Hello guys. I am a JEE aspirant. Can I join in with my physics doubts?

@πtimese Welcome brother :-) We'd be very happy to help you.

@user8718165 thanks a lot brother

@JohnRennie, morning sir, Alesha here.

@yuvrajsingh morning:-)

I think sir today there is election in UK?

@JohnRennie are you going for giving vote?

@yuvrajsingh the election is tomorrow, and yes I will be voting :-)

@JohnRennie, Hi sir. Good Morning :-)

@M.GuruVishnu morning :-)

Are you free now sir?

Yes

@JohnRennie hello sir...good morning

@user8718165 morning :-)

@JohnRennie yeah... I have a little question sir...

@user8718165 yes?

Sir, in YDSE when the source of light is changes from red to violet, how does this affect the intensity of fringes? I reasoned that, the intensity increases as the same amount of light is limited to thinner fringes (fringe width decreases on moving from red to violet because it is given by $D\lambda /d$ where $D$ is the distance of slit from screen and $d$ is the separation between slits)

Assuming you have the same power incident on the slits the intensity of the maxima is the same for all wavelengths.

@JohnRennie Why will it not increase sir? I think the same power is now confined to a narrower band instead of a broader one. So intensity increases. Why is this reasoning incorrect sir?

Each maximum is accompanied by a minimum, so we can add the two to get an average intensity. Yes?

@JohnRennie Yes sir.

The minima have zero intensity, so the average will be half the maximum intensity.

So if the maxima are higher for blue light than for red light the average intensity must be higher fr blue light than for red light.

But if the average intensity is higher then when we integrate over the diffraction angle $\theta$ we'll get a higher total power.

But we started out saying the incident power is the same for both. So where is the extra power for the blue light coming from?

@JohnRennie Ok sir. I thought accumulating the same amount of energy in less area will increase the intensity.

I'll read your messages once again. I'll try to understand based on that and will ask further if I have any doubts sir. Thank you sir.

@JohnRennie hello sir

@user8718165 hi

@JohnRennie sir, suppose an object of mass m is moving with constant velocity v upwards. I'm doing it with hand

OK ...

@JohnRennie after moving up to height h, I suddenly reduce my applied force by (mg-2) N...so now net force is downwards 2N...right sir?

OK ...

@JohnRennie how will the PE and KE change now sir...? Please help me

@JohnRennie I think PE will increase and KE will decrease slowly....but now sure how :(

We'll take time zero to be the moment you reduce the force, and we'll call the velocity at this moment $u$. The force is now -2N (downwards is negative) so the acceleration is $a = -2/m$. Then the equations of motion are:

$$ v(t) = u - \frac{2t}{m}$$

@JohnRennie yes sir...

$$ h(t) = ut - \tfrac12 \frac{2}{m} t^2 $$

@JohnRennie okay sir... got it...so mgh is the PE stored...right?

Yes, relative to your starting point the PE change is just $mgh(t)$.

And the KE is just $\tfrac12 m v(t)^2$

@JohnRennie okay sir...

This seems straightforward. Were you expecting a more complicated answer?

@JohnRennie Sir, I was confused thinking that after I reduced my force, it will fully be used to increase the PE of the object as long as I was applying it till the velocity of the object was 0. Also , the KE will also be converted to PE...There I got confused sir.

@JohnRennie Sir can you please tell me if my thinking is wrong?

I don't understand what you are suggesting ...

@JohnRennie sir suppose at that time I just removed my hand instead of reducing the force...then the KE the object had would be converted to PE very quickly...right sir?

At any height $h$ the PE has changed by $-mgh$ and your hand has done work $2h$. So the change in KE will be $\Delta KE = 2h - mgh$

@JohnRennie That's it sir...Thank you so much

@JohnRennie Now I understand...Did you get my problem by the way :)

@user8718165 no ...

@JohnRennie okay Sir...no worries....I can't explain in words... But I understood the problem properly now Sir...

Cool :-)

@JohnRennie Thank you very much for helping me sir :-)

@JohnRennie hello sir...

@user8718165 hi

@M.GuruVishnu what did you want to ask?

@JohnRennie hello sir, just for clarification....isn't the work done by me $(mg-2)\Delta h$ ?

Ah, yes, sorry. You reduced the force you are applying by 2N not to 2N.

@JohnRennie no worries Sir...got it... Thanks :-)

@JohnRennie Are you free now sir?

@M.GuruVishnu yes

Actually, I asked this question on the main side day before yesterday when you were not here. I got an answer there. I understood one part of it. But the follow up questions were not answered in the comment. Could you please clarify them? Thank you.

@M.GuruVishnu The path difference between the rays from the two slits is $\Delta s = d\sin\theta$. And we get a maximum when the path difference is equal to $n\lambda$. Yes?

@JohnRennie Yes sir

The central fringe has zero path difference and $n=0$, because both rays travel the same distance to reach the central point.

Now suppose we put a plate of thickness $h$ and refractive index $\mu$ in front of one of the slits.

@M.GuruVishnu Do you know how to calculate the extra optical path length when the light passes through this plate?

@JohnRennie $\mu d$ in vacuum is same as $d$ in medium.

Were you referring to this sir?

Yes. So in passing through the plate the light in effect travels an extra distance $\Delta s = h(\mu-1)$. Yes?

@JohnRennie Yes sir. I understood this part.

So that means when the light passes through the slits one beam has already travelled a greater distance than the other. So when the two beams reach the centre point they won't have travelled the same distance and that means there won't be a maximum there.

To make the distance travelled by the beams equal we have to move away from the centre slightly. Specifically we have to move to the angle given by:

$$ h(\mu-1) = d\sin\theta $$

That is we change the angle until the path difference between the beams cancels out the extra path difference one beam got from passing through the plate.

@JohnRennie Ok sir. I understood till this. But when we gradually increase the thickness of the sheet, will the fringes oscillate with fringe width as their amplitudes or they just shift more and more and more?

The spacing between the fringes doesn't change.

All that happens is the whole pattern shifts by an angle given by $h(\mu-1) = d\sin\theta$

i.e. with the plate in place the position of the maxima changes from:

$$ n\lambda = d\sin\theta $$

to:

$$ h(\mu-1) + n\lambda = d\sin\theta $$

Yes sir. I understand fringe width doesn't change. I was referring to the fact that the central maximum (say) oscillates up and down the screen. When the path difference is $\lambda$ (as we discussed before) there must be no change at all. It means the central fringe will be in the same place when there was no sheet at all. My doubt is will the net shift in pattern be visible? (I think it will not be visible sir)

Correct. If $h(\mu-1) = m\lambda$ then our equation becomes:

$$ (m + n)\lambda = d\sin\theta $$

@JohnRennie Yes sir.

And of course that is indistinguishable from the original equation because $m+n$ is just an integer.

This makes sense, because if you consider the incident light and imagine shifting the incident light along by one wavelength it would be impossible to tell.

@JohnRennie Ok sir. Now this is where the problem starts. If we cannot see any difference, then why should the author introduce such formulas? Is that incorrect? So the shift in pattern must be $0$ fringe length, right sir?

Right, but we can change the thickness of the plate smoothly from zero, and as we do this we will see the central fringe shift sideways.

It is true that once the central fringe has shifted sideways by exactly the fringe spacing we wouldn't be able to tell it had moved, but that doesn't mean the pattern hasn't shifted.

After all, we saw it shift as we increased the plate thickness ...

@JohnRennie Will the central fringe have net shift sir? Earlier we said a difference in pathlength equal to wavelength means no change at all. So the central fringe stays happily in the centre of the screen. Is our conclusion contradictory?

Let us assume, the central fringe is comparatively brighter than other fringes. So we could say whether there is net shift or not

Suppose we stare at the central fringe, then introduce a plate that makes it move by 0.1 of the fringe spacing. Then we'd say the central fringe has shifted. Yes?

@JohnRennie Yes sir.

And if we increased the plate thickness to make the shift 0.2 times the spacing we'd still all agree the central fringe has shifted. Yes.

@JohnRennie Yes sir.

Now we increase the plate to make the shift 0.9 times the spacing. Do we still say the central fringe has shifted?

@JohnRennie Yes sir.

Ok, now make the shift 0.9999999999999999 times the spacing.

Or 0.999999999999999999999999999999999999999999999999999999999999999

At what point do you decide that the central fringe hasn't shifted?

Yes sir :-) But when it's 1 will it abruptly shift to the normal position (centre of the screen)?

That's only because of the way you are defining the term central fringe

So really this is just a matter of terminology.

Suppose we say the central fringe is the one closest to the centre.

Then we'd have to say the central fringe jumps back when the shift is 0.5 i.e. it jumps from +0.5 to -0.5.

@JohnRennie Fine sir. If the central fringe (initially) has the highest intensity. Will the centre of the screen (new central fringe) has highest intensity even after the shift takes place?

Yes.

@JohnRennie So ultimately the final pattern looks similar to the initial interference pattern? Right sir?

@JohnRennie a question same as you are discussing.

@M.GuruVishnu yes

@yuvrajsingh hi :-)

@JohnRennie So as the central fringe moves from the centre it looses its intensity?

@M.GuruVishnu yes

@JohnRennie That's great. Interesting. Thank you very much sir :-)

Actually, I have doubt, let say I passed a light through slots then in a medium 1,and then medium 2,and again medium 1,will the central maxima will shift. @JohnRennie

@yuvrajsingh That gets complicated because the light rays will refract at the boundaries between the media ...

I'll post after yuvraj's doubt sir. Please ping me when you're done sir :-)

But some light will transmit too?.

@yuvrajsingh refract not reflect i.e. the angle of the light ray will change according to Snell's law.

Forget about ray optics?

Because we pass the light through ydse skits.

I think I need to see a diagram ...

Diagram is simple we immersed the ydse in water, then

Air, and then gain water. Then screen.

Yes.

The change in optical path length would be the same for both rays, so the position of the maxima wouldn't change.

OK so if I understand correctly the phase change will be canceled.?

@JohnRennie

The phase change would be the same for both rays. The interference depends on the difference between the phases of the beams, and since the phase change for both rays is the same it does not affect the difference.

@JohnRennie, May I ask my doubt now sir?

@M.GuruVishnu yes

Yesterday, I tried to do YDSE with my mouse's laser. But I didn't see any fringes, only saw the slits. I know lasers are coherent. But why didn't I see fringes? Any ideas, or suggestions sir?

Is the light from an LED coherent ... ?

@JohnRennie No. It was from my computer's mouse? So computer's mouse has red LED?

not a laser?

Yes. The light in mice is from an LED not a laser and I don't think it is coherent.

@JohnRennie Ok sir. I though it was a laser. Thank you.

You can get coherent LEDs, but they are more expensive and wouldn't be used in a mouse. If you buy a cheap laser pointer it probably uses a coherent LED.

You could do diffraction experiments with that.

Huh. Looks like @JohnRennie simply can't catch a break. First it was us, years ago and now you have a new batch of students😆

@Sid :-) It gives me something useful to do! :-)

@M.GuruVishnu I'm 99% sure laser pointers use a special type of LED that is specifically designed to emit coherent light.

@JohnRennie Ok sir. To see diffraction effects, may I use incoherent sources? (I'll be glad if you say yes)

@M.GuruVishnu you need to use a coherent source, but you can create a coherent source by passing incoherent light through a pinhole.

So to create a YSDE pattern from incoherent light you shine the light though a pinhole then through the slits.

Yes sir. I didn't try that because I felt they will cause a significant loss of intensity. But I'm going to give it a try :-)

In about 1.5 hours, PSLV will lift off. Unfortunately skies are cloudy, so I will not be able to see the launch directly :(

@JohnRennie, Thank you sir. Bye :-) Will catch you after the launch

@M.GuruVishnu bye :-)

@JohnRennie, Hi sir. Are you there?