Number of maxima in Young's double slit experiment

[Ayesha02]

Homework Statement

A plane wavefront (wavelength = ) is falling on YDSE apparatus as shown in the
figure. The lower half of it is filled with water (refractive index = 1.5) How many
maximas can be theoretically obtained on the screen if d = 98lambda and D = 20000lambda
(include the maxima at infinity, assume coherence length to be infinite)

Relevant Equations

All i know is y coordinate of nth maxima is y=(n* lambda* D)/d,
where D= Distance between source and slits
and d=distance between slits

I honestly do not have any idea regarding this sum.
Any help will be appreciated:)

 

[etotheipi]

Start off by considering the extreme case, where the rays emerging from each slit are vertical and converge on the screen at infinity. Also, what is the condition for constructive interference?

Note that when you have media of two different refractive indices you must either count wavelengths or use optical path lengths.

 

[haruspex]

All i know is y coordinate of nth maxima is y=(n* lambda* D)/d,
where D= Distance between source and slits
and d=distance between slits

Isn't that just an approximation for small angles?
Draw a diagram showing rays from the slits going at angle theta. (If the screen is a long way off, rays to the same point will be near enough parallel.)
If the path lengths differ by a whole number of wavelengths, what equation can you write relating to d and theta?

 

[Ayesha02]

Isn't that just an approximation for small angles?
Draw a diagram showing rays from the slits going at angle theta. (If the screen is a long way off, rays to the same point will be near enough parallel.)
If the path lengths differ by a whole number of wavelengths, what equation can you write relating to d and theta?

Yes it surely is an approximation for small angles.

and as for the equation, will it be like d* sin theta=n* lambda?

I still don't see how is this helping

 

[etotheipi]

and as for the equation, will it be like d* sin theta=n* lambda?

That's not taking into account the different refractive indices. That construction is only valid if the refractive index is uniform. If you have two paths from each slit to a point on the screen, and want constructive interference, you need to make sure that the difference in number of wavelengths is a whole number.

That's a bit more work in the general case, so it might be a good idea at least initially to consider the case where the rays travel vertically. How many more wavelengths are there in the path from the lower slit than in the path from the upper slit?

 

[Ayesha02]

That's not taking into account the different refractive indices. That construction is only valid if the refractive index is uniform. If you have two paths from each slit to a point on the screen, and want constructive interference, you need to make sure that the difference in number of wavelengths is a whole number.

That's a bit more work in the general case, so it might be a good idea at least initially to consider the case where the rays travel vertically. How many more wavelengths are there in the path from the lower slit than in the path from the upper slit?

Ahh buddy, I am a bit weak in this topic. I have no idea how to find out the difference in number of wavelengths. Could you help me on that?

 

[etotheipi]

Ahh buddy, I am a bit weak in this topic. I have no idea how to find out the difference in number of wavelengths. Could you help me on that?

Okay, say both rays are traveling vertically upward. The ray from the lower slit has to travel an extra distance $49 \lambda$ up through the water, and $49 \lambda$ up through the air.

If the wavelength in air is $\lambda$, what is the wavelength in water?

Consequently, how many wavelengths are in each section of this bit of the path?

 

[Ayesha02]

Okay, say both rays are traveling vertically upward. The ray from the lower slit has to travel an extra distance $49 \lambda$ up through the water, and $49 \lambda$ up through the air.

If the wavelength in air is $\lambda$, what is the wavelength in water?

Consequently, how many wavelengths are in each section of this bit of the path?

Ummm if wavelength in air is $\lambda$, wavelength in water is (V*$\lambda$)/c
where v is velocity of light in water and c is velocity of light in air

Am i right?

 

[etotheipi]

Ummm if wavelength in air is $\lambda$, wavelength in water is (V*$\lambda$)/c
where v is velocity of light in water and c is velocity of light in air

Am i right?

Right, though also note that $\frac{c}{v_w}$ is the definition of the refractive index of water, $n = 1.5 = \frac{c}{v_w}$. So in water you have a wavelength of $\frac{\lambda}{n}$.

So consider the two stretches each of length $49\lambda$, one in air and one in water; how many wavelengths are in each?

 

[Ayesha02]

Right, though also note that $\frac{c}{v_w}$ is the definition of the refractive index of water, $n = 1.5 = \frac{c}{v_w}$. So in water you have a wavelength of $\frac{\lambda}{n}$.

So consider the two stretches each of length $49\lambda$, one in air and one in water; how many wavelengths are in each?

Ohh so number of wavelengths in air should be 49 $\lambda$/ $\lambda$ and in water should be 49 $\lambda$/( $\lambda$/n) isn't it?

Am I right?

 

[etotheipi]

Ohh so number of wavelengths in air should be 49 $\lambda$/ $\lambda$ and in water should be 49 $\lambda$/( $\lambda$/n) isn't it?

That's correct, so how many total wavelengths does that turn out to be?

And if that is the maximum number of wavelengths by which the lower slit can lead the upper slit, can you figure out how many fringes you get in total (hint: don't forget about the cases where the light from the upper slit has more wavelengths in its path than the lower slit!)

 

[Ayesha02]

That's correct, so how many total wavelengths does that turn out to be?

And if that is the maximum number of wavelengths by which the lower slit can lead the upper slit for constructive interference, can you figure out how many fringes you get in total (hint: don't forget about the cases where the light from the upper slit has more wavelengths in its path than the lower slit!)

Ohh finally!
Itll be 245/2 *2 for the second case. Hence 245 is the answer:)

Thank you so much for your time and help:)

 

[etotheipi]

Yeah that's it, you have 122 cases where the path from the upper slit contains more wavelengths than that from the lower, another 122 from the vice versa approach, and a further 1 case where both paths contain the same number of wavelengths (i.e. the "central" fringe, which won't be central here...).

 

[Ayesha02]

Yeah that's it, you have 122 cases where the path from the upper slit contains more wavelengths than that from the lower, another 122 from the vice versa approach, and a further 1 case where both paths contain the same number of wavelengths (i.e. the "central" fringe, which won't be central here...).

Yeah! Thanks dude:)

 

[haruspex]

Yeah! Thanks dude:)

Glad you chaps have it sorted, but I still don't understand the set-up. Can we have a diagram, please?

 

[etotheipi]

This is the configuration I was picturing, with slit spacing $98\lambda$, and water in the lower half/air in the upper half:

 

[haruspex]

This is the configuration I was picturing, with slit spacing $98\lambda$, and water in the lower half/air in the upper half:
View attachment 262026

Hmm.. that's all I could think of, but then I don't understand the solution you arrived at. Below the waterline it is going to get complicated with reflections.

 

[etotheipi]

Hmm.. that's all I could think of, but then I don't understand the solution you arrived at. One of the rays is going to be partly in each medium.

My reasoning was that for constructive interference here, the number of wavelengths in each ray must differ by an integer. And both rays will be partially in each medium. When the rays are traveling vertically, the difference in the number of wavelengths in each ray to some common point is 122.5 wavelengths. This means that no matter what, for an arbitrary point on the screen at some angle $\theta$ less than $90^o$, the difference in the number of wavelengths in each path is going to be less than 122.5 wavelengths.

So if we slide the point on the screen upward from the central fringe, we will necessarily pass through 122 further fringes, i.e. one for where the difference in wavelengths is 1, then 2, ... and all the way up to 122.

 

[haruspex]

My reasoning was that for constructive interference here, the number of wavelengths in each ray must differ by an integer. And both rays will be partially in each medium.

When the rays are traveling vertically, the difference in the number of wavelengths in each ray to some common point is 122.5 wavelengths. This means that no matter what, for an arbitrary point on the screen at some angle $\theta$ less than $90^o$, the difference in the number of wavelengths in each path is going to be less than 122.5 wavelengths.

So if we slide the point on the screen upward from the central fringe, we will necessarily pass through 122 further fringes, i.e. one for where the difference in wavelengths is 1, then 2, ... and all the way up to 122.

Yes, I eventually got that - very neat - and edited my post, but I am still concerned about reflections. Does your reasoning cover that?

 

[etotheipi]

Yes, I eventually got that - very neat - and edited my post, but I am still concerned about reflections. Does your reasoning cover that?

That's interesting! I hadn't even considered that. It does make things a little bit spicy... since there might well be more/less of them.

 

[haruspex]

That's interesting! I hadn't even considered that. It does make things a little bit spicy... since there might well be more of them.

For a distance $D\cot(\theta_c)-\frac 12d$ below the water line, two rays from the lower slit will reach the screen, as well as one from the upper slit.

 

[etotheipi]

For a distance $D\cot(\theta_c)-\frac 12d$ below the water line, two rays from the lower slit will reach the screen, as well as one from the upper slit.

Of the fringes we've counted so far, the 122 formed by the ray from the lower slit traveling below the horizontal should be safe, as well as the central-not-so-central one. For the 122 above, some adjustment might need to be done. Where the ray from the lower slit has an incidence angle of less than $\theta_c$, those fringes should be safe. However there is a possibility of extra fringes from certain partial reflections.

Where the incidence angle from the lower slit exceeds $\theta_c$, we'd need to get rid of those fringes above the water line. I think all of these fringes are preserved below the waterline, however, so they should still be there.

Needless to say it's a much more devilish problem than I initially signed up for . It appears to be a matter of counting the number of possible successful partial reflections.

 

[haruspex]

Of the fringes we've counted so far, the 122 formed by the ray from the lower slit traveling below the horizontal should be safe, as well as the central-not-so-central one. For the 122 above, some adjustment might need to be done.

Isn't it the other way around?
Above the water line, only one ray from each slit can reach the screen. From just below the water line down a distance $D\cot(\theta_c)-\frac 12d$ there will be one from the upper slit but two from the lower. These will arrive in three different phases, mostly, so probably extra local maxima.

 

[etotheipi]

Isn't it the other way around?
Above the water line, only one ray from each slit can reach the screen. From just below the water line down a distance $D\cot(\theta_c)-\frac 12d$ there will be one from the upper slit but two from the lower. These will arrive in three different phases, mostly, so probably extra local maxima.

I was meaning to say that, initially ignoring reflections, the existing fringes formed by rays from the lower slit traveling below the horizontal will still be there. And likewise, existing fringes above the water line will still be there, except where the angle of incidence of the ray from the lower slit exceeds the critical angle these will actually end up forming fringes below the water line. But it doesn't make a difference to our counting so far, and the 245 existing ones are safe.

I think there even more extra cases with partial reflections where $\theta < \theta_c$, where two rays from the lower slit reach the screen for distances further down from the water level than $D\cot(\theta_c)-\frac 12d$. In addition to the extra cases you mention where $\theta > \theta_c$ which converge above that depth. `

 

[haruspex]

the existing fringes formed by rays from the lower slit traveling below the horizontal will still be there.

Well, they won't be in the same places. We will have a different set of fringes, and it is not meaningful to say that some of them are the original ones.

fringes above the water line will still be there, except where the angle of incidence of the ray from the lower slit exceeds the critical angle these will actually end up forming fringes below the water line

No, they all will be there, as before.

where two rays from the lower slit reach the screen for distances further down from the water level than

Either my geometry is wrong or that can’t happen.

 

[etotheipi]

Either my geometry is wrong or that can’t happen.

It was my understanding that you can still obtain partial reflections when the angle of incidence is less than the critical angle. This might result in rays from the lower slit being reflected to depths lower than that critical depth, in addition to those above that critical depth which you mention.

I agree with your geometry, but it holds for the cases of total reflection only. I have the feeling you are right, however, since I am unsure if the planes of polarisation differ on whether TIR or PIR occurs.

 

[haruspex]

It was my understanding that you can still obtain partial reflections when the angle of incidence is less than the critical angle. This might result in rays from the lower slit being reflected to depths lower than that critical depth, in addition to those above that critical depth which you mention.

View attachment 262029

I agree with your geometry, but it holds for the cases of total reflection only. I have the feeling you are right, however, since I am unsure if the planes of polarisation differ on whether TIR or PIR occurs.

Yes, I was only counting full reflections. Partial reflections would not be very strong, would they?