Problem Solving Strategies

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?

yuvraj singh

@JohnRennie

John Rennie

@yuvrajsingh you've done a question like that haven't you?

M. Guru Vishnu

5:59 AM

@JohnRennie, Sir, you may mute the audio before watching that animation :-)

John Rennie

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 $W = \int_0^L F(x)dx$

yuvraj singh

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.

John Rennie

@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 ...

M. Guru Vishnu

@JohnRennie Ok sir. Thank you for clarifying.

John Rennie

@yuvrajsingh

I've shown the block displaced a distance $x$ from the starting position.

yuvraj singh

6:08 AM

Yes.

John Rennie

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?

yuvraj singh

Now I understand so the equation would be kxmg =$kxmgdx/l$ right?

From x to l.

@JohnRennie

John Rennie

The friction varies as $\mu = kx$, so the friction at my element $dy$ is $\mu = k(x+y)$. Yes?

yuvraj singh

OK.

John Rennie

So the frictional force on the element $dy$ is:

$$ dF(x,y) = k(x+y) \frac{Mg}{L} dy $$

yuvraj singh

6:13 AM

OK.

John Rennie

And we get the total friction by integrating from $y=0$ to $y=L-x$

yuvraj singh

Yes.

John Rennie

$$ F(x) = \int_0^{L-x} k(x+y) \frac{Mg}{L} dy $$

yuvraj singh

Yes

John Rennie

And once you have $F(x)$ the total work done is then just:

$$ W = \int_0^L F(x) dx $$

yuvraj singh

6:15 AM

Sir one question?

John Rennie

Yes?

yuvraj singh

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

John Rennie

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.

yuvraj singh

OK.

John Rennie

Actually I could make the integrals a bit simpler by defining my variable $y$ differently ...

It would still give the same result though.

yuvraj singh

6:20 AM

Yes I understand now I have one more question.? @JohnRennie

John Rennie

If we define $y$ like that the friction is just $\mu = ky$ and the first integral is $\int_x^L dy$

@yuvrajsingh yes?

yuvraj singh

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

John Rennie

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.

yuvraj singh

Thanks sir I got my point.

yuvraj singh

7:00 AM

@JohnRennie hi yuvraj here.

John Rennie

@yuvrajsingh hi :-)

yuvraj singh

Recently I was reading a column in science magazine says Einstein was always oppose the uncertainty principle.

Why?

@JohnRennie

John Rennie

Einstein didn't believe quantum mechanics was the correct theory.

yuvraj singh

Why?

In colum it is also mentioned Copenhagen interpretation

John Rennie

@yuvrajsingh You'd have to ask him.

skullpatrol

7:11 AM

Bohr–Einstein debates

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^ :-)

yuvraj singh

@JohnRennie, quantum theory was later prove right and Einstein wrong?

John Rennie

@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.

yuvraj singh

Ah,

skullpatrol

have you read anything about what Ed Witten has to say about this, sir?

John Rennie

@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.

skullpatrol

7:23 AM

I see.

user8718165

7:40 AM

@JohnRennie Good Morning sir :-)

John Rennie

@user8718165 hi :-)

M. Guru Vishnu

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?

John Rennie

@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.

M. Guru Vishnu

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.

John Rennie

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.

M. Guru Vishnu

Yes sir. So we could use an incoherent source after a slit. I think it will work.

John Rennie

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.

M. Guru Vishnu

@JohnRennie Could you please explain how Lloyd's mirror is related to this?

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.

M. Guru Vishnu

10:13 AM

@JohnRennie Yes, then they'll become coherent!

yuvraj singh

@JohnRennie Alesha here, I have a question regarding ydse.

John Rennie

@M.GuruVishnu yes, so that's why we don't need any extra measures to ensure coherency.

@yuvrajsingh hi, what's the question?

M. Guru Vishnu

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?

John Rennie

@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.

yuvraj singh

OK, actually, if we change the slit width of ydse,sinosidally. will we see a new pattern on screen

M. Guru Vishnu

10:15 AM

@JohnRennie Ok sir. Thank you :-)

John Rennie

@yuvrajsingh the slit width? Not the distance between the slits?

yuvraj singh

Sorry, distance between the slitsm

John Rennie

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.

yuvraj singh

OK.

Will the phase be same.

Can we locate of central maxima.

M. Guru Vishnu

@yuvrajsingh Yes. The central maxima doesn't shift its position, only its width varies. So we can find it.

John Rennie

10:24 AM

@yuvrajsingh the central maximum won't move because it is always exactly in between the two slits regardless of their spacing.

yuvraj singh

@JohnRennie oops.

I mean next bright fringe up side of central maxima.

John Rennie

@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 $\lambda = d(t)\sin\theta = d_0\sin(\omega t)\sin\theta$

yuvraj singh

OK.

@JohnRennie

John Rennie

@yuvrajsingh hi

yuvraj singh

Yes

M. Guru Vishnu

10:36 AM

@JohnRennie Sir, I think you meant user8718165 :-)

John Rennie

@M.GuruVishnu I see you've now asked a question on the main site :-)

M. Guru Vishnu

@JohnRennie Yes, I have deleted that.

Do you wish to answer it?

I'll undelete it then.

John Rennie

@M.GuruVishnu why delete it. It's a perfectly good question. I upvoted it! :-)

@M.GuruVishnu I was going to let Farcher answer it.

M. Guru Vishnu

@JohnRennie Thank you for the upvote :-)

@JohnRennie Why not you could answer sir?

@JohnRennie For why delete: My doubt was cleared.

John Rennie

Yes, but other people might also be interested in your question.

M. Guru Vishnu

10:39 AM

@JohnRennie Fine sir. Do you know about colour filters?

John Rennie

What about colour filters?

M. Guru Vishnu

0

Q: How does a violet colour filter work in the sense of RGB colour model?

I 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.

John Rennie

> 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.

M. Guru Vishnu

@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!

John Rennie

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.

M. Guru Vishnu

10:44 AM

@JohnRennie So magenta has red and blue components right sir?

John Rennie

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.

M. Guru Vishnu

@JohnRennie Interference?

John Rennie

@M.GuruVishnu No. The cells that detect colour are called cone cells:

Cone cell

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.

M. Guru Vishnu

@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?

John Rennie

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".

M. Guru Vishnu

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?

John Rennie

Violet or magenta?

M. Guru Vishnu

@JohnRennie Sorry sir. Magenta only. I'm making the same mistake again.

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.

M. Guru Vishnu

@JohnRennie But I think, we can decompose the magenta light into red and blue components. What are your opinions on this sir?

John Rennie

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.

M. Guru Vishnu

10:56 AM

@JohnRennie Isn't colour a result of frequency (as we discussed 4 days ago)?

John Rennie

@M.GuruVishnu there are colours perceived by your brain that do not correspond to light of a single wavelength.

Magenta is one of them.

M. Guru Vishnu

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?

John Rennie

Or put another way, a rainbow contains only a subset of all the colours that the eye can see.

user8718165

@JohnRennie hello sir

M. Guru Vishnu

@JohnRennie Or visible spectrum is just a random mix of RGB, sir?

John Rennie

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.

M. Guru Vishnu

@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.

John Rennie

@M.GuruVishnu Correct.

M. Guru Vishnu

@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.

John Rennie

@user8718165 hi

user8718165

@JohnRennie hello sir.. I want to ask a qn :)

John Rennie

11:04 AM

@user8718165 yes?

user8718165

@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?

John Rennie

@user8718165 Do you know the equation for the capacitance of a sphere?

user8718165

@JohnRennie Ah I forgot :( Please wait a while sir

John Rennie

@user8718165 there's an easy way to get it ...

user8718165

@JohnRennie is it kQ/r ?

John Rennie

11:08 AM

A charged sphere has exactly the same electric field as a point charge.

user8718165

@JohnRennie yeah sir

John Rennie

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:

$$ V = \frac{kQ}{r} $$

user8718165

@JohnRennie yeah sir

John Rennie

So you can immediately see that for constant $V$ we get $Q \propto r$

user8718165

@JohnRennie yeah sir...

John Rennie

11:11 AM

So for a constant surface voltage the large sphere has a higher charge.

user8718165

@JohnRennie sir but objects with higher curvature have more charges for same potential...right?

@JohnRennie okay sir

John Rennie

No, a higher curvature means a greater potential for the same charge.

user8718165

@JohnRennie yeah sir...

@JohnRennie yeah sir...I was confused about that point... Now I got it

John Rennie

@user8718165 cool :-)

user8718165

@JohnRennie Thank you very much sir :-)

@JohnRennie sir, could you please come to our general chat room? :-)

M. Guru Vishnu

11:15 AM

@JohnRennie, Can we continue sir?

John Rennie

@M.GuruVishnu yes

M. Guru Vishnu

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

John Rennie

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.

M. Guru Vishnu

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.

John Rennie

OK ...

M. Guru Vishnu

11:20 AM

So our eyes cannot detect pure magenta's wavelength but can only detect red+blue combination. Am I right sir?

John Rennie

What does "pure magenta's wavelength" mean? There is no light of a single wavelength that is magenta.

M. Guru Vishnu

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?

John Rennie

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.

M. Guru Vishnu

@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)?

John Rennie

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.

M. Guru Vishnu

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.

John Rennie

@M.GuruVishnu Correct

M. Guru Vishnu

@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?

John Rennie

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.

M. Guru Vishnu

@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!!

John Rennie

That's a lot of chatting :-)

M. Guru Vishnu

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.

user8718165

@JohnRennie hello sir

John Rennie

@user8718165 hi

user8718165

@JohnRennie can I ask a last qn sir ?

John Rennie

@user8718165 yes

user8718165

@JohnRennie sir how does this work?

M. Guru Vishnu

11:44 AM

@user8718165 Corona discharge?

user8718165

@M.GuruVishnu sort of...but not here :(

M. Guru Vishnu

@user8718165 Hmm.

John Rennie

user8718165

@JohnRennie hello sir

John Rennie

9

Q: Why does charge accumulate at points?

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

electromagnetism

user8718165

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

John Rennie

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.

user8718165

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

John Rennie

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.

user8718165

@JohnRennie okay sir...got it..but why is a sphere different sir?

John Rennie

A sphere is spherically symmetric i.e. every point on its surface is just like every other pint.

user8718165

@JohnRennie yeah sir

John Rennie

That symmetry guarantees the charge density is the same everywhere (in the absence of external influences)

user8718165

12:12 PM

@JohnRennie okay sir

@JohnRennie so that's why it's different?

John Rennie

@user8718165 the pointed rod doesn't have the same symmetry so the charge doesn't have to be the same everywhere on the surface.

user8718165

@JohnRennie okay sir...got it

M. Guru Vishnu

@JohnRennie, Hi sir. Are you free now?

John Rennie

@M.GuruVishnu yes

M. Guru Vishnu

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.

John Rennie

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.

M. Guru Vishnu

Ok sir.

John Rennie

Then pure yellow light has a spectrum like this i.e. the intensity is only non-zero at the wavelength of yellow light.

M. Guru Vishnu

Yes sir.

John Rennie

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.

M. Guru Vishnu

@JohnRennie Yes and will the average of the two give yellow?

John Rennie

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.

M. Guru Vishnu

Ok sir. How can we represent two waves of different colour as a single wave? Is this simply not possible?

John Rennie

It is simply not possible.

M. Guru Vishnu

@JohnRennie Ok sir. If it had worked I thought of implementing the same to interference in non-monochromatic light.

Thank you very much sir :-)

John Rennie

:-)

M. Guru Vishnu

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

John Rennie

12:37 PM

@M.GuruVishnu I'll be around next year :-)

M. Guru Vishnu

@JohnRennie :-)

user8718165

12:51 PM

@JohnRennie hello sir :-)

John Rennie

@user8718165 hi

user8718165

@JohnRennie sir I worked it out. I thought about it for a while... now it got it about the spheres...