Solving for ##\theta## in a Trigonometric Equation

[phantomvommand]

TL;DR Summary

To observe missing fringes, must slit separation d be an integer multiple of slit width b?

Can a situation at angle $\theta$ happen, where:

$d sin \theta = 3 \lambda$
$b sin \theta = 2 \lambda$
$d/b = 3/2$

 

[sophiecentaur]

Are your "missing fringes" something you have observed?
The two interference patterns will be 'multiplicative' so the zeros / minima of, say the slit pattern will impose themselves on the maxes of the grating pattern. That can break up the simple pattern that you'd expect from the grating.
This Hyperphysics link shows you what I'm talking about. You can see where the diffraction pattern stamps out some of the array pattern maxima. The earlier page in the series shows basic two slit interference.

 

[phantomvommand]

Are your "missing fringes" something you have observed?
The two interference patterns will be 'multiplicative' so the zeros / minima of, say the slit pattern will impose themselves on the maxes of the grating pattern. That can break up the simple pattern that you'd expect from the grating.
This Hyperphysics link shows you what I'm talking about. You can see where the diffraction pattern stamps out some of the array pattern maxima. The earlier page in the series shows basic two slit interference.

Thanks for this. This is exactly what I am asking about. My teacher told me that for this to happen, the slit separation must be an integer multiple of slit width. Is that necessarily true? My original post contains an example of why I think it may not be true.

 

[sophiecentaur]

This link gives a formula which shows where the nulls occur for a single slit. This link has a formula showing where the nulls occur of double slits (implicitly with the slits of near zero width).
If the minima of the single slit coincide with the minima for the two slits then all the maxima will show (albeit of varying levels). If one minima coincides with a maximum then it will be squashed and not be visible (at least, it will have two lobes of very reduced amplitude on either side of the null)

Thus, to obtain destructive interference for a single slit,
b sinθ = mλ

and for double slits the condition for minimum intensity at P, when the path difference is a multiple of half wavelengths, is given by,
d sin θ = (n+¹/₂)λ

You can do some fiddling round with those two. This implies that, for the minima to be at the same angle
d/b = (n+¹/₂)/m
or d = b(n+¹/₂)/m
Does your 3:2 ratio fit in with this?
(I think my maths is ok there)

I gained some insight into this from the derivation of the single slit interference pattern which is done by splitting the slit into two halves, separated by b/2, which gives the first minimum at twice the angle of a double slit with separation d.

 

[phantomvommand]

This link gives a formula which shows where the nulls occur for a single slit. This link has a formula showing where the nulls occur of double slits (implicitly with the slits of near zero width).
If the minima of the single slit coincide with the minima for the two slits then all the maxima will show (albeit of varying levels). If one minima coincides with a maximum then it will be squashed and not be visible (at least, it will have two lobes of very reduced amplitude on either side of the null)

Thus, to obtain destructive interference for a single slit,
b sinθ = mλ

and for double slits the condition for minimum intensity at P, when the path difference is a multiple of half wavelengths, is given by,
d sin θ = (n+¹/₂)λ

You can do some fiddling round with those two. This implies that, for the minima to be at the same angle
d/b = (n+¹/₂)/m
or d = b(n+¹/₂)/m
Does your 3:2 ratio fit in with this?
(I think my maths is ok there)

I gained some insight into this from the derivation of the single slit interference pattern which is done by splitting the slit into two halves, separated by b/2, which gives the first minimum at twice the angle of a double slit with separation d.

My teacher told me that d/b must be an integer for the missing fringe to be seen, where the minima of the diffraction causes the expected maxima to go missing.

I think what you have done here is to find the condition such that the minima of the diffraction meets the minima of the interference, and thus all maxima are retained.

The 3/2 ratio I mentioned in post 1 is just a random ratio I set, to explain why I think the missing fringe can be observed even if d/b is not a whole number.

May I know how what you have done proves that my teacher is right? I think I am not seeing something!

 

[sophiecentaur]

Unless all the nulls coincide you can get a maximum somewhere that’s squashed by a min in the other pattern(?). I thought that should be enough, although it’s a sort of negative argument.
Those graphs in the hyperphysics site (maybe on a different page from the link show how you can get holes in fringe patterns with some ratios. There’s an interactive diagram in the section.
What level are you at? How happy are you with tinkering about with algebra? You’re looking for values of d and b where you get m coinciding with n?
It’s a sort of beat pattern between the two sets of zeros and only works for integer values of m and n.
At this stage I would go and ask ‘teech’ to give a bit more detail.

 

[phantomvommand]

Unless all the nulls coincide you can get a maximum somewhere that’s squashed by a min in the other pattern(?). I thought that should be enough, although it’s a sort of negative argument.

I was thinking that the null of the diffraction could fall somewhere in between the max and min of the interference, so not all maximas will necessarily be squashed.

Those graphs in the hyperphysics site (maybe on a different page from the link show how you can get holes in fringe patterns with some ratios. There’s an interactive diagram in the section.
What level are you at? How happy are you with tinkering about with algebra? You’re looking for values of d and b where you get m coinciding with n?

Yes, I am looking for d and b where m coincides with n. Ultimately, I'm trying to find out if d/b must be an integer. (in which case, m/n must also be an integer)

 

[sophiecentaur]

I was thinking that the null of the diffraction could fall somewhere in between the max and min of the interference, so not all maximas will necessarily be squashed.

You are probably right about that BUT the criterion that your teacher gave you was that all the maxes / fringes would get through. The 'beat pattern' I referred to would arrive somewhere where the max is squashed by a null unless that exact ratio was achieved. If he's a teacher like me, he would want to be a bit smartarse about it and the 'precise' statement he gave you was almost certainly right - but, remember, that squashed maximum could occur way out from the centre. You should definitely go and challenge him with all that - after having OKed it with yourself first, of course.
He is probably right but it would probably be easy to violate his test and yet the fringes that are squashed out could be way off to the side and you could miss them.

There is a practical note here. Even with 'monochromatic' light there are experimental factors that will mess with the remote fringes and you could well find that you can't see any fringes at 60 degrees from the boresight , in any case.

Good luck matey!