It's been a few years since uni physics and I was glad to get out of there. Crap, I barely made it through the electricity stuff. Those diagrams made my head hurt.

You're a better woman than I, Es.

We'll see. I've forgotton so much. But I knew I could count on Mike to help, now i just have to work through what he said! :D

#26. Centre of mass is half way up the rod, the gravitational potential energy of that when initially upright is

M * g * L / 2

[ g is acceleration due to gravity ]

when laid down that gravitational potential energy is zero. Now one can start with half the mass but at twice the height and the energy change will be no different during the falling process. Also the torque ( force times distance ) will also be the same whether we have all the mass half way up vs. half the mass all the way up. So the rate of change of angle will be the same, specifically the velocity of the tip when it hits the dirt.

So I've converted the stated problem to an identical but simpler one. Now the change in gravitational potential energy equals the change in kinetic energy.

(M/2) * g * L = M * V * V / 2

V^2 = g * L

V = SQRT(g * L)

Option B.

#27. After some research, option A - real. I cheated and looked it up! Some long forgotten stuff about diagonalising a matrix to get eigenvalues. :-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter.Blaise Pascal

Wow, Mike, you're doing a terrific job! I wanted to help with at least one and thought I'd jump right to the last one since you're blazing through them so fast :)

#100. The distance that the mirror moves, d_m, will be proportional to the number of fringes counted, N, and proportional to the wavelength, Î» (lambda), of the laser light, and also proportional to a factor of 2 (since the light goes out and reflects back, traveling double the distance that the mirror moves, but the factor of 2 drops out in the next step):

d_m = 2 * N * Î»

The distance that the mirror moves will be the same for both laser beams, which means:

2* N * Î» = 2 * N * Î»

The wavelength of the green laser is unknown, so solve for Î» (where the factor of 2 drops out):

Î» = ( N * Î» ) / N

Substituting the values given in the problem:

Î» = 85,865 * 632.82 nm / 100,000

Î» = 543.37093 nm

So the answer is Option (B) 543.37 nm

- - -

It's a good test â€“ note that someone who doesn't know the proper way to think it through might notice that the number of fringes for the red light can be divided by 100 (maybe loosing three orders of magnitude in the number of green fringes when going from meters to nanometers?) and viola: Option (E) 858.65 nm.

Even trickier, note the wording of the problem, â€œ ...so that a ratio of fringe counts may be used to compare the wavelengths of two lasers with high precision.â€? If a person tries A is to B as C is to D:

Î» / 100,000 = 632.82 nm / 85,865

Î» = 736.994... nm

That would be Option (D) 736.99 nm. It's a very good test :)

I'm not a physicist and I had to research it â€“ a few days ago I would have guessed a wavemeter is what's used to give the surfing conditions. So Mike or anyone please feel free to add/correct anything â€“ couldn't be a better forum for asking about interferometry :)

I'll get to that one Chipper! You can work back if you like while I go forward. T'is fun! :-)

Update on #27. Hermitian operator means the square matrix which represents that is it's own conjugate transpose ie:

H(i,j) = [H(j, i)]*

so you swap the rows for columns, and then take the complex conjugate of all entries ( or vice versa if you like ) . This implies the elements in the leading diagonals are real as

H(i,i) = [H(i, i)]*

means the imaginary parts are zero. Now any square matrix when diagonalised ( by pre and post multiplying by a certain unitary matrix and it's transpose ) will remain with those real diagonals - and these entries are thus interpreted as the eigenvalues. You need at least a first course in linear algebra to catch up on this stuff - Google 'Gilbert Strang', he's the man for that! :-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter.Blaise Pascal

Hey, I've just noticed the answers are at the back! :-)

#1. Doh! Read the question properly Mike...... :-)

#2. They're wrong by a factor of ten.

#10. They've combined the capacitances incorrectly!

#22. My mistake, if you have the semi-major axis and the orbital period you can get the central mass - as the proportionality between R and T involves it.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter.Blaise Pascal

#31. The antimatter angle is a red herring - this is a Hydrogen atom with a ground state ( n = 1 ) of -13.6eV. Levels go like 1/ n^2. From n = 3 to n = 1

#33. The slick way around this one is to recall ( and by aligning the x-axis along the path of the particle's travel ), that the spacetime distance/metric

c^2 * dt^2 - dx^2

is the same value in both frames ( where dt and dx are the time and space separations of two events in the one frame ). In the rest frame this is

c^2 * [10^(-8)]^2 - 0^2

NB. it didn't go anywhere in it's rest frame! For the lab frame :

#34. This is an implicit question about light cones. How can two non co-incident events in one frame be seen as co-incident in another? Basically if they are separated by a sublight speed path. Option B. [ I disagree with their answer ]

#35. For black bodies the energy model goes like T for the oscillators and a T for each degree/dimension of freedom in the cavity ie. 3 of those. So four powers in all. 3^4 = 81. Option E.

#36. Quasi-static ( succession of states close to equilibrium, thus ~ reversible ) and adiabatic ( no heat flow in or out ) means options A, C and D are true. I'm not sure you can firmly make comment either way on B. The temperature will go down however, so Option E.

#37. As the path BC is isothermal then PV = nRT can be used to deduce that P ~ 1/V for that path. Going from B to C has P going from 200 to 500 then V must have started at 5 ( ie. point B ) to then go to 2. The area under the curve is less than for a triangle ( area = base * height / 2 ), and so energy/work = area bounded < (500 - 200) * ( 5 - 2 ) / 2 = 300 * 3 / 2 = 450. If you look at the cycle, the gas is being pressurised under constant volume ( CA, ie. heated ), allowed to expand under constant pressure ( AB, ie. it will cool ) and then being compressed while losing heat to maintain constant temperature ( BC ). This shows work is being done on the gas ( not by it which would be the reverse way around the loop ). Given the phrasing of the question I'd opt for -300J and hence Option D.

#38. This a driven oscillator with damping. The AC source drives, the resistor converts current enrgies to heat, and the charge slops back and forth between the capacitor and the inductor. The capacitor soaks up the extremes of charge build up ( voltage ), the inductor the extremes of charge movement changes ( current fluctuations ). The maximal current amplitude is when C & L are tuned for full oscillation. This implies the impedances of C and L must be equal ( though acting 90 degrees of phase apart ). To much C lowers the voltage to then later kick the current back through L, while too little C raises the voltage and inhibits the current. Hence ( with w the angular frequency ) :

Option D. The exact values of the AC voltage and the resistance don't matter.

#39. OK, capacitors prevent DC current ( frequency zero ), and inductors suppress large current changes ( ie. AC ). In I at high frequencies the inductor acts as an open ciruit and the signal does not pass. In II at high frequencies the the inductor acts as an open ciruit and the signal does pass via the resistor. In III at high frequencies the capacitor becomes a dead short and ditto. In IV at high frequencies the capacitor becomes a dead short across the output. Thus Option D.

#40. In the steady state the inductor won't have a voltage across it ( it will be a dead short, unlike a capacitor which would ), thus options A, B and C are out. So how fast would the voltage decay? 10mH doesn't sound like much, so I'll go for a quick transition - Option D.

#41. Equation II reflects that we have no magnetic monpoles ie. any enclosing surface will not have a nett flux in or out. So that equation would change to resemble I. III would also change as we could then have currents of monpoles similiar to the J term in IV. So Option E.

#42. Before any movement the flux from the middle loop passes through the enclosed area of loops A and B towards the observer ( right hand rule with the thumb along the current and the curled fingers identifying the field ). As the middle loop slides toward loop A, the flux through A will rise and induce a current. The flux produced by this induced current must oppose the original change ( due to conservation of energy, amongst others ). Thus A has current in the clockwise sense ( right hand rule .... ). Now for loop B the flux is diminishing, similiar logic applies ie. any induced current must produce a flux that opposes the original flux change. ( that's the minus sign in equation III of Q#41 ). Thus a counterclockwise current in loop B results. So that's Option C. Now I'll just untangle my fingers ...... :-)

Oh crap .... some hard-boiled QM questions next! :-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter.Blaise Pascal

Basically whatever interference feature repeats in a given distance for one wavelength will repeat with the other. A shorter wavelength is a higher wavenumber ( number of oscillations per meter ) - while similiar to a time frequency, this is a spatial frequency. We know green light has a shorter wavelength ( higher time frequency ) than red light & also ought to be in inverse proportion to those fringe counts.

Wow, Mike, you're doing a terrific job! I wanted to help with at least one and thought I'd jump right to the last one since you're blazing through them so fast :)

#100. The distance that the mirror moves, d_m, will be proportional to the number of fringes counted, N, and proportional to the wavelength, Î» (lambda), of the laser light, and also proportional to a factor of 2 (since the light goes out and reflects back, traveling double the distance that the mirror moves, but the factor of 2 drops out in the next step):

d_m = 2 * N * Î»

The distance that the mirror moves will be the same for both laser beams, which means:

2* N * Î» = 2 * N * Î»

The wavelength of the green laser is unknown, so solve for Î» (where the factor of 2 drops out):

Î» = ( N * Î» ) / N

Substituting the values given in the problem:

Î» = 85,865 * 632.82 nm / 100,000

Î» = 543.37093 nm

So the answer is Option (B) 543.37 nm

- - -

It's a good test â€“ note that someone who doesn't know the proper way to think it through might notice that the number of fringes for the red light can be divided by 100 (maybe loosing three orders of magnitude in the number of green fringes when going from meters to nanometers?) and viola: Option (E) 858.65 nm.

Even trickier, note the wording of the problem, â€œ ...so that a ratio of fringe counts may be used to compare the wavelengths of two lasers with high precision.â€? If a person tries A is to B as C is to D:

Î» / 100,000 = 632.82 nm / 85,865

Î» = 736.994... nm

That would be Option (D) 736.99 nm. It's a very good test :)

I'm not a physicist and I had to research it â€“ a few days ago I would have guessed a wavemeter is what's used to give the surfing conditions. So Mike or anyone please feel free to add/correct anything â€“ couldn't be a better forum for asking about interferometry :)

Good work Chipper. That's actually one of the few questions I got right. I'm still fighting off the flu at the moment, but when I'm feeling better I'll come back and start going through all the solutions posted here.

It will probably take several months to do that :D

## RE: Were those questions in

)

We'll see. I've forgotton so much. But I knew I could count on Mike to help, now i just have to work through what he said! :D

Physics is for gurls!

## A top up :-) #26. Centre

)

A top up :-)

#26. Centre of mass is half way up the rod, the gravitational potential energy of that when initially upright is

M * g * L / 2

[ g is acceleration due to gravity ]

when laid down that gravitational potential energy is zero. Now one can start with half the mass but at twice the height and the energy change will be no different during the falling process. Also the torque ( force times distance ) will also be the same whether we have all the mass half way up vs. half the mass all the way up. So the rate of change of angle will be the same, specifically the velocity of the tip when it hits the dirt.

So I've converted the stated problem to an identical but simpler one. Now the change in gravitational potential energy equals the change in kinetic energy.

(M/2) * g * L = M * V * V / 2

V^2 = g * L

V = SQRT(g * L)

Option B.

#27. After some research, option A - real. I cheated and looked it up! Some long forgotten stuff about diagonalising a matrix to get eigenvalues. :-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## Wow, Mike, you're doing a

)

Wow, Mike, you're doing a terrific job! I wanted to help with at least one and thought I'd jump right to the last one since you're blazing through them so fast :)

#100. The distance that the mirror moves, d_m, will be proportional to the number of fringes counted, N, and proportional to the wavelength, Î» (lambda), of the laser light, and also proportional to a factor of 2 (since the light goes out and reflects back, traveling double the distance that the mirror moves, but the factor of 2 drops out in the next step):

d_m = 2 * N * Î»

The distance that the mirror moves will be the same for both laser beams, which means:

2* N * Î» = 2 * N * Î»

The wavelength of the green laser is unknown, so solve for Î» (where the factor of 2 drops out):

Î» = ( N * Î» ) / N

Substituting the values given in the problem:

Î» = 85,865 * 632.82 nm / 100,000

Î» = 543.37093 nm

So the answer is Option (B) 543.37 nm

- - -

It's a good test â€“ note that someone who doesn't know the proper way to think it through might notice that the number of fringes for the red light can be divided by 100 (maybe loosing three orders of magnitude in the number of green fringes when going from meters to nanometers?) and viola: Option (E) 858.65 nm.

Even trickier, note the wording of the problem, â€œ ...so that a ratio of fringe counts may be used to compare the wavelengths of two lasers with high precision.â€? If a person tries A is to B as C is to D:

Î» / 100,000 = 632.82 nm / 85,865

Î» = 736.994... nm

That would be Option (D) 736.99 nm. It's a very good test :)

I'm not a physicist and I had to research it â€“ a few days ago I would have guessed a wavemeter is what's used to give the surfing conditions. So Mike or anyone please feel free to add/correct anything â€“ couldn't be a better forum for asking about interferometry :)

## I'll get to that one Chipper!

)

I'll get to that one Chipper! You can work back if you like while I go forward. T'is fun! :-)

Update on #27. Hermitian operator means the square matrix which represents that is it's own conjugate transpose ie:

H(i,j) = [H(j, i)]*

so you swap the rows for columns, and then take the complex conjugate of all entries ( or vice versa if you like ) . This implies the elements in the leading diagonals are real as

H(i,i) = [H(i, i)]*

means the imaginary parts are zero. Now any square matrix when diagonalised ( by pre and post multiplying by a certain unitary matrix and it's transpose ) will remain with those real diagonals - and these entries are thus interpreted as the eigenvalues. You need at least a first course in linear algebra to catch up on this stuff - Google 'Gilbert Strang', he's the man for that! :-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## Hey, I've just noticed the

)

Hey, I've just noticed the answers are at the back! :-)

#1. Doh! Read the question properly Mike...... :-)

#2. They're wrong by a factor of ten.

#10. They've combined the capacitances incorrectly!

#22. My mistake, if you have the semi-major axis and the orbital period you can get the central mass - as the proportionality between R and T involves it.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## OK, bangin' on ....

)

OK, bangin' on .... :-)

#31. The antimatter angle is a red herring - this is a Hydrogen atom with a ground state ( n = 1 ) of -13.6eV. Levels go like 1/ n^2. From n = 3 to n = 1

1/1^2 - 1/3^2 = 8/9

Hence 13.6 * (8/9) ~ 12.08

Option C [ their answer is wrong ]

#32. Start with

E^2 = E0^2 + p^2 * c^2

(2 * m * c^2)^2 = (m * c^2)^2 + p^2 * c^2

p^2 = [4 * m^2 * c^4 - m^2 * c^4]/c^2 = [3 * m^2 * c^4]/c^2 = 3 * m^2 * c^2

p = SQRT(3) * m * c

Option D

#33. The slick way around this one is to recall ( and by aligning the x-axis along the path of the particle's travel ), that the spacetime distance/metric

c^2 * dt^2 - dx^2

is the same value in both frames ( where dt and dx are the time and space separations of two events in the one frame ). In the rest frame this is

c^2 * [10^(-8)]^2 - 0^2

NB. it didn't go anywhere in it's rest frame! For the lab frame :

c^2 * dt^2 - 30^2

ie. dt is to be found. So ....

c^2 * [10^(-8)]^2 - 0^2 = c^2 * dt^2 - 30^2

dt^2 = [900 + c^2 * 10^(-16)]/c^2 = 900/9 * 10^(-16) + 10^(-16) = 101 * 10^(-16)

dt ~ 10.049 * 10^(-8)

Thus in the lab frame

speed = distance/time = 30/dt ~ 2.98 * 10^8

Option D

#34. This is an implicit question about light cones. How can two non co-incident events in one frame be seen as co-incident in another? Basically if they are separated by a sublight speed path. Option B. [ I disagree with their answer ]

#35. For black bodies the energy model goes like T for the oscillators and a T for each degree/dimension of freedom in the cavity ie. 3 of those. So four powers in all. 3^4 = 81. Option E.

#36. Quasi-static ( succession of states close to equilibrium, thus ~ reversible ) and adiabatic ( no heat flow in or out ) means options A, C and D are true. I'm not sure you can firmly make comment either way on B. The temperature will go down however, so Option E.

#37. As the path BC is isothermal then PV = nRT can be used to deduce that P ~ 1/V for that path. Going from B to C has P going from 200 to 500 then V must have started at 5 ( ie. point B ) to then go to 2. The area under the curve is less than for a triangle ( area = base * height / 2 ), and so energy/work = area bounded < (500 - 200) * ( 5 - 2 ) / 2 = 300 * 3 / 2 = 450. If you look at the cycle, the gas is being pressurised under constant volume ( CA, ie. heated ), allowed to expand under constant pressure ( AB, ie. it will cool ) and then being compressed while losing heat to maintain constant temperature ( BC ). This shows work is being done on the gas ( not by it which would be the reverse way around the loop ). Given the phrasing of the question I'd opt for -300J and hence Option D.

#38. This a driven oscillator with damping. The AC source drives, the resistor converts current enrgies to heat, and the charge slops back and forth between the capacitor and the inductor. The capacitor soaks up the extremes of charge build up ( voltage ), the inductor the extremes of charge movement changes ( current fluctuations ). The maximal current amplitude is when C & L are tuned for full oscillation. This implies the impedances of C and L must be equal ( though acting 90 degrees of phase apart ). To much C lowers the voltage to then later kick the current back through L, while too little C raises the voltage and inhibits the current. Hence ( with w the angular frequency ) :

w * L = 1 / [w * C]

C = 1 / [L * w^2] = 1 / [25 * 10^(-3) * 10^6] = 0.04 * 10^(-3) = 40 * 10^(-6) = 40 uF

Option D. The exact values of the AC voltage and the resistance don't matter.

#39. OK, capacitors prevent DC current ( frequency zero ), and inductors suppress large current changes ( ie. AC ). In I at high frequencies the inductor acts as an open ciruit and the signal does not pass. In II at high frequencies the the inductor acts as an open ciruit and the signal does pass via the resistor. In III at high frequencies the capacitor becomes a dead short and ditto. In IV at high frequencies the capacitor becomes a dead short across the output. Thus Option D.

#40. In the steady state the inductor won't have a voltage across it ( it will be a dead short, unlike a capacitor which would ), thus options A, B and C are out. So how fast would the voltage decay? 10mH doesn't sound like much, so I'll go for a quick transition - Option D.

#41. Equation II reflects that we have no magnetic monpoles ie. any enclosing surface will not have a nett flux in or out. So that equation would change to resemble I. III would also change as we could then have currents of monpoles similiar to the J term in IV. So Option E.

#42. Before any movement the flux from the middle loop passes through the enclosed area of loops A and B towards the observer ( right hand rule with the thumb along the current and the curled fingers identifying the field ). As the middle loop slides toward loop A, the flux through A will rise and induce a current. The flux produced by this induced current must oppose the original change ( due to conservation of energy, amongst others ). Thus A has current in the clockwise sense ( right hand rule .... ). Now for loop B the flux is diminishing, similiar logic applies ie. any induced current must produce a flux that opposes the original flux change. ( that's the minus sign in equation III of Q#41 ). Thus a counterclockwise current in loop B results. So that's Option C. Now I'll just untangle my fingers ...... :-)

Oh crap .... some hard-boiled QM questions next! :-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## That's right for #100

)

That's right for #100 Chipper! :-)

Basically whatever interference feature repeats in a given distance for one wavelength will repeat with the other. A shorter wavelength is a higher wavenumber ( number of oscillations per meter ) - while similiar to a time frequency, this is a spatial frequency. We know green light has a shorter wavelength ( higher time frequency ) than red light & also ought to be in inverse proportion to those fringe counts.

wavelength(green) / wavelength(red) = fringes(red) / fringes(green)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## RE: Wow, Mike, you're doing

)

Good work Chipper. That's actually one of the few questions I got right. I'm still fighting off the flu at the moment, but when I'm feeling better I'll come back and start going through all the solutions posted here.

It will probably take several months to do that :D

Physics is for gurls!