/0's physics phread

[TrueTears]

1. Why is it that in double-slit experiments we don't consider the contributions from the top and bottom halves of each slit as we do in single-slit diffraction?


2. In single-slit diffraction, you know how they take parallel rays of light and use to find the path difference... well the common reasoning is that the screen is so far away and the slits are so close that the rays are virtually coincident.
But if we make approximations like that, then doesn't that mean that the path difference should also be 'insignificant'? I would think that saying two parallel rays converge at infinity is such an impractical estimate that it completely negates the accuracy of , which must be VERY accurate if we consider light waves.

I think you should imagine a single slit not as really just one slit, but a continous domain of infinitely small slits the size of those found in double slit experiments. The reasoning is consistent then. As far as approximations try derive it without approximations, I think you have all the analytical geometry you need at your disposal

True, I imagine a single slit having 2 "corners" which the wave has to bend around hence cause diffraction. I think that is a good way of thinking about it.

[kamil9876]

Whoa, looked like a mathematical inconsistency haha.





Equating the constant term in the end with what they give you gives you're first method. Notice that this requires knowledge of only, and does not require knowledge of the gradient they have given you.

Your second method does use the gradient they have given you hence any errors that have crept in must have been caused by this. Notice that that gradient must be this is not actually the case according equal to the values of and that you have assumed.

[TrueTears]

Thanks TT, it is a tricky proof

Also, I'm having trouble with VCAA 2003 AOS4 Q5

"In a photoelectric effect experiment, light of various frequencies falls on a metal surface in a photocell. The photoelectrons are decelerated across a retarding voltage, and the stopping potential, , is measured for each frequency. Determine a value for the work function of this metal surface. Include the unit in your answer."

There is a graph which, using regression is approx where is in Volts and in Hz.

The intercept of this graph is .

From , when ,

Therefore,

But if I let then

And

Which method is wrong and Why? thanks

Your first method is wrong.

it is





Your 2nd method is fine.

The reason 1.8eV does not match with your ~1.7eV answer is because of the estimated intercepts, both answers are fine.

[TrueTears]

Whoa, looked like a mathematical inconsistency haha.





Equating the constant term in the end with what they give you gives you're first method. Notice that this requires knowledge of only, and does not require knowledge of the gradient they have given you.

Your second method does use the gradient they have given you hence any errors that have crept in must have been caused by this. Notice that that gradient must be this is not actually the case according equal to the values of and that you have assumed.

/0's inconsistency in his/her answers was not due to the use of gradient or whatever but it was because he/she did not recognize that he/she got mixed up with eV and Joules.

ok.

[kamil9876]

Whoa, looked like a mathematical inconsistency haha.





Equating the constant term in the end with what they give you gives you're first method. Notice that this requires knowledge of only, and does not require knowledge of the gradient they have given you.

Your second method does use the gradient they have given you hence any errors that have crept in must have been caused by this. Notice that that gradient must be this is not actually the case according equal to the values of and that you have assumed.

/0's inconsistency in his/her answers was not due to the use of gradient or whatever but it was because he/she did not recognize that he/she got mixed up with eV and Joules.

ok.

Notice that that gradient must be \frac{h}{q_e} this is not actually the case according equal to the values of q_e and h that you have assumed.

This pretty much says that or is wrong. Because even if he/she/it derived the gradient using WRONG values of q_e or h, he/she/it would find no inconsistency in the two methods. But the occurence of an inconsistency suggests a mistake in either the given gradient or the two constants. Hence I pointed him/her/it to the sauce of the error.

[/0]

lol thanks guys, gah I had the wrong units for q

[TrueTears]

lol thanks guys, gah I had the wrong units for q

took you a while to figure that out.

[/0]

lol thanks guys, gah I had the wrong units for q

took you a while to figure that out.

yes i am quite stupid

/0's inconsistency in his/her answers

[/0]

In Nelson there's







Where 'L' is the distance from the slits to the screen.

However, in the derivation the set up attached is used. This does not have 'L' as the perpendicular distance.

Did the book assume to be small, so that ?

However, isn't this made very inaccurate by the assumption that the screen is very far away?

[TrueTears]

Yes, I had the exact same problem, I think book assumes theta is very small hence L is the vertical distance

[/0]

Cool
also http://en.wikipedia.org/wiki/Planck_length






Apparently that's the smallest length possible. But what if you have rock of moving at , then

.

Moderator action: post edited by Mao

[TrueTears]

Eh what's all that spam for???

All I see is boxes.

Stop spamming. Your trolling needs to stop it is getting annoying.

[/0]

nevermind, I thought you would get it (also afaik i haven't trolled in a while)

A very large pipe in a science museum is open at both ends. It is large enough for people to walk inside it. A large loudspeaker faces inwards on the right end. Jin is walking along the pipe.

Later it goes on to say "the two ends of the pipe are nodes".

If the pipe is open at both ends then shouldn't there be antinodes at both ends?

[NE2000]

nevermind, I thought you would get it (also afaik i haven't trolled in a while)

A very large pipe in a science museum is open at both ends. It is large enough for people to walk inside it. A large loudspeaker faces inwards on the right end. Jin is walking along the pipe.

Later it goes on to say "the two ends of the pipe are nodes".

If the pipe is open at both ends then shouldn't there be antinodes at both ends?

Sound is reflected out of phase at open ends of a pipe and in the same phase at closed ends of a pipe. That is, at an open end a compression will turn into a rarefaction and vice versa. Therefore at the open end there will be destructive interference occurring. Hence a node.

[/0]

nevermind, I thought you would get it (also afaik i haven't trolled in a while)

A very large pipe in a science museum is open at both ends. It is large enough for people to walk inside it. A large loudspeaker faces inwards on the right end. Jin is walking along the pipe.

Later it goes on to say "the two ends of the pipe are nodes".

If the pipe is open at both ends then shouldn't there be antinodes at both ends?

Sound is reflected out of phase at open ends of a pipe and in the same phase at closed ends of a pipe. That is, at an open end a compression will turn into a rarefaction and vice versa. Therefore at the open end there will be destructive interference occurring. Hence a node.

thanks NE2000