Final Exam - PY106 Spring 1997

## Final Exam - PY106 Spring 1997

This test is 120 minutes long.

[10 points] 1. A moving charge in a magnetic field.

[5 points] (a) In a certain region there is a uniform magnetic field in the -y direction as shown below. A positive charge with a velocity in the +x direction experiences a force in which direction because of this field?

Using the right hand rule, the force is into the page.

[5 points] (b) For the situation described above, the positive charge would follow a circular path; we now want the charge to move along a spiral around a line parallel to the y axis. Draw a vector diagram (similar to the one above) showing the directions of v and B necessary to accomplish this.

The charge circles around the magnetic field, so if the axis of the spiral is parallel to the y axis then the magnetic field must be parallel to the y axis. To make it go in a spiral, the velocity must be something other than perpendicular or parallel to the field. An arrangement like the one below would work nicely, for instance.

[25 points] 2. DC Circuits.

You have some wire, one 6 V battery, and three 8 ohm resistors.

[5 points] (a) Draw a circuit with the battery and resistors so that each resistor receives the same current. Note that there is more than one solution; you only need to draw one.

Two arrangements are possible. You can connect the resistors in series, or you can connect the resistors in parallel.

[7 points] (b) Calculate the current through one of the resistors in your circuit in part (a).

If you drew the series circuit, the overall resistance is 24 ohms, so the current would be 0.25 A.
If you have the resistors in parallel, each one has V = 6 volts, so the current through each resistor is 6 /8 = 0.75 A.

[5 points] (c) Now draw a circuit with the battery and resistors arranged so the current through one resistor is twice as large as the current through each of the other two. Again there is more than one solution; you only need to draw one.

Again, there are two solutions. The first possibility is to have the current come out of the battery, pass through one resistor, and then split going through the other two, which are in parallel. The current then comes back together and returns to the battery.
The second possibility is to have two parallel branches, one with one resistor and one with two. Twice as much current passes through the branch with only one resistor.

[8 points] (d) How much power is provided by the battery in your circuit in part (c)?

If you went with the first possibility above, the overall resistance is 12 ohms. The current is 0.5 A, and the power is 3.0 W.
For the second possibility, the overall resistance is 5.33 ohms. the power in this case is 6.75 W.

[25 points] 3. Two charged spheres.

Two hollow conducting spheres, one with radius 5 cm and one with radius 10 cm, are arranged about a common center. The inner sphere has a charge of +3.00 microcoulombs uniformly distributed over its surface; the outer sphere has a uniformly-distributed charge of -2.00 microcoulombs.

[6 points] (a) On the diagram, sketch the electric field produced by the spheres.

The diagram has no field lines inside the smaller sphere; plenty of field lines going from the inner sphere to the outer sphere; and a few field lines going out from the outer sphere.

[5 points] (b) The point P in the diagram is 12 cm from the center of the spheres. Using a dashed line, draw the equipotential line that passes through P.

The equipotential line is a circle a little bigger than the outer sphere, passing through P.

[6 points] (c) What is the electric potential at a distance of 12 cm from the center of the spheres?

V = kQ / r. Because 12 cm is outside both spheres, Q can be the net charge, so it's +1.00 microcoulombs.

V = 8.99 x 109 (+1.0 x 10-6) / 0.12 = 7.49 x 104 volts.

[8 points] (d) A charged particle is fired directly at the spheres from a point a long way to the right. It slows down as it approaches the spheres, comes to a stop at point P, and is then repelled away from the spheres. The particle has a mass of 2 x 10-4 kg, and a speed, when it is very far from the spheres, of 60 m/s. What is the sign and magnitude of the charge on the particle?

First get the sign. The charge is obviously repelled, so it must have the same sign as the net charge on the two spheres. The net charge is positive, so the particle must be positive.

The charge can be found by conserving energy. Very far away, all the energy is kinetic. At point P, where it turns around, the energy is entirely potential. Conserving energy and setting these things equal gives:

qV = 1/2 mv2

V is the value from part (c). Plugging all these things in and solving for the magnitude of the charge gives q = 4.81 x 10-6 C.

Therefore the charge is +4.81 microcoulombs.

[25 points] 4. Induction.

A coil consists of 128 turns of wire wrapped around a square frame 0.3 m on a side; the coil has a total resistance of 1.8 ohms. The coil is located in a region where there is a constant magnetic field B = 0.8 T. Initially, the field is perpendicular to the plane of the coil.

[5 points] (a) What is the magnetic flux through one turn of the coil in the starting orientation?

The flux is BA = 0.8(0.32) = 0.072 T m2

[10 points] (b) The coil is rotated until the magnetic field is in the plane of the coil. The rotation is done in such a way that the emf induced in the coil is constant during the motion, which takes 3.0 seconds. What is the magnitude of the induced emf during this rotation?

The induced emf has a magnitude of N (change in flux) / time. The final flux is zero.
This gives emf = 128 (0.072) / 3 = 3.07 volts.

[10 points] (c) The wires from the coil are connected to a meter that measures the total charge that flows through it during the rotation. How much charge flows?

The charge is the current multiplied by the time. The current can be found from V = IR, so that gives:

I = V / R = 3.07 / 1.8 = 1.71 A.

Q = It = 1.71 (3.0) = 5.12 C.

One interesting thing about the charge is that you get the same charge no matter how fast you rotate the coil. If you increase the time, the current decreases, and vice versa.

[25 points] 5. Refraction.

A light ray traveling through some plastic has a frequency of 5.5 x 1014 Hz. It is incident upon a flat slab of glass, making an angle of 32° with the normal to the interface; it passes through the glass and into vacuum, as shown in the sketch. The refractive index of the plastic is 1.50. The refracted ray in the glass has a wavelength of 321 nm.

[5 points] (a) What is the refractive index of the glass?

You might be tempted to write down Snell's law here, but you don't have enough information. What you do know is the wavelength of the light in glass, 321 nm, and you know the frequency. The frequency is the same all the way through - it doesn't change when the light enters another medium.

Multiplying the wavelength by the frequency gives you the speed in the glass:
v = (5.5 x 1014) (321 x 10-9) = 1.766 x 108 m/s.

The index of refraction is n = c / v = 3.00 x 108 / 1.766 x 108 = 1.70.

[10 points] (b) What is the angle of refraction (measured from the normal) for the light ray in vacuum?

Now apply Snell's law. Applying it at the plastic-glass interface gives:

n1 sinQ1 = n2 sinQ2

Note that I'm using Q's here instead of the Greek letter theta!

At the glass-vacuum interface, Snell's law applies again. Because the two sides of the piece of glass are parallel to each other, the angle between the ray and the normal is the same ( Q2 ) at both interfaces. This gives:

n2 sinQ2 = n3 sinQ3

Combining the two equations allows us to ignore the glass (only because the two faces of the glass are parallel) and write:

n1 sinQ1 = n3 sinQ3

The index of refraction in vacuum is 1.00. Plugging in the numbers gives:

1.50 sin32° = 1.00 sinQ3

So the angle is Q3 = 52.6°.

[10 points] (c) What is the minimum angle of incidence (in plastic) for which the ray will be totally internally reflected at the glass-vacuum interface?

We're looking for the critical angle here. A neat way to find it is to just plug n1 and n3 into the critical angle equation and solve. Again, you can do this only because the sides of the glass are parallel to each other. To see why you can do this, remember that at the critical angle, the light ray refracts at 90° to the normal (along the interface, in other words). Applying Snell's law at both interfaces gives:

n1 sinQ1 = n2 sinQ2 = n3 sin90°.

This gives 1.50 sinQ1 = 1.00 sin90°. This gives the critical angle equation:

sinQ1 = 1.00 /1.50

The angle is 41.8°.

[10 points] 6. Refraction and reflection.

Three parallel rays are incident on a converging lens; the lens has a focal length of 3.0 cm. In each of the two cases below, draw the path of each light ray and state the x and y coordinates of the point where the rays converge.
Assume the lens lies at the origin of an x-y coordinate system, and the grid has a spacing of 1 cm by 1 cm.

[5 points] (a) The rays are parallel to the principal axis, and shine on the upper half of the lens.

The lens brings all these paraxial rays (rays parallel to the principal axis) to the focal point. The diagram should show each ray going through the focal point, 3 cm to the right of the lens.

The rays converge at: x = 3 cm, y = 0 cm

[5 points] (b) The rays are parallel to the principal axis; a plane mirror is placed at x = 2 cm, parallel to the y-axis.

Again, the rays head off towards the focal point. The mirror intercepts them, and they reflect back obeying the law of reflection. Instead of converging 1 cm behind the mirror, they will converge 1 cm in front of the mirror (after reflecting!). In some sense, the focal point is reflected in the mirror.

The rays converge at: x = 1 cm , y = 0 cm

[25 points] 7. The double slit.

Two very narrow parallel slits separated by a distance d1 = 1.0 mm are illuminated with monochromatic light with a wavelength of 400 nm. An interference pattern is observed on a screen 1.3 m away.

[8 points] (a) What is the distance between the central and first interference maxima on the screen?

First find the angle using d sinQ = m (wavelength)

Using m = 1, and converting everything to meters, the angle works out to 0.2292°.

The distance between the central and first interference maxima is then found from
tanQ = y / L, where y is the distance we're looking for and L is the distance to the screen. Solving for y gives y = 0.00052 m.

[6 points] (b) Suppose the distance between the slits is increased slightly to a value d2. The separation between the central and first interference maxima will:

[ ] increase
[ x ] decrease
[ ] stay the same

The separation is inversely related to d.

[6 points] (c) The wavelength of light is now changed to 630 nm; the slit separation is still at the unknown increased distance d2. It is found that the separation between the central and first interference maxima is the same value obtained in part (a) (i.e., when the wavelength was 400 nm, and the slits were 1.0 mm apart). What is the value of the current slit separation d2?

If the separation is the same then the angle must be the same, 0.2292°. Plugging this into d2 sinQ = m (wavelength), and again using m = 1, gives a slit separation of d2 = 1.575 mm.

[5 points] (d) Suppose the entire apparatus is now immersed in a transparent medium with refractive index greater than 1. The separation between the central and first interference maxima will:

[ ] increase
[ x ] decrease
[ ] stay the same

The wavelength decreases, so the angles decrease.

[25 points] 8. The photoelectric effect.

The work function of a metal surface is 2.48 eV. The metal surface is illuminated by a light source.

[5 points] (a) Find the minimum frequency that light must have to eject electrons from the surface of this metal.

The minimum frequency is found when the energy of a photon, hf, equals the work function. Converting the work function to joules gives:
W = 2.48 eV (1.6 x 10-19 J/eV) = 3.968 x 10-19 J

Minimum f = W / h = 3.968 x 10-19 / 6.63 x 10-34 = 5.98 x 1014 Hz.

[5 points] (b) The metal surface is illuminated by a light source with a wavelength of 600 nm. Are any photoelectrons emitted from the metal surface?

This wavelength can be converted to frequency using:

f = c / wavelength = 3.00 x 108 / 600 x 10-9 = 5.0 x 1014 Hz.

This is less than the minimum frequency calculated in (a), so these photons don't have enough energy to eject electrons from the metal. No electrons are emitted.

[10 points] (c) The metal surface is now illuminated by a light source with a wavelength of 400 nm. What is the maximum kinetic energy of the emitted photoelectrons?

The energy of the photons is E = hf = hc / wavelength. In this case that works out to:
E = 4.972 x 10-19 J.

The maximum KE of the electrons is KEmax = E - W. That works out to
KEmax = 1.00 x 10-19 J (or 0.628 eV )

[5 points] (d) The intensity of the light source is doubled, without changing its wavelength. What is the maximum kinetic energy of the emitted photoelectrons now?

Same as in part (c). Doubling the intensity means twice as many photons strike the metal plate, but it doesn't change the energy of each photon. If the photon energy is the same, the electrons have the same energy.

[20 points] 9. Light.

[6 points] (a) A light source emits 100 W of red light. A second light source emits 1 W of green light.
Which of these light sources is emitting photons of greater energy?

[ ] red
[ x ] green

The smaller the wavelength, the higher the energy.

Which of these light sources is emitting the greater number of photons in a second?

[ x ] red
[ ] green

Two reasons for this. First, each red photon has less energy than each green photon, so the red light must throw out more photons to produce a certain power. Also, the red light is 100 times more powerful, so it must be emitting many more photons than the green light. If the green light sends out N photons per second, the red light emits 100 N x (red wavelength) / green wavelength.

[9 points] (b) A hydrogen atom emits light when an electron jumps from the n = 2 Bohr energy level to the ground state. Find the wavelength of the emitted light.

The energies of the n = 2 and n = 1 levels can be found from the Bohr energy equation.

For n = 2, E2 = -13.6 eV / 22 = -3.4 eV

For n = 1, E1 = -13.6 eV

The photon has an energy equal to the difference. The photon energy is thus
E = -3.4 - (-13.6) = 10.2 eV.

Converting this to joules gives:
E = 10.2 (1.6 x 10-19 J/eV) = 1.632 x 10-18 J

Solving for the wavelength gives:

wavelength = hc / E = 1.22 x 10-7 m (or 122 nm).

[5 points] (c) A glass plate (refractive index n = 1.60) is coated with a thin film of magnesium fluoride (refractive index 1.38) to make it nonreflecting for normally incident light of wavelength 550 nm in air. What is the minimum thickness t of the film?

The wave reflected off the top surface is shifted by half a wavelength because of the phase change that occurs at a low n - high n interface. The wave reflected off the bottom surface is shifted by 2t + half a wavelength (it travels an extra distance of 2t, and also has a phase change).

The net shift is the difference, so that's 2t. Bring in the condition for destructive interference, and you get:

2t = (m + 1/2) x (wavelength in the film).

The wavelength in the film is 550 / n = 398.55 nm.

The minimum thickness is found by using m = 0. This gives
t = 398.55 / 4 = 99.6 nm

[10 points] 10.

[5 points] (a) A wooly mammoth bone found in an archeological dig is sent to a laboratory to determine its age. Analysis shows that there are an average of 2 decays per minute per gram of carbon, from the radioactive decay of carbon-14 in the bone. Normal living organisms exhibit 16 decays per minute per gram of carbon. Assuming that the bone had the normal concentration of carbon-14 when the mammoth died, what is the approximate age of the bone? The half-life of carbon-14 is 5700 years.

This one you can do in your head. If you start at 16 decays per minute, after 1 half-life you're down to 8. Two half-lives gets you to 4, and three half-lives get you to the 2 decays per minute observed with this bone.

The bone is three half-lives old. 3 x 5700 = 17,100 years.

[5 points] (b) Within the core of a nuclear reactor there are 3 x 1019 nuclei fissioning each second. The energy released by each fission is about 200 MeV. Determine the power (in watts) being generated.

This is really just a problem in converting between units. The power in watts is the energy in joules per second. 200 MeV = 2 x 108 eV.

Converting eV to joules gives an energy / fission of 3.2 x 10-11 J.

Multiplying by the number of fissions per second gives the power in watts, so:

P = 3.2 x 10-11 (3 x 1019 ) = 9.60 x 108 W