Solving an Optics Exam Problem: Interference Fringes

[Cynosura]

So, I'm about to take an Optics exam, and while I can easily do most of the last years' exam problems, there are some which I cannot solve for different reasons (perhaps some gross oversight?). Yesterday I stumbled upon this:

1. Homework Statement
(my own translation from Catalan)
We have a rectangular prism with an small angle of α=0.1 rad and an index of refraction n=1.5 at distance d=20 mm to the right of a point source emitting at λ₀=500 nm. 1m to the right of the prism lies a screen, as seen in the drawing.

(had to draw it myself, it wouldn't let me use the original image)

a) Calculate the angle of deviation of a light beam in normal incidence.
b) Draw where on the screen will be interference fringes, and calculate the width of the zone in which they can be observed.
c) Calculate the separation between two consecutive fringes.
d) How many minima and maxima can be observed?

Homework Equations

Two-beam interference, Snell's law.

The Attempt at a Solution

a) I applied Snell's law on the second surface of the prism, since a prependicular surface does not refract light. It gave me an angle between the axis and the beam (which I will call β) of 0.05032 radians.

b) The interference zone will lie between the axis (since above it there is only one incident beam and the point where the normally incident beam strikes the screen. To calculate that, I applied the definition of the tangent of an angle, and obtained that the lowermost point on the screen where fringes may be observed is x=50.36 mm. But I cannot solve the uppermost point, since it seems to depend on the size of the prism (which is not there!). I could slap "let's assume that the prism is big enough so that there is incident light just above the axis" in there, but I've gotten bad marks for that type of thing before, or I could try and calculate the height of that point as a function of one of the prism's dimensions, but that looks downright nasty having in mind the timeframe of an exam. Did I miss something?

c) My sketch here consists of finding the point where two beams from the prism cross, use that as a second source and apply the equation that s·sin(θ)=mλ, where s is the separation between sources, θ is the angle between the point we are evaluating and the rect which is always equidistant to both sources, and m is an integer. I have not plugged the values yet because I'm doubtful of my assumptions being sensible.

d) I cannot solve it because it depends on b).Any measure of help will be greatly appreciated.

 

[blue_leaf77]

b) Why do you think the fringe region will span until somewhere in the upper half of the screen?

 

[Cynosura]

That is in the lower half, I had forgotten to indicate it with a minus :)

My reasoning is based on Snell's Law. The beam on normal incidence will not be refracted by the left surface, and the second surface will send it at a certain angle such that the beam ends up at x distance below the axis. A beam which incides with a small angle will get refracted by the first surface. Now, the first surface is vertical and, since the beam was traveling somewhat upwards, it will continue going upwards (n is positive), making the angle of incidence with the second surface a little bit smaller than the normal case, and thus the angle of exit will be a little lower as well. Applying that recursively leads me to the conclusion that, the higher the entrance angle, the higher the beam will end up on the screen.

And I think the fringe zone must be somewhere on the lower half of the screen because that's where two beams from the same source will possibly interfere. In the upper half, there is only the light deviated by the prisma, whereas from 0 to x in the lower half there are both the beam coming straight from source and the one coming from the prism, like this:

where the green beams come straight from the source and the red ones were deviated by the prism. a is the beam which enters the prism on normal incidence, and b is the beam which *just* misses it. Hope I made myself a little bit clearer :)

 

[blue_leaf77]

But I cannot solve the uppermost point, since it seems to depend on the size of the prism (which is not there!)

Is this still your problem? I thought you have clearly shown that you have found the fringe region.

 

[Cynosura]

Yes. It turns out that, if the prism is small enough, there will be no refracted ray reaching the upper part of the screen, and then the fringe region would span from x to some point below the axis, not from x to the axis. And thus the answer may not be technically correct.

Also, I don't know if my sketch to c)'s solution would be correct, or I have to calculate the OPD as a function of the position/angle/... , not pretty.

 

[blue_leaf77]

the fringe region would span from x to some point below the axis, not from x to the axis.

Why do you think so? The region where interference fringe will be observed is that where the two waves overlap. Are you familiar with Fresnel biprism? This problem is actually a modified version of Fresnel biprism.

Also, I don't know if my sketch to c)'s solution would be correct, or I have to calculate the OPD as a function of the position/angle/... , not pretty.

The trick here is, when you draw the back extension of the red lines in your last picture, you should find that they all intersect at a same point. This point may be regarded as a virtual point source.

 

[Cynosura]

This is why I think that could happen:

As you can see, the fringe region for the blue (smaller) prism f is smaller than that of the bigger prism f'. If the prism in the problem is bigger than the clear one, the fringe region is still f', so it *imposes* a minimum size of the prism. That's what I meant with "I can slap an assumption in there...".

Also, on c), I had already suspected that it was some sort of Fresnel's biprism, and drawing the extensions was exactly my intention here, but I had to be sure :)

PS: the path of the middle beam across the clear prism is not that which is shown! It would strike the screen a bit higher.

 

[blue_leaf77]

I don't see your point of introducing two prisms with differing size, there is only one prism in the problem and from the above picture, it seems to me that you are thinking that the interference occurs between the waves due to the two prisms you draw there. I will just give you a hint here, the problem here can be translated into that where there are two point sources emitting waves which interfere with each other. The problem then becomes the interference between two point sources, however unlike the usual two point sources problem, here the interference region is limited. This region is what you want to find in b). Part c) follows naturally from b).

 

[Cynosura]

No! You missed the point entirely!

What I meant is, that if the prism is small enough, the fringe zone will be smaller! Sorry if I have not been clear. I introduced a second prism just to "demonstrate" my statement, perhaps I should have made two pictures...

 

[blue_leaf77]

I think I begin to get your point. I agree that, in a strict sense you need to know the highest point on the screen a ray passing through the prism will hit. But in most Fresnel biprism experiment, I think the main source is close enough to the prism such that the ray which has passed through the upper tip of the prism hits a point in the upper half of the screen.