Interference of light by splitting the wavefront and recombining

[ItsKurious]

In the above diagram, I have illustrated what is written in the summery. So, if the thickness and refractive index of the material are chosen such that the part of the wave that travels through the slab acquires a path difference of λ/2 and at the right end if I keen another convex lens so as to converge the incoming parallel wavefronts to a point, will I be able to see a dark spot due to the fact that the rays will interfere destructively? Also, it would be really helpful if you could tell how is the wavefront of the light is going to look after it passes through the slab.

Thank you!

 

[sophiecentaur]

@ItsKurious Hi and welcome to PF

I am not sure what your diagram represents. Is it a source on the left with 'rays' spreading out towards the right? This is not a fruitful way to approach wave phenomena like diffraction because the ray model cannot show the phases of a wavefront.
The best way to describe the effect of interference is to use the standard representation of a wavefront. https://opentextbc.ca/universityphysicsv3openstax/chapter/youngs-double-slit-interference/explains the simplest two slit interference pattern but what you need is more than just two point sources but an infinite number of point sources and that has to be described in terms of Diffraction Integrals.
You could look up Fresnel Diffraction at an edge, which shows the effect at a single opaque edge. The effect of a wavelength shift of half the incident flux could be calculated in a similar way.

 

[hutchphd]

This in essence is how any birefringence-based optical device works. Light polarized differently travels with different speeds. The effect of the interference causes the plane of polarization of light to "corkscrew" as it traverses the material. If you are looking at an LCD screen you are seeing the process in action. Rather than recapitulate it here, I suggest

http://web.media.mit.edu/~stefan/liquid-crystals/node3.html

 

[sophiecentaur]

if I keen another convex lens so as to converge the incoming parallel wavefronts to a point, will I be able to see a dark spot due to the fact that the rays will interfere destructively?

This may be the key part of your post and I can see how @hutchphd introduced the idea of birefringence.
I will assume that you take my point about your version of the situation. What you have is, essentially, a wide source with one phase and another source with another phase. You can treat each source separately. You will get two edge diffraction patterns (Fresnel or far field, whichever) and the resultants will add (vectorially) . With half wave phase difference, you will get a null on axis but off axis, the diffraction patterns are not in step so you will get other maxima increased. The energy is conserved so it has to go somewhere - I liken it to stamping on an air bed in one place and the rest of the surface goes up. So I would expect (no details, because you'd have to do the sums for where you are looking at the pattern) a dark line on axis and bright stripes on either side. Look at the pattern for fresnel diffraction at an edge (in above link) and draw a mirror image of the pattern on top. The patterns are not symmetrical about the edge.You cannot just arithmetically 'subtract' one from the other because the path distances (i.e. relative phases) are different but the result of that crude subtraction will show you where there will be no nulling (a max with no corresponding max). But you may not want to get into that. (Me neither) Suffice to say that the total energy will be the same.

 

[hutchphd]

Let me add the way I understand the process you describe: perhaps it will help further:
The use of converging lenses just changes a spherical wavefront into a planar wavefront (it moves a point source out to infinity...similarly for a detector...and is not really part of the question at hand).
From the source to the detector, the light travels all possible paths, weighted according to the source geometry. Each path will accrue a phase which depends upon the path length and the local wavelength, determined by local speed. Each accrued phase (or path integral) is added into the final weighted sum. Usually we can make the problem tractable by choosing a few representative paths (i.e. two slits=two paths)
Because the resulting phases can give a positive or negative result the sum can be large or, in fact, zero. The primary difficulty of such calculations is choosing the appropriate weight for each representative path (i.e. normalizing the path integral ) but this result is quite general and was the basis of Richard P Feynman's doctoral thesis.

 

[sophiecentaur]

The use of converging lenses just changes a spherical wavefront into a planar wavefront (it moves a point source out to infinity...similarly for a detector...and is not really part of the question at hand).

This is an important point. A lens will not make diffraction fringes any 'sharper' - only closer or further apart, or perhaps brighter at a distance. What happens with the two mutually interfering sets of waves from the two halves is a function of the wavelength of the waves, for a large enough width of setup. The diffraction patterns are due to the Integral from 0 to ∞. (0 to -∞ for the other side) and the answer is in flux per unit angle.

Returning to the OP and the question about what the resulting wavefront 'looks like'. It depends where you are looking at it. At a great distance, there will be the original plane wavefront, modified by the fact that the resultant of the Integral varies in amplitude and phase according to the sums of the two resultants (see the Cornu Spiral that's in the reference). A variation in amplitude and a variation in phase relative to the straight distance from the inserted glass section. But a 'dip' on boresight.
If the idea of path distances and phases is not intuitively OK with the OP then that's not surprising. Diffraction is a upside down world and it takes a bit of getting used to.

 

[hutchphd]

Absolutely. It always amazes me how differently people process stuff. I figure a slightly different viewpoint can't hurt, and a reminder that the two slit diffraction question has direct import for a startling variety of modern physics is always good.

 

[ItsKurious]

You could look up Fresnel Diffraction at an edge, which shows the effect at a single opaque edge. The effect of a wavelength shift of half the incident flux could be calculated in a similar way.

That indeed gives a better picture of the situation and I think it actually explains a large part of my doubt.

I will assume that you take my point about your version of the situation. What you have is, essentially, a wide source with one phase and another source with another phase. You can treat each source separately. You will get two edge diffraction patterns (Fresnel or far field, whichever) and the resultants will add (vectorially) .

Yes, this is what I was actually thinking about. So, now I believe, since each point source at the left creates its own Fresnel diffraction pattern after encountering the edge of the glass, too many such point sources would destroy the contrast of the pattern on the screen, if I keep one at the right end.

 

[sophiecentaur]

too many such point sources would destroy the contrast of the pattern on the screen,

The source plane wave 'can be looked upon as' an array of point sources, which is the Huygens approach. If you take an aperture, rather than just an edge, the wider the aperture, the bigger proportion of the energy goes in the straight through direction. In the limit, a small aperture becomes an omnidirectional (point) source.
The inverse of an aperture is a small obscuring line and the fringes are still there (the two patterns add to give a uniform illumination) but with very low contrast.
Your mention of the term 'destroyed the contrast' is perfectly right. The fringes that occur behind the obscuring edge will appear against a black background (shadow) but the same amplitude fringes outside of the shadow are indeed diluted. In your case of two transparent paths, there will be a lot of dilution except on the boresight where, depending on how good the equipment is, the null could be very deep. Everything is working against you - flare, other light paths and the finite curvature of the edge of the glass AND quality / purity of the glass.
For interest, see this.
PS The other thing I forgot is that there will be reflections at the two interfaces and that will mean the intensities are not the same in both halves - so you cannot expect perfect cancellation anywhere.