@ManasDogra You can calculate the speed in the lab frame using the equation for the relativistic addition of velocity.

But it isn't clear to me what the question means. Does it mean the velocity as the particle moves downwards along the left side of the square, or the velocity when it's moving horizontally away from the start point?

If (a) is the correct answer that's very odd as it's only about 0.19c.

I don't know I'm afraid. If you ask in the hbar chat room someone there will know.

Can you migrate the chats like last time?...I don't know how to do that..

→ 3 messages moved to The h Bar

@ManasDogra done!

@JohnRennie One last...I guess this is related to Newton's rings somehow...just cant understand how.

I have no idea what is going on there ...

Oh wait ...

Maybe the filter paper just scatters the light isotropically, so you're just calculating the angles where you get constructive interference i.e. a path length of nλ

Hmm, but what pair of rays would be interfering ...?

Something of this sort?

That's what I was thinking. The paper scatters the incoming light so it has a continuous range of values of θ₁

And the first ring appears at the smallest value of θ₁ that can give constructive interference.

You need the path length in the glass to be λ/2. It's half λ because the ray reflecting at the top surface undergoes a phase shift of π when it relects.

@JohnRennie @JohnRennie Can you help me with the handout henceforth?

@JohnRennie That gives me $2*\mu*d=\lambda/2$...From there how should we introduce the radius of the rings(and also eliminate the wavelength).

The path length in the glass is $2\mu d/\cos\theta_2$ not $2\mu d$

But I don't see how we eliminate the wavelength ...

Maybe it isn't interference after all ...

If it was interference the radius R must depend on the wavelength.

And what's the significance of the filter paper being wet?

@user586228 Can you upload the handout somewhere? The small part you've posted is too little for me to work out what it means.

ok

showard.sdsmt.edu/Met321/04_SolnThermo/ChangeInSS.pdf

@JohnRennie

@ManasDogra this problem

@ManasDogra ah!

It's the critical angle for total internal reflection!

Of course :-)

For small angles of scattering the light exits the far side of the slab, but when the angle in the glass reaches the critical angle for TIR we get a sudden increase in brightness because all the light reflects off the far side of the glass.

@user586228 got it

@JohnRennie I get it now completely...Thank you so much.

@ManasDogra I didn't think it would be that simple :-)

@JohnRennie Tell me why the minus sign vanished

I'm not getting this on a first read through, so I'll have to put it aside and try again later when I have more time.

ok no problem

@JohnRennie But which one are you replying me to

The previous question or this?

@user586228 I wonder if it's a misprint and the minus sign was omitted by accident. As far as I can tell the minus sign should still be present.

Ok

So Delta G nought is -RT ln K

K=activity of products divided by reactants right?

Not the reverse by any chance?

K is definitely products/reagents not the reciprocal.

I wondered if the missing minus sign could be explained by taking the reciprocal of the expression inside the log, but that doesn't seem to be the case.

ok

@JohnRennie How do you know chemistry so well

I don't know chemistry very well :-)

?

I remember bits and pieces of the chemistry courses I did, but only bits and pieces.

That's great nice to hear.

@JohnRennie Can you tell me such chats for discussing chemistry questionsother thanChemistry stackexchange?

I don't know of any others. Sorry :-(

@JohnRennie In this problem tell me something...They have constructed tie lines to arrive at the composition of alpha and beta.In the solution to the Pb-Sn phase diagram they write,"A tie line has been contructed at 175 C ;its intersection with the alpha-alpha+Beta phase boundary is at 16 wt% Sn,which corresponds to the composition of alpha phase.Similarly the tie line intersection with the alpha+beta -beta phase boundary occurs at 97 wt % Sn,which is composition of Beta Phase".

How are they finding compositions of each individual phases by drawing straight lines

?

The curved line in the red box is the composition of the α phase as a function of temperature, and the line in the blue box is the composition of the β phase.

ok

In the region in between the two lines the material is a mixture of the α and β phases.

Then

The horizontal line is a line of equal temperature, so where it intersects the curved lines gives the compositions of the α and β phases at that temperature.

@JohnRennie Very true but tell me something....While you move along the x axis then the compistion is supposed to reamin the same isn't it?

'composition*

Since temperature is not changing

Huh? As you move horizontally the overall tin/lead ratio is changing.

The compositions of the α and β phases remain constant, but the relative amounts of the two phases changes.

At the left end we have all α phase. As we move right along the line we get more and more β phase present until at the right end we have only the β phase.

@JohnRennie Correct me if I am wrong the wt % of Sn is fixed for a phase at a given temperature.It is only the contect of alpha or beta that changes and proportionately the weight percent that changes as well.Yes or no?

Yes. As you move along the tie line the composition of the α and β phases is constant, but you get different amounts of those two phases.

ok

Thanks a lot @JohnRennie

@JohnRennie I just found out that this problem is from Hecht's optics book problem no. 4.91 in 5th Pearson edition.

Ah. I still have the copy of Hecht's book that I used in 1980 :-)

It's a bit tattered now :-)

@JohnRennie Hello sir I am unable to unvite you to engineering probs I would be greatful if you could just come there

@Jasmine Can you post the link to the room?

Sorry Sir I have to leave :'(

OK. Ping me when you're back.

Hi @JohnRennie sir

@Rover hi :-)

Sir , in this question I got the answer.., but I want to know whether they will collide if condition is not satisfied...

OK ... ?

What's the condition?

These are options and option c comes correct...

But , if that condition is not satisfied, then will they collide I want to know ... I.e will the instant will come such that at a particular same time they will collide...?

Presumably you just calculate the diameters of the circles that the particles move in, and the sum of the diameters has to be equal to or greater than the distance between the particles. Yes?

Yes , I did the same ...

That is a necessary condition, but not a sufficient condition.

Why?

Just because the trajectories overlap that doesn't mean the particles will collide. For that the two particles have to arrive at the point where the circles intersect at the same time.

Yes , exactly and how can we find that condition...?

I think that's hard.

You would have to solve for the position of the intersection and also the times taken for the two particles to reach that position.

How?

The two times would have to be identical for the particles to collide.

Yes

Offhand I don't know how to do that calculation. I'd have to go away and think about it ...

I suspect it would be messy.

Yep a bit , I am not getting...

I have to confess I feel no great enthusiasm to attempt the calculation ...

And I have to go now anyway.

Ok , but if you tell me the approach I will do...

I have to go, but we can talk about it later.

Ok

I'll be out until after 4 p.m. UK time - 9:30 p.m. Indian time.

We can discuss it then.

Ok , sir ...

@JohnRennie Sir this is another question I am having problems with...It would be great if we can discuss once.

@JohnRennie sir can you tell me the difference between geometrical and optical paths of l feel confused about it. . .

@JohnRennie

Why "electrochemical" and nothing else?

Hi @JohnRennie

@Rover hi :-)

Sorry for the delay but I'm running late today. I'm around for about half an hour more if you want to discuss the problem.

@JohnRennie yes sir ...

@JohnRennie ,are you there ....

@Rover hi :-)

Give me a moment to draw a diagram ...

Ok

@Rover This shows the two particles with velocities $v_1$ and $v_2$ initially parallel.

The trajectories overlap, so they could collide, but will they collide?

The angular velocity of particle 1 is $\omega_1 = v_1/r_1$, and likewise for particle 2. OK so far?

Yes ok

So the angle that particle rotates is $\theta_1 = \omega_1 t$, and for particle 2 $\theta_2 = \omega_2 t$.

Yes

With a bit of geometry we could figure out what the angles are from the starting point to the point of intersection.

How , just hint...?

And for the collision to happen we require that particle 1 travels through this angle $\theta_1$ in exactly the same time particle 2 travels through the angle $\theta_2$. Yes?

Yes.

That's going to give us $\theta_1/\theta_2 = \omega_1/\omega_2$

Yes

And the angular velocity $\omega$ depends on the particle speed and mass.

As indeed does the radius of the circle.

Yes...

So you are going to end up with conditions for the particle mass and velocity that have to be satisfied for the collision to happen.

How we got theta 1 and theta2...

That's basic geometry. If you know the spacing of the centres of the circles, and you know their radii you can work out the angles by drawing a few triangles.

distance between their centres ?will it be d/2,but only when r1=r2..

Ok

I got it sir..

OK :-)

Bye ,thanks for your time ....

I have to go now, but I'll be around tomorrow as usual.

Ok you do great work sir...