Hard physics question not for the faint of heart.

TimeGnome · Nov 9, 2010

Okay im sitting here doing my home work and im just really happy i finished this question its truely tough for me... im in high school but its a great question involving lasers. Here it is.

A very thin sheet of plastic ( n = 1.60 ) covers one slit of a double slit apperatus illuminated by a 640 nm light. The center point on the screen, instead of being a maximum, is dark. What is the (minimum) thickness of the plastic?

The answer is 533 nm

Anyone here think they can do this its a really good question.

Elektrotechniker · Nov 9, 2010

Does not seem to be an easy one.

bobhaha · Nov 9, 2010

Hmmm....

If a wave is delayed by time ∆T, the phase change is ∆λ = c*∆T

Since the film thickness is much less than the distance to the slit, the slowing of the wave in the film produces a ∆T = n*tp/c (∆T = distance/velocity = tp / (c/n), since the velocity in the plastic is c/n.)

In order for the max to be turned into a min, the phase must change by λ/2 (half wave shift).

λ/2 = n*tp

tp = λ/(2*n)

tp = 640/(2*1.6) = 200 nm

This is the minimum thickness, The actual thickness can be any integer multiple of that.

0.5 = 2*π*n*tp

Benm · Nov 9, 2010

I suppose the purpose of the plastic is to invert the resulting pattern?

I would go about it by making a delay of 0.5 wavelength in one of the slits, as is suggested by the exercise. Obviously the wave that doesnt go through the plastic does go through air/vacuum, so its all about the relative delays here.

Light moves 1.6 times slower (or 0.625 times as fast) in plastic. I guess i'd plot the minima and maxima of both waves and find the first place where they are in antiphase. This would be the minimum required distance, but you could also use ticker plastic at the second and subsequent points of antiphase.

Benm · Nov 9, 2010

... cross posted with the answer above, i think i'd have to agree on 200 nm being the answer, not 533.

Bluefan · Nov 9, 2010

Sounds fairly simple, the minimum thickness for it to give a half wave retardation compared to air. Index DIFFERENCE is 0.6, wavelength 640nm, so we need an OPL (optical path length) difference of 320nm (half wave for full extinction). OPL = n*L = 0.6*L = 320nm
L = 320/0.6= 533.3nm

You guys forgot that a plastic plate takes length the other path also has travelled in air (n~1). Also, bobhaha, other thicknesses causing full extinction aren't an integer multiple of this. The next extinction point is 3/2 wave retardation, needing a plate thickness of 1600nm.

TimeGnome · Nov 9, 2010

Actualy the 200nm is incorrect here is my solution to the problem 533nm is correct.

First you find Delta V by going 3x10^8-3x10^8/1.6 which gets you 1.125x10^8 m/s
Then using lamda/2 you get a wavelegth of 320nm
320x10^-9/1.125x10^8 you get 2.84x10^-15 s
Now we sub into the equation d=vt
(3x10^8/1.6)(2.844x10^-15)=d
d=533 nm

Sorry if my explenation sucks but this is how i did the question and the answer has been labled correct by the textbook answer key and by my teacher. But yeah if you guys want more questions about optic/interference patterns or electromagnatizm and stuff i can look up the tough ones in my text and set out a weekly or daily question.

Bluefan · Nov 10, 2010

The difference velocity works, difference in time travelled works, but considering a waveLENGTH is given, and a LENGTH is asked, why convert twice? But your answers and logic are correct, so it's not really a problem.

An interesting question for you: I have a parallel light beam, and I block the center of the beam by an object. But right behind this object (again in the center of the beam) I still find light. What effect is causing this?
The name of the physicist that thought this was impossible is actually the name of our student union. This spot is named after the physicist that first observed it.

Meatball · Nov 10, 2010

I can't remember what it is called, but I think it's a form of diffraction... is that right?

These spots are observed all the time from what I remember..

EDIT: The Fresnel Bright Spot or Arago's (Poisson's) spot

TimeGnome · Nov 10, 2010

Hmmm i have no idea let me do some research and ill try to have a answer for you by tommorow. But first i woudnt mind some clarafication on the question being asked such whats being asked i am interperating this as what is the light behind a object or the leaked light hitting a semi tansparint (ie thin paper) object. Is it still froma diffraction grating? Or is it just parallel light?

Benm · Nov 10, 2010

Bluefan said:
An interesting question for you: I have a parallel light beam, and I block the center of the beam by an object. But right behind this object (again in the center of the beam) I still find light. What effect is causing this?

There are multiple eplanations for this phenomon. The most sensible invole scattering from the object obscuring the beam, but even if all of those options would be exhausted, you still have gravity.

The object placed in the center of the beam would have mass, and act as a gravitational lens, bending some light that just doesnt hit it on a path that ends up in the middle of the interrupted beam. The intensity of that light is debatable, but it would hold true for any distance beyond the obstruction in theory.

Meatball · Nov 10, 2010

By scattering, do you mean diffraction? Does the distortion of the wavefront result in scattering behind the object?

Bluefan · Nov 10, 2010

Meatball said:
I can't remember what it is called, but I think it's a form of diffraction... is that right?

These spots are observed all the time from what I remember..

EDIT: The Fresnel Bright Spot or Arago's (Poisson's) spot

Arago spot - Wikipedia, the free encyclopedia
No gravity needed, no semi-transparent object, no scattering, just diffraction. I did get the names of the physicist mixed up though, but the theory is pretty easy.
I see this spot regularly if I don't keep my optics clean

Next question: I have an ideal fabry perot interferometer with 99% reflectivity for both mirrors. If the cavity is on resonance, the transmission through the interferometer is 100%. How is this possible if 99% already reflects back of the first surface?

Any backreflection is considered lost, before you worry about those. If this is too easy, calculate it's finesse, cavity length and free spectral range needed to resolve two lines from a 15cm long HeNe laser (those lines are considered to be infinite narrow).

Edit: post 473nm

Benm · Nov 10, 2010

Meatball said:
By scattering, do you mean diffraction? Does the distortion of the wavefront result in scattering behind the object?

On that, i suppose the terminology depends on how big the object is compared to the wavelength. I suppose the difference between the mechanisms isn't all that important here.

As for the Arago spot: That would require the object to be circular, wouldn't it? This was not mentioned in the original question.

Bluefan · Nov 10, 2010

A square would work too, it just changes the shape diffraction pattern behind it. You can calculate how the pattern would look like.

Meatball · Nov 11, 2010

Bluefan said:
Next question: I have an ideal fabry perot interferometer with 99% reflectivity for both mirrors. If the cavity is on resonance, the transmission through the interferometer is 100%. How is this possible if 99% already reflects back of the first surface?

Any backreflection is considered lost, before you worry about those. If this is too easy, calculate it's finesse, cavity length and free spectral range needed to resolve two lines from a 15cm long HeNe laser (those lines are considered to be infinite narrow).

Edit: post 473nm

I'm stumped. I thought parrots were for pirates.

Hard physics question not for the faint of heart.

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