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@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

Are you free now sir? I'm having a doubt in a problem in light waves.

I guess you're busy right now. Sorry for disturbing. But could you ping me once you're free, sir?

@GuruVishnu Yes I'm free. What's the problem?

Oh great! Thank you sir. I'll post it now. Please wait till (End of message).

If a minima in YDSE occurs directly in front of one of the slits, then the wavelength of the radiation used is:

(a) 2 cm only

(b) 4 cm only

(c) 2 cm, 2/3 cm, 2/5 cm and so on

(d) 4 cm, 4/3 cm, 4/5 cm and so on

Given, D=12 cm and d=5 cm; where D and d are the distance between the slit plane and the screen and the distance between the two slits respectively.

I figured out the correct answer (c) by simply figuring out the path difference for a point opposite to a slit and applying the condition for minima. But I want to know why my alternate method as described below failed:

$$\frac{nD\lambda}{d}=d$$

Where $n$ is an integer. However, on solving the above, I don't even get close to any of the options. So why does this method fail? I'm interested in knowing the reason this fails because this was the method which came to my mind first.

(End of message)

@GuruVishnu Where does the slit width come into this?

@JohnRennie Ah. Sorry sir, that was a typo. It's fringe width not slit width sir.

You mean the spacing between fringes?

Yes sir. It gives the distance between consecutive dark or bright fringes.

The equation I arrived at is simply based on fitting as many as bright fringes within the distance between the two slits on the screen, ensuring dark fringes always appear opposite to the slits.

The spacing between consecutive fringes is $D\lambda/d$, and as you argue, the spacing between fringes has to be equal to $d/n$ for some integer $n$ i.e. an integral number of fringe spacings equals $d$.

$$ \frac{D\lambda}{d} = \frac{d}{n} $$

Yes sir. Isn't that exactly the same equation I mentioned?

So I get the same result as you ...

Ok sir. So I guess there's no mistake in this argument.

Writing it for $\lambda$, we get:

$$\lambda=\frac{d^2}{Dn}$$

where we know the values of $d$ and $D$. But on substituting this and different values of $n$, we don't get any of the options, sir.

Ah wait, the equation $n\lambda = d\sin\theta$ only applies in the Fraunhofer limit.

Oh! Ok sir. So the equation for fringe width is invalid in this case as $d$ is not much smaller than $D$. Am I right sir?

D=12 and d=5; (both in centimetres)

In this case the slit spacing and screen distance are comparable so we'll be in the Fresnel region and the equation won't apply.

@GuruVishnu yes

Ok sir. Thank you for the clarification :-)

:-)

Otherwise, is this method logically right?

Yes, that's the way I immediately thought to do it.

Ok sir. Then that's a sign I've started to think like you :-)

@GuruVishnu whether that's a good thing is debatable :-)

@JohnRennie: Hi sir. If you're free after Yuvraj's doubt, could you ping me?

@GuruVishnu you can ask now. Yuvraj and I are just chatting not dealing with a specific question.

Ok sir. Thanks!

I'm having a doubt in the following question on optical path difference:

Two parallel rays are travelling in a medium of refractive index $\mu_1=4/3$. One of the rays passes through a parallel glass slab of thickness $t$ and refractive index $\mu_2=3/2$. The path difference between the two rays due to the glass slab will be _______.

My approach: A distance of $t$ in $\mu_1$ and $\mu_2$ in media 1 and 2 corresponds to an optical length of $t\mu_1$ and $t\mu_2$ in vacuum and hence the difference between these gives the optical path difference. The answer must be $t(\frac{3}{2}-\frac{4}{3})=\frac{t}{6}$. But my book says it is $\frac{t}{\color{red}{8}}$

I don't find any mistakes in my approach. Can we conclude the key is incorrect?

I think it depends on how you define path difference ...

I think I don't get your statement. Till one of the two rays meets the slabs there's no path difference. And even after emerging from the slab there'll not be any. And so I'm calculating only the difference due to the thickness of the glass slab for one and in the medium for another.

For the first ray the number of wavelengths in the distance $t$ is $n_1 = \tfrac43 t/\lambda$

And for the second ray it is $n_2 = \tfrac32 t/\lambda$

(where $\lambda$ is the wavelength in vacuum)

Ok sir. I can understand so far.

So the difference in the number of wavelengths is $(\tfrac32 - \tfrac43)t/\lambda = \tfrac16t/\lambda$.

That was what I also got. Six in the denominator.

To convert this path difference back to a distance we have to multiply by the wavelength $\lambda$, but what wavelength should we use? The wavelength in vacuum or the wavelength in the medum?

If we multiply by the wavelength in vacuum we get $\tfrac16t$ and that's what you got.

If we multiply by the wavelength in the medium, $\tfrac34\lambda$, we get $\tfrac18 t$.

Which is what your books says.

Ok sir. So here we're not considering the vacuum as the reference standard instead using the initial medium of both the rays as standard? What's the reason for this choice?

Good question, and that's why I said it depends on how you define the path difference. I'm not sure either answer is definitively correct.

The point is that in diffraction we are really interested in the phase difference not the physical path difference.

Ok sir. I think I have got some ideas on this. At first on seeing the following in the book:

$$\Delta x=\left(\frac{\mu_2}{\mu_1}-1\right)t=\left(\frac{3/2}{4/3}-1\right)t=\frac{t}{8}$$

I didn't get anything. But based on your method it seems understandable to me.

Yes. He is doing the same as me but wrapping it in a single impenetrable equation :-)

Thank you sir. Let me think about this for some time. This seems bit different than what I initially thought about path differences :-)

@JohnRennie A small clarification, by 'physical' path difference, did you mean geometrical path difference or optical path difference? I guess it's the former.

I probably mean geometrical path difference though I'm not sure exactly how you are defining those terms.

Ok sir. Geometrical path difference is the actual distance in space from the source to the sink along the path of the ray. Whereas optical path length takes the difference in media into account. For example, different rays from the focal point of a lens travel different geometrical lengths to reach the other side of the convex lens. Whereas, all the rays travel equal optical path lengths and arrive in phase at the point of image formation.

This was how I learnt. So far it worked well in all cases except this one.

Hmm, I would normally consider the optical path difference to be the difference in wavelengths i.e. a phase difference. But if the JEE books say otherwise than go with their definition.

Oh. Ok sir. That was why you took how many wavelengths are in a particular thickness instead of translating the geometrical path lengths to optical path lengths in vacuum?

Yes

Ok sir. If it's not a problem for you, will it be possible for you to explain the book's answer? The usage of the common medium's wavelength in the method is something I'm not used to. So could you provide an alternate route to that?

I'm not feeling as confident with this method as I did with vacuum translation of optical path differences sir.

I can't explain the book's answer because it depends on the definition of path distance and I don't know how the book is defining that.

I guess the definition is:

How much farther would the original ray have to travel in order to have the same optical path length as the ray that passed through the block

That's why the answer is to take the number of wavelengths in the block and multiply it by the wavelength of the other ray.

Actually, now I think about it, that does make sense as a definition.

Ok sir. Even as per your definition, I think, when we consider some kind of unitary method (using vacuum as an intermediate step), we'll get $t/6$ again. I guess both of them must give the same result. Do you have any ideas on the "lens makers' formula" like term as provided in the book? Is there any special name for the formula?

My definition will give t/8 directly.

You define the distance $t_1$ as $\lambda_1 \times t/\lambda_2$

i.e. $t_1$ is the distance the first ray has to travel in order to have the same optical path length as the ray that went through the block. Yes?

The the path difference is $t_1 - t$

Please wait sir. Let me think about it.

@JohnRennie: Eureka! Thank you very much sir. Now, I got totally on what I was missing.

:-)

Putting $\mu_1=4/3$,

$$t'=\frac{t}{8}$$

Yes

Ok sir. Thank you :-)

Good bye sir.

Bye :-)

@JohnRennie: Hi sir :-)

Did you finish having your lunch?

@GuruVishnu I'm eating now. I was a bit late making lunch today.

Oh. Ok sir. Sorry for disturbing. Shall we have a general conversation after your lunch, if possible?

OK. I'll ping you when I've finished eating.

Thanks! No hurries.

@GuruVishnu hi, lunch is officially eaten! What do want to ask?

Great! I don't have anything to ask.

But I wanted to inform you that my order on Lenovo was cancelled and the refund process has been initiated. They say this is due to 'transit damage' even though the estimated dispatch date was 27th of this month :-(

That's a shame.

But then it gives you the chance to look again at what laptops are available.

What's the status of the lockdown in your area? Kerala isn't it?

@JohnRennie Yes sir. Outcomes have both positive and negative impact. Now, I'm trying to mitigate the eye irritation problem by using stuff in dark mode.

Are you finding that helps?

Hmm, Googling suggests the situation in Kerala isn't good.

Actually, I'm from Chennai sir. Nor Kerala. But somewhat close.

@JohnRennie I find dark mode to be better. But it hasn't solved the problem completely.

Chennai is Tamil Nadu isn't it. I knew someone who lived there.

Yes sir. It's the capital of TN. I guess you are referring to me. I had told you once in past.

No, it was a JEE student from several years ago. She lived in Chennai when she took the exam but then moved to Kochi to go to a college there. That's probably why I was thinking of Kerala.

Oh! Ok sir. I remember that you said someone introduced you to dosas before me :-)

Yes, that would have been her :-)

Wow there are lots of offers on the Lenovo India web site at the moment.

Yes. But the price for the one I ordered has increased by 10k. And Microsoft Office Professional is no longer free. Comes at an extra cost.

I might be able to help with Office Pro :-)

Thanks for the offer! I don't know what method you follow. I actually copy some text from a webpage into a notepad and then set the file type as .cmd(?) or something like that and run it. Then Office gets activated somehow. That's how it's still running.

I guess you'd have a better, formal approach.

I have loads of old Office keys from PCs that have been decommissioned and no longer used. Strictly speaking you're not allowed to reuse the keys, but if you don't mind it being an older version of Office the reality is that Microsoft aren't going to care.

So it would be a perfectly legal key - not some hack.

Oh. Ok sir.

I'm probably going to buy only after everything returns to normal condition. Till then I need to manage with dark mode and loads of H$_2$O :-)

It's obviously nice to have the latest version of Office if you can afford to include it with the laptop, but if the cost puts it out of reach then bear in mind I can give you an older copy like Office 2014.

Actually I'm still using Office 2010 because I've never bothered to change. It still seems to work fine.

I can give you a (perfectly legal) key for Windows 10 as well if it's necessary.

@JohnRennie Yes sir. It's more costlier than the OS itself. If needed, I'll seek your help in future. Thank you :-)

I'd agree with you that it's best to wait and order after things have settled down a bit.

Though that does mean putting up with the old screen for longer than you wanted :-(

Yes sir. That's the main issue in continuing with this display. I don't know what exactly is wrong with this one because my mom's mobile (AMOLED) doesn't give the same irritation and also the TV;

At least I can be sure it's not a fault on my side as I'm totally fine with other devices.

Just a though, can you find a cheap external display and connect it to the laptop?

Actually you could probably connect the laptop to your TV!

I like the second idea. I need to look for a HDMI cable. It's a long time since I used it. The first idea again involves purchase which I need to avoid now. Otherwise it's the perfect choice considering accessibility of the keyboard and touchpad.

I wish I had an wireless keyboard.

I'd say it was worth a try.

Yes sir.

It's getting late here. Let's see tomorrow. Thank you for the conversation. Good bye sir :-)

Bye :-)