
a) 5 x 10^-5T
b) West
c) Net B-field is of magnitude 7.07 x 10^-5T directed at 45 degrees North of West. (It is not enough to say 7.07 x 10^-5T at 45 degrees!)

2. a = 0.0076 m/s^2

3. 48 A

4.
a) negative
b) Electric field is directed towards the top plate.
c) 2280 V
d) 9.5 x 10⁶ C/kg

5.
a) perpendicular to the plane of the page, directed OUT of the page.
b) 2 x 10^-7T
c) negative
d) 5 x 10^-7C
e) 0.2 N/C directed to the left

6.
a) perpendicular to the page, out of the page
b) 1.9 x 10^-15N
c) 19 cm
d) 1.2 x 10⁴N/C
e) The electric field is in the plane of the page, directed towards the top of the page.

7. This problem is very similar to #55 from yesterday. Important points: What is R in terms of given quantities? If the charge q crosses a potential difference V, what is the final speed if it starts from rest?

Additionally, some help on #67:
Many of you are finding the speed of the electrons, and are then able to figure out that you need to find the radius of curvature of the path. The image below suggests a way to find the deflection of the electrons from their intended path.