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5:29 AM
@JohnRennie, Good Morning Sir 😀
 
@JohnRennie hi, Alesha here.
 
@M.GuruVishnu @yuvrajsingh morning :-)
 
I have a question.
 
@yuvrajsingh yes?
 
I'm having a doubt from our yesterday's discussion sir. I'll ask after yuvraj's.
 
5:45 AM
OK sir let me upload screens hot of question paper.
@JohnRennie
 
@yuvrajsingh That looks straightforward. What's the problem with it?
 
I did it wrong, let me explain what I have tried.
Or the part which confuses me.
I know x=ut.
And u and the time is
Root 2h/g.
But, that, s when the cannon does not move., but if I apply momentum conservation when bullet fired the cannot must go back.
So it also travel a relative distance.
@JohnRennie
 
You are told that the shell has velocity $v_0$ with respect to the cannon. Not with respect to the ground.
So the relative horizontal velocity of the shell and the cannon/wagon is $v_0$.
You do not need to know what speeds the cannon and shell are moving with respect to the ground.
 
Ah, I misread the question.
 
@JohnRennie, Sir, may I ask my doubt now?
 
5:54 AM
@M.GuruVishnu yes, go ahead. @yuvrajsingh the answer should be obvious now. Yes?
 
Sorry to disturb one last a short question
@JohnRennie
 
@yuvrajsingh Yes?
 
My doubt is related to introducing a thin transparent film in one of the slits in YDSE. Yesterday, I understood that the net shift shouldn't be visible. I looked for some demonstrations online to understand this better. I got one in this video - youtube.com/watch?v=dg5Ch0blQOI&t=338s - at t=5m38s, the central fringe shifting is visible and the pattern shift is visible unlike what we discussed yesterday.
Could you please tell whether the video is incorrect sir?
 
@JohnRennie
 
@yuvrajsingh you've done a question like that haven't you?
 
5:59 AM
@JohnRennie, Sir, you may mute the audio before watching that animation :-)
 
If you consider the rod displaced a distance $x$ you can calculate the total frictional force $F(x)$ by integrating from $x$ to $L$. Then just integrate $W = \int_0^L F(x)dx$
 
Problem is that.
Latest say I have displace block x.
This consist of small x, let say x1+x2+x3,now different part of the block would experience different friction.
 
@M.GuruVishnu the video is incorrect. I think they are showing the brightest fringe displaced just to make it obvious to the viewer that te pattern is moving.
@yuvrajsingh give me a moment to draw a diagram ...
 
@JohnRennie Ok sir. Thank you for clarifying.
 
@yuvrajsingh
I've shown the block displaced a distance $x$ from the starting position.
 
6:08 AM
Yes.
 
So the part of the block that experiences friction is the part from $x$ to $L$. I've shown where the friction exists by the red line. OK so far?
 
Now I understand so the equation would be kxmg =$kxmgdx/l$ right?
From x to l.
@JohnRennie
 
The friction varies as $\mu = kx$, so the friction at my element $dy$ is $\mu = k(x+y)$. Yes?
 
So the frictional force on the element $dy$ is:
$$ dF(x,y) = k(x+y) \frac{Mg}{L} dy $$
 
6:13 AM
OK.
 
And we get the total friction by integrating from $y=0$ to $y=L-x$
 
Yes.
 
$$ F(x) = \int_0^{L-x} k(x+y) \frac{Mg}{L} dy $$
 
And once you have $F(x)$ the total work done is then just:
$$ W = \int_0^L F(x) dx $$
 
6:15 AM
Sir one question?
 
Yes?
 
There will be two type of force which are doing work.
And friction and f.
So we assume that change in k. E will be zerom
 
The question just asks for the total heat generated.
And that is the work done by the friction. You don't need to worry about the external force.
 
Actually I could make the integrals a bit simpler by defining my variable $y$ differently ...
It would still give the same result though.
 
6:20 AM
Yes I understand now I have one more question.? @JohnRennie
 
If we define $y$ like that the friction is just $\mu = ky$ and the first integral is $\int_x^L dy$
@yuvrajsingh yes?
 
Sir I have disc which is rotating with constant angular velocity $ \ omega $after some time angular velocity changes from w1 to w2, without torque change question ask me to find ratio of radius of gyrations so I apply energy conservation, 1/2Iw^2=1/2iw_2^2
Is that correct?
Because I think it is wrong, since net torque given is same , there no information about force.
@JohnRennie
 
I would conservation of angular momentum.
You can't use conservation of energy because work will be done by whatever force is changing the shape of the object.
 
Thanks sir I got my point.
 
7:00 AM
@JohnRennie hi yuvraj here.
 
@yuvrajsingh hi :-)
 
Recently I was reading a column in science magazine says Einstein was always oppose the uncertainty principle.
Why?
@JohnRennie
 
Einstein didn't believe quantum mechanics was the correct theory.
 
Why?
In colum it is also mentioned Copenhagen interpretation
 
@yuvrajsingh You'd have to ask him.
 
7:11 AM
The Bohr–Einstein debates were a series of public disputes about quantum mechanics between Albert Einstein and Niels Bohr. Their debates are remembered because of their importance to the philosophy of science. An account of the debates was written by Bohr in an article titled "Discussions with Einstein on Epistemological Problems in Atomic Physics". Despite their differences of opinion regarding quantum mechanics, Bohr and Einstein had a mutual admiration that was to last the rest of their lives.The debates represent one of the highest points of scientific research in the first half of the twentieth...
or, read this^ :-)
 
@JohnRennie, quantum theory was later prove right and Einstein wrong?
 
@yuvrajsingh I don't think Einstein ever said QM didn't work, because he could see that it agreed with experiment.
His objection was that QM seemed to be founded on some rather arbitrary principles.
He believed there was a deeper underlying theory that would explain QM and make it clearer what it was based on.
There are still a few physicists who believe something like this. The Nobel prize winner Gerald 't Hooft has proposed similar ideas, though he accepts that it's a speculative idea that will probably be proved wrong.
 
have you read anything about what Ed Witten has to say about this, sir?
 
@skullpatrol I don't think Witten has ever commented on it. The trouble with this area is that ultimately it comes down to a matter of opinion and there's no real proof either way. Witten generally stays out of this sort of debate.
 
7:23 AM
I see.
 
7:40 AM
@JohnRennie Good Morning sir :-)
 
@user8718165 hi :-)
 
 
2 hours later…
9:30 AM
@JohnRennie, Hi sir. A quick doubt. In YDSE, for what reasons in a laser light preferred? Are there reasons other than its monochromatic nature?
It's a good coherent source, but I don't think it's need. As we could use a primary slit to illuminate the double slit to produce interference pattern.
Is that because, the fringes could be more clearly seen?
 
@M.GuruVishnu The reason we use lasers is just that they are bright.
As you say we can use a single slit as a source of coherent light, but that severely reduces the intensity of the light.
 
10:04 AM
Ok sir. Thank you. (Sorry for the late reply)
But in Lloyd's mirror, it isn't necessary to have laser light, right sir?
@JohnRennie sir.
 
The thing about Lloyd's mirror is that the two sources are both derived from the same single source.
In the YSDE we have two sources (one at each slit) and if the light is not coherent the two sources will have a phase offset that varies randomly.
 
Yes sir. So we could use an incoherent source after a slit. I think it will work.
 
It's like that experiment where you place a sheet in front of one of the slits, but where the thickness of the sheet varies randomly with time.
So you still get a diffraction pattern, but that pattern shifts around so fast that to your eye it just looks like a blur.
 
@JohnRennie Could you please explain how Lloyd's mirror is related to this?
 
In the Young's mirror the (virtual) second source is a reflection of the first source, so of course it always has exactly the same phase as the first source.
 
10:13 AM
@JohnRennie Yes, then they'll become coherent!
 
@JohnRennie Alesha here, I have a question regarding ydse.
 
@M.GuruVishnu yes, so that's why we don't need any extra measures to ensure coherency.
@yuvrajsingh hi, what's the question?
 
Sir could you please explain this:
2 mins ago, by M. Guru Vishnu
@JohnRennie Could you please explain how Lloyd's mirror is related to this?
 
@M.GuruVishnu I just did
2 mins ago, by John Rennie
In the Young's mirror the (virtual) second source is a reflection of the first source, so of course it always has exactly the same phase as the first source.
 
OK, actually, if we change the slit width of ydse,sinosidally. will we see a new pattern on screen
 
10:15 AM
@JohnRennie Ok sir. Thank you :-)
 
@yuvrajsingh the slit width? Not the distance between the slits?
 
Sorry, distance between the slitsm
 
If you change the spacing between the slits it changes the spacing between the fringes in the diffraction pattern. If you change the spacing sinusoidally with time all that happens if the fringe spacing oscillates sinusoidally with time.
 
OK.
Will the phase be same.
Can we locate of central maxima.
 
@yuvrajsingh Yes. The central maxima doesn't shift its position, only its width varies. So we can find it.
 
10:24 AM
@yuvrajsingh the central maximum won't move because it is always exactly in between the two slits regardless of their spacing.
 
@JohnRennie oops.
I mean next bright fringe up side of central maxima.
 
@yuvrajsingh if the fringe spacing varies as $d(t) = d_0\sin(\omega t)$ then the position of the first fringe will be given by $\lambda = d(t)\sin\theta = d_0\sin(\omega t)\sin\theta$
 
OK.
@JohnRennie
 
@yuvrajsingh hi
 
10:36 AM
@JohnRennie Sir, I think you meant user8718165 :-)
 
@M.GuruVishnu I see you've now asked a question on the main site :-)
 
@JohnRennie Yes, I have deleted that.
Do you wish to answer it?
I'll undelete it then.
 
@M.GuruVishnu why delete it. It's a perfectly good question. I upvoted it! :-)
@M.GuruVishnu I was going to let Farcher answer it.
 
@JohnRennie Thank you for the upvote :-)
@JohnRennie Why not you could answer sir?
@JohnRennie For why delete: My doubt was cleared.
 
Yes, but other people might also be interested in your question.
 
10:39 AM
@JohnRennie Fine sir. Do you know about colour filters?
 
What about colour filters?
 
0
Q: How does a violet colour filter work in the sense of RGB colour model?

M. Guru VishnuI learnt after reading How do color filters placed in front of a light source,change the color of light that passes through? colour filter work by permitting only one colour and absorbing or reflecting the rest. In this sense, a violet colour filter works by allowing only violet colour to pass th...

I tried to use this for interference. That's why I had that doubt sir.
 
> It can be seen that violet colour is obtained as a result of red and blue colours
That colour is magenta not violet.
Violet normally means the colur in the rainbow to the short wavelength side of blue.
 
@JohnRennie Oops. It looked violet in my monitor. But everything else remains the same sir.
Seems I don't know to identify colours, need to get help from kindergarten friends!
 
The eye doesn't measure the wavelength of the light and deduce the colour from that.
The eye has three different colour receptors and it judges the colour based on how strongly those receptors get triggered.
 
10:44 AM
@JohnRennie So magenta has red and blue components right sir?
 
So a single wavelength light and light that was a mixture of two wavelengths could look the same if they stimulated the receptors in your eye in the same way.
 
@JohnRennie Interference?
 
@M.GuruVishnu No. The cells that detect colour are called cone cells:
Cone cells, or cones, are photoreceptor cells in the retinas of vertebrate eyes (e.g. the human eye). They respond differently to light of different wavelengths, and are thus responsible for color vision and function best in relatively bright light, as opposed to rod cells, which work better in dim light. Cone cells are densely packed in the fovea centralis, a 0.3 mm diameter rod-free area with very thin, densely packed cones which quickly reduce in number towards the periphery of the retina. There are about six to seven million cones in a human eye and are most concentrated towards the macula...
You have three different types of cone cell (assuming you aren't colour blind :-) and they detect light of different wavelengths.
 
@JohnRennie That was funny!
@JohnRennie I know only basics about rods and cones, as I don't have biology in my syllabus. But, can we consider pure violet light as a result of constructive interference of red and blue lights?
 
It isn't interference. The red light triggers the red cone cells and the blue light triggers the blue cone cells, and the brain goes "aha, that means magenta".
 
10:52 AM
Ok sir. Now we pass white light through a violet filter. Then pass the resulting light through a red filter. Will the final light beam will be red in colour?
 
Violet or magenta?
 
@JohnRennie Sorry sir. Magenta only. I'm making the same mistake again.
 
I don't think there is a single wavelength light that appears to us as magenta i.e. that colour can only be produced by mixing two different wavelengths.
 
@JohnRennie But I think, we can decompose the magenta light into red and blue components. What are your opinions on this sir?
 
It's important to remember that colour is an invention of the human brain not a real thing. It's just a way the brain encodes various mixtures of light.
 
10:56 AM
@JohnRennie Isn't colour a result of frequency (as we discussed 4 days ago)?
 
@M.GuruVishnu there are colours perceived by your brain that do not correspond to light of a single wavelength.
Magenta is one of them.
 
So, when we see light dispersed by a prism, the different colours are not of different wavelength but of a combination of different wavelengths interpreted as a different colour by our brain?
 
Or put another way, a rainbow contains only a subset of all the colours that the eye can see.
 
@JohnRennie hello sir
 
@JohnRennie Or visible spectrum is just a random mix of RGB, sir?
 
10:59 AM
@M.GuruVishnu no, light dispersed by a prism is a range of light of a single varying wavelength.
But if you try it you'll find the colour magenta isn't in that range.
 
@JohnRennie Oh. So a continuous visible spectrum like rainbow or prism, doesn't have all colours we see? I though the other way round sir.
 
@M.GuruVishnu Correct.
 
@JohnRennie Ok sir. Thank you :-) I think @user8718165, is waiting for a doubt. I'll ask further after theirs. Meanwhile I'll think about this further.
 
@user8718165 hi
 
@JohnRennie hello sir.. I want to ask a qn :)
 
11:04 AM
@user8718165 yes?
 
@JohnRennie sir if there are two spheres of radius $r_1$ and $r_2 (r_1>r_2)$ for the same potential $r_1$ should have less charge...is it sir?
 
@user8718165 Do you know the equation for the capacitance of a sphere?
 
@JohnRennie Ah I forgot :( Please wait a while sir
 
@user8718165 there's an easy way to get it ...
 
@JohnRennie is it kQ/r ?
 
11:08 AM
A charged sphere has exactly the same electric field as a point charge.
 
@JohnRennie yeah sir
 
So the field at the surface of a charged sphere of radius $r$ is exactly the same as the field a distance $r$ from the same point charge.
So the potential is just:
$$ V = \frac{kQ}{r} $$
 
@JohnRennie yeah sir
 
So you can immediately see that for constant $V$ we get $Q \propto r$
 
@JohnRennie yeah sir...
 
11:11 AM
So for a constant surface voltage the large sphere has a higher charge.
 
@JohnRennie sir but objects with higher curvature have more charges for same potential...right?
@JohnRennie okay sir
 
No, a higher curvature means a greater potential for the same charge.
 
@JohnRennie yeah sir...
@JohnRennie yeah sir...I was confused about that point... Now I got it
 
@user8718165 cool :-)
 
@JohnRennie Thank you very much sir :-)
@JohnRennie sir, could you please come to our general chat room? :-)
 
11:15 AM
@JohnRennie, Can we continue sir?
 
@M.GuruVishnu yes
 
Coming back to our colour filters. Could you please explain the final answer for the following? (neglecting violet, kindly assume it to be magenta)
24 mins ago, by M. Guru Vishnu
Ok sir. Now we pass white light through a violet filter. Then pass the resulting light through a red filter. Will the final light beam will be red in colour?
I think no light must come out just like a crossed polaroid
 
A magenta filter must allow through red light and blue light and block green light, because for the light to appear magenta to us it has to be a mixture of red and blue light.
 
Ok sir. I have a doubt from your previous message:
24 mins ago, by John Rennie
I don't think there is a single wavelength light that appears to us as magenta i.e. that colour can only be produced by mixing two different wavelengths.
 
OK ...
 
11:20 AM
So our eyes cannot detect pure magenta's wavelength but can only detect red+blue combination. Am I right sir?
 
What does "pure magenta's wavelength" mean? There is no light of a single wavelength that is magenta.
 
Ok sir. For me magenta is confusing. Can we switch to yellow? A combination of red and green and which is also present in the rainbow?
 
Yes, there is a light of a single wavelength that appears yellow. That's because single wavelength yellow light triggers both the red and the green receptors in our eye.
A mixture of red light and green light also triggers the red and green receptors in our eyes, and the brain can't tell the difference.
 
@JohnRennie Ok sir. What about two colour filters placed in the path of white light? Will they act as crossed polaroids (pass axes oriented perpendicularly)?
 
Suppose you have a filter that only allows light of a wavelength $\lambda_1$ and a second filter that only allows light with a wavelength $\lambda_2$.
If you put the two filters together then they will block all light.
 
11:27 AM
@JohnRennie Thank you sir. So our eyes cannot tell whether something yellow is purely yellow or a combination of red and green.
 
@M.GuruVishnu Correct
 
@JohnRennie Thank you sir. If possible could you please copy all your messages to the answer box in my question, I could upvote and accept?
 
It would take considerable rewriting to convert our conversation about into a coherent answer, and right now I'm doing other stuff so I don't have that time. I'll look at doing it later.
@M.GuruVishnu you should probably edit your question to make it refer to yellow not violet.
 
@JohnRennie Fine sir. However, I always store our conversation in a word document for my future reference. I also have identification codes for them. Now it's about 65!!
 
That's a lot of chatting :-)
 
11:31 AM
@JohnRennie Thank you sir. I'll make the changes accordingly.
@JohnRennie Yes sir. Sometimes chats extend till 4 pages, each page has unique code.
 
@JohnRennie hello sir
 
@user8718165 hi
 
@JohnRennie can I ask a last qn sir ?
 
@user8718165 yes
 
@JohnRennie sir how does this work?
 
11:44 AM
@user8718165 Corona discharge?
 
@M.GuruVishnu sort of...but not here :(
 
@user8718165 Hmm.
 
@JohnRennie hello sir
 
9
Q: Why does charge accumulate at points?

user24082This is a very trivial question, but I can't seem to reason it out, again, as to why charges gather at points and edges.

 
11:47 AM
@JohnRennie thank you sir...I'll look at it
@JohnRennie sir can we apply "greater potential for same charge" to this rod?
@JohnRennie hello sir
 
The trouble is that the charge moves around on the surface of the rod, so the charge density is different at the pount.
With a sphere the symmetry means the charge density is the same everywhere.
So the increased potential at the point is combination of both the increased curvature and the increased charge density.
 
12:03 PM
@JohnRennie okay sir, since the surface is at the same potential, the pointed ends should have more charge to match the same potential as the less curved area....right sir?
@JohnRennie okay sir
 
Yes. If the charge density was the same at the pointed end the potential would be lower.
So charge flows onto the pointed end until the potentials are equal.
 
@JohnRennie okay sir...got it..but why is a sphere different sir?
 
A sphere is spherically symmetric i.e. every point on its surface is just like every other pint.
 
@JohnRennie yeah sir
 
That symmetry guarantees the charge density is the same everywhere (in the absence of external influences)
 
12:12 PM
@JohnRennie okay sir
@JohnRennie so that's why it's different?
 
@user8718165 the pointed rod doesn't have the same symmetry so the charge doesn't have to be the same everywhere on the surface.
 
@JohnRennie okay sir...got it
 
@JohnRennie, Hi sir. Are you free now?
 
@M.GuruVishnu yes
 
Previously, we concluded a pure yellow light wave activates both red and green cones in our eyes. Also we know red and green light waves in combination give yellow as the result. So can we say pure yellow light is a result of superimposing (I used interference incorrectly previously) green and red light waves?
@JohnRennie sir.
 
12:20 PM
No.
Suppose we draw a diagram to show the spectrum of light.
On the x axis we have wavelength and on the y axis we have intensity.
 
Ok sir.
 
Then pure yellow light has a spectrum like this i.e. the intensity is only non-zero at the wavelength of yellow light.
 
Yes sir.
 
A mixture of red and green light would look like this i.e. it would be non-zero at red wavelengths (about 700nm) and green wavelengths (about 550nm).
Our eye sees the two as the same, but they are not the same.
 
@JohnRennie Yes and will the average of the two give yellow?
 
12:28 PM
No.
Suppose you pass pure yellow light through a red filter. No light will get through.
But if you pass your mixture of red+green light through a red filter than red light will get through.
So the two give different results when you do the same experiment.
 
Ok sir. How can we represent two waves of different colour as a single wave? Is this simply not possible?
 
It is simply not possible.
 
@JohnRennie Ok sir. If it had worked I thought of implementing the same to interference in non-monochromatic light.
Thank you very much sir :-)
 
:-)
 
I do wish to discuss even more about using non-monochromatic light, colour filters, and related phenomenon with you, after both JEEs, next year sir. Now I'm forced to prepare for the exams. @JohnRennie
 
12:37 PM
@M.GuruVishnu I'll be around next year :-)
 
@JohnRennie :-)
 
12:51 PM
@JohnRennie hello sir :-)
 
@user8718165 hi
 
@JohnRennie sir I worked it out. I thought about it for a while... now it got it about the spheres...
@JohnRennie Now I fully agree with you sir... Thank you very much
 
OK ...
 

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