Suggested Solutions (Including Photonics as Detailed Study)

[DuBistKomisch]

Electric Power

1. Left through the inside of the solenoid and looping back around to the right on the outside.
2. Flux is zero, as the area is parallel to the magnetic field.
3. Downwards using the right hand rule (B left, I out of page).
4. F= BIln = 0.024 N
5. Force is zero as the current and magnetic field are parallel.
6. A; emf is negative time derivative of flux and slip rings collect AC.
7. Slip rings maintain contact between ends of loop and same terminal whereas a split-ring commutator alternates contact with each terminal each half-revolution. Slips rings for AC with generator/motor and split-ring for DC (expand though).
8. 0-1 below the axis, 1-2 on the axis, 2-4 above the axis but only half the magnitude of 0-1. Shouldn't matter whether you drew a line on the axis after 4 seconds or not.
9. Faraday's law (of induction).
10. 3.0V.
11. Q -> P using Lenz's law and right hand screw rule.
12. P = V^2/R = 120^2/48 = 1200W.
13. Connect in series to obtain 120^2/96 = 600W.
14. Power is lost but current must be the same throughout so voltage is less at the globe, hence less energy delivered and not as bright.
15. Current is 2.0A, voltage drop across lines is 2.0*4.0=8.0V, so 2.0+8.0=10V initially.
16. P_loss = I^2*R = 2.0^2*4.0=16W.
17. AC can be stepped up and down using transformers to reduce current and minimise power loss in lines.
18. D: 20.8*sqrt(2)*2 = 58.8V peak to peak.
19. 10:1 = 1460:x -> x=146 turns.
20. V2=2.0V and I2=2.0A, 10:1 leads to V1=20V and I2=0.2A, then P_loss=I^2*R=0.2^2*4.0=0.16W.

For questions 8 and 10 I am not sure how significant signs are. I am thinking they may accept any combination with each part opposite in sign since no reference polarity for the voltmeter was given.
For question 9, refer to the introduction of http://en.wikipedia.org/wiki/Lenz's_law and also http://en.wikipedia.org/wiki/Electromagnetic_induction#Technical_details.

Light and Matter

1. Young's double slit. Talk about diffraction and interference. Explain why only wave model explains these two phenomena.
2. Observation 2 is the only one that only the particle model supports. Explain how frequency = energy of packet and intensity = number of packets and the consequences.
3. E = hc/lambda = 2.14eV.
4. 2.5*580=1450nm.
5. D: intensity has no effect on threshold frequency or stopping voltage, only the current.
6. A: Magnesium has lower W, hence lower f_0 so its intercept on the horizontal axis will be to the left.
7. lambda = h/mv = 0.049nm.
8. A: greater velocity -> smaller wavelength -> smaller pattern.
9. Very similar wavelength ->  very similar pattern.
10. Convering 600eV kinetic energy to joules and then finding the de Broglie wavelength, equating to the X-ray wavelength and finding E=hc/lambda=2.5*10^4eV.
11. E=hc/lambda=2.6eV which corresponds to n=4 (12. -> n=2 (10.2).

Photonics

1. C: Lasers are coherent so C was the only option with only incoherent sources.
2. C: Switch-on voltage of 2.0V indicates band gap of 2.0eV which corresponds to lambda=hc/E=621nm.
3. B: Voltage drop across LED is 2.0V so voltage drop across the 400 ohm resistor is 10V and current must be 10/400=25mA.
4. C: Blue light has a higher frequency and hence a blue LED must have a larger band gap energy. Therefore a larger voltage drop occurs over the LED and smaller over the resistor which means a smaller current through the circuit.
5. D: arcsin(1.38/1.44) = 73 degrees.
6. B: arcsin(sqrt(1.44^2-1.38^)) = 24 degrees.
7. B: When you bend it, the light is more likely to hit at an angle closer to the normal. Therefore angle of incidence becomes below the critical angle and is not reflected.
8. C: Water's refractive index is higher than the plastic so TIR cannot occur.
9. D: Absorption line is clearly higher on the graph at 2000nm.
10. C: Taking into account both lines (addition of ordinates), 1200nm had the lowest total attenuation. The only tricky question.
11. B: Graded index is an attempt to make the different modes arrive at the same time despite the different distances taken. Hence B modal dispersion.
12. C: Rayleigh scattering is caused by slight variations in the optical density in the core of an optical fibre caused by imperfections.

Anyone have reliable Sound or Synchrotron answers?

[Linkage1992]

FUCK FUCK FUCK i worked out the flux in question 2 by doing BA, i forgot it was parallel FUCK

... but that appears to be the only thing i've lost so far.

[rubiks]

I didn't halve the magnitude for question 8. I'd still get a mark though for putting 0-1 under and 2-4 over wouldn't I?

[715761]

Wait doesn't a smaller wavelength mean wider fringe spacing?

[3Xamz]

1. Left through the inside of the solenoid and looping back around to the right on the outside.

Can you expand on this?
I had my lines going from the left of the page to the right of the page.
iTute says right to left =='

[DuBistKomisch]

I didn't halve the magnitude for question 8. I'd still get a mark though for putting 0-1 under and 2-4 over wouldn't I?

I don't remember how many marks it was worth, but I presume they'd look for the sign and the magnitudes to be correct, so you would get some marks but not full.

Wait doesn't a smaller wavelength mean wider fringe spacing?

Distance between fringes is proportional to wavelength.

[Linkage1992]

The south pole would have been on the right using the right hand grip rule, meaning the the lines through the centre would go from the right to the left.

[DuBistKomisch]

1. Left through the inside of the solenoid and looping back around to the right on the outside.

Can you expand on this?
I had my lines going from the left of the page to the right of the page.
iTute says right to left =='

The current comes up the front from the right side of the solenoid. Using the right hand screw rule, the magnetic field through the solenoid goes from the right end to the left end, then out and back around from left to right.

[Linkage1992]

I didn't halve the magnitude for question 8. I'd still get a mark though for putting 0-1 under and 2-4 over wouldn't I?

i uh, didn't do that either *cries*
this is getting worse and worse, i shouldn't have come here to look at the solutions, it's just making me depressed now.
That's at least 3 marks gone already.

[coletrain]

FUCK FUCK FUCK i worked out the flux in question 2 by doing BA, i forgot it was parallel FUCK

... but that appears to be the only thing i've lost so far.

dw i did this too, so far only 3 marks lost

[Chavi]

for 10 I pretty sure voltage was -3V (negative)

[kyzoo]

I agree with everythign but light and matter Q2. I think its option 2, because I saw in a trial paper solution I did once that "intensity proportional to photoelectric current" is an explanation that is supported by both wave and particle model.

[Chavi]

For light and matter question 2, I thought that only 2 was acceptable - as the wave theory does stipulate that photo current depends on the intensity of incident light. . .
but it CANNOT explain why Ek is solely dependent on frequency. . .

[DuBistKomisch]

for 10 I pretty sure voltage was -3V (negative)

Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...

[Blakhitman]

isn't question 2 of Light observation 2???

cause wave model suggests that higher intensity means higher energy, which isn't what the observation suggested, it suggested that intensity did not affect the energy emitted...only frequency.