thin-film interference

[erskine]

This is example 10.9b in the BR physics books.

A thin hair is placed between two microscope slides. When laser light is shined down onto the slides, periodic intensity maxima are seen to span the slides. Which of the following changes will decrease the distance between the maxima?

I. Decrease the wavelength of the laser light.
II. Fill the air gap between the slides with carbon disulfide (n=1.6)
III. Move the hair to the left, while keeping the slides in contact with each other at their left edges.

The correct answer is I and III.

I'm pretty much clueless about everything here- what equations are necessary for this problem?

This is as far as my reasoning goes:
We are observing the reflected light rays off of the bottom layer of the top microscope slide (which doesn't have a phase change) and the reflected light rays off of the top of the bottom microscope slide (which DOES have a phase change). Thus for constructive interference 2d= m+0.5 (wavelength/n) where n is the index of refraction and m is any integer and d is the gap of air between the two slides. I'm entirely confused, though, as to how to determine the distance between the maxima- any help would be greatly appreciated.

---

Members don't see this ad.

 

[Isoprop]

Not a physics expert, but here's my input:

As the laser passes through the hair, it will create a diffraction pattern. It turns out to be similar to the single slit experiment. Even if you didn't know that, the wavelength is almost always proportional distances between the maxima. I is correct.

Changing the gas may change the thin film problems, but the diffraction pattern is caused by the hair not by thin film inteference. Even though we are changing the index of refraction of the gas between the slides, the light waves are always parallel to each other. That is why II is eliminated.

As for III, I'm not sure. Are we moving the hair out of the way of the laser? Perhaps someone else can clarify this part. I would have picked I only.

 

[erskine]

thanks for the quick response.

There's a picture included with the problem, which explains a lot. Basically it shows the two microscope slides with the hair wedged in between the two of them at the far right side. There's a small gap of air all along the length of the two slides, and it tapers off towards the left side. There's also a diffraction pattern that extends the entire length of the microscope slides.

 

[Isoprop]

Sorry, I'm either not picturing it correctly or I'm just as lost as you are.

Hmm, the "triangle" or "wedge" between two pieces of glass is actually a classic thin film application in physics. But I'm pretty certain that the diffraction pattern is coming from the light passing through the hair and not from the thin film. But if that reasoning was correct, answer III would still be wrong, so I'm missing something.

 

[Isoprop]

I drew out a picture, and I think I get it now. What's happening when you move the hair to the left is that the distance between the hair and the slide decreases. The angle of diffraction will be the same but the distances between the maxima will decrease because the light didn't travel as far.

This is essentially a single slit problem. The hair is the single slit, and the microscope slide is the viewing screen. Decreasing the distance to the screen or the wavelength will decrease the distances between the maxima.

 

[schlieren]

Well both of you understand but I am confused. I have the book at my hand, but why is II not correct? You wrote the equation d=(N+0.5)*(wavelength/n), so should not II (increase n) have the same effect as I (decrease wavelength) ???

 

[Isoprop]

This isn't a thin film problem. This is a single slit problem. So you don't use the thin film equation.

Let me try to explain everything step by step from the beginning.

Instead of a hair think of a single slit between the slide. Light goes into the first slide, then exits to hit the hair, which acts like a single slit. Then light diffracts and we have maxima that "spread." These maxima will hit the second microscope slide and produce a diffraction pattern.

I. The distances between the maxima is directly proportional to the wavelength. Look at the equation here.

II. If we fill the gap with a gas that has an n = 1.6, then the diffraction maxima should increase, not decrease. When the diffracted rays of light enter the microscrope slide, they will bend away from each other according to Snell's law (glass has an n = 1.33). This is contrary to what I said above because I misunderstood the problem.

III. Moving the hair/slit to the left will likely increase the distance between it and the second glass. The diffraction pattern will have more space to spread out so the maxima should increase. In other words, if you use the same equation in part I found on the wikipedia page, we find distance inversely proportional to the distance between the maxima. So increasing the distance of the hair/single slit and the slide will decrease the distance between each maximum.

Now, I'm waiting for another physics person to confirm/deny what I said. Optics isn't my greatest physics topics.

Edit: the equation on the page calculates the minima. The maxima is approximately half-way between each min.

 

[schlieren]

No there is no way for this to be a slit diffraction problem, first because diffraction requires the slit's size comparable to wavelength and a hair is too large for that, second because on Berkeley Review this question appears right after section about interference but before diffraction. This has to be a thin film problem.

 

[Isoprop]

No there is no way for this to be a slit diffraction problem, first because diffraction requires the slit's size comparable to wavelength and a hair is too large for that

Yes you can produce a diffraction pattern with a laser and a piece of hair.

Edit: google results-- http://www.google.com/#hl=en&q=hair+diffraction

second because on Berkeley Review this question appears right after section about interference but before diffraction. This has to be a thin film problem.

Well, I don't know what to say, I don't have the book. If someone else can come up with an alternative explanation, I'd be happy to hear it. I'm sure there's bound to be a physics expert to shed some light on this problem.

 

[schlieren]

Well I retract my first point, this is not a human hair, so diffraction is possible with a very thin hair.
I wish this forum could allow me post pictures. If you see the picture in that question, it draws the maxima as parallel and equally bright stripes. Single-slit diffraction pattern only has one brightest maximum at center and the other maxima are much darker.
But thanks a lot for discussing around this. I appreciate your time very much.

 

[Econ2MD]

look up diffraction grating.

 

[erskine]

first of all, thanks everyone for your help (especially isoprop!) This is obviously a confusing problem and I finally figured out how to upload an image to the internet. Probably should've done this earlier and I apologize for the confusion.

The first picture is the side view of the hair wedged between the two microscope slides.
The second picture is the periodic intensity maxima as seen from the top of the slides with the hair on the right side. The maxima should be equal in size (i'm just horrible at paint)

 

[Isoprop]

Okay, I think I finally understand the problem. I thought the laser was being shone onto the hair! The diagram finally clears things up for me.

This is, in fact, a thin film problem. My apologies.

For a thin film where the phase change is 180 degrees, the maxima exists when this equation is true:

2nt = (m+0.5)λ or t = λ(m+0.5)/(2n)
(where n = index of refraction of the film; m = 0, 1, 2..., and λ = wavelength of light at the source)

The second form of the equation tells us that maxima will occur at λ/(4n), 3λ/(4n), 5λ/(4n)...

Since the maximum thickness of the film is determined by the diameter of the hair (d), we can come up with this inequality.

d > λ(m+0.5)/(2n)

This makes sense because if we plug in numbers of n, m, and λ, the calculated t should always be less than d.

So how do we decrease the length of the maxima? By counting how many m's will make the above inequality true. The more times the inequality is true, the more maxima we have, and the smaller the distances between the maxima.

I. Decrease the wavelength? If we decrease the wavelength, we have more m's that will make the inequality true. Thus, decreasing the wavelength will increase the number of maxima and decrease the lengths between them.

II. Increasing the index of refraction of the gap? Since we still get a 180 degree phase change between the two rays, the inequality still holds true. When we increase the index of refraction, we should also have more m's that fit the inequality, which should also decrease the lengths between each maxima.Not sure why this is false because the above inequality shows it to be true.

III. Moving the hair? Moving the hair to the left increases the maximum thickness by increasing the angle between the two slides. Thus, we are increasing d and also increasing the number of m's that make the inequality true. This will then decrease the lengths between all the maxima.

Sooooo, now I got I, II, and III to be true. Sorry, I tried my best.

 

[erskine]

thanks dude, at least i understand I and III, now.

I also don't understand why II is not correct. The n of glass is usually 1.5, right?

 

[Isoprop]

thanks dude, at least i understand I and III, now.

I also don't understand why II is not correct. The n of glass is usually 1.5, right?

Yes, but the n of glass shouldn't matter. The above equation holds true whether or not n = 1 or n = 1000 because one reflection is always a 180 degree phase shift and the other is not.

 

[BerkReviewTeach]

Here is the answer explanation from the proof copy I have of the new book. Please forgive any typos I make in the transcription.

• Solution
  A shorter wavelength will decrease the distance between adjacent maxima. This is shown in equations (10.13) and (10.14), where lambda is directly proportional to the distance between adjacent maxima. This makes Statement I valid.
  
  Filling the gap with carbon disulfide, as opposed to air, would lead to a smaller difference between the index of refraction of the medium and that of the glass. A smaller difference in n values results in less bending of light as it passes from one medium into the next, so the light would need to travel farther to get the same maxima as it got when air filled the gap. This would spread the bright spots, thereby increasing the distance between adjacent maxima. Statement II is invalid.
  
  For Statement III, it is perhaps best to visualize what happens when the hair is moved to either the left or right. If the hair moves to the left, then the glass slides grow farther apart. If the hair moves to the right, then the glass slides grow closer together. Taken to an extreme, if the two slides touch one another, then all of the bright spots are gone (pushed off of the plate if you will). Opening the gap creates the spots and closing the gap eliminates the spots, so widening the gap must bring adjacent maxima closer together. This makes Statement III a valid statement.
  
  The best answer is choice B.

 

[erskine]

Thank you!!

 

[cycloethane]

Don't you increase L when you spread the slides apart? Doesn't this increase y (spacing between the intensity maxima) according to this formula: y=lambda*L/d

 

[MrNeuro]

heres the question

Example 10.12a TBR


I still have an issue with Statement II

wikipedia states that for constructive interference

2 n2 d cosϴ2=(m+0.5)λ

as m increases the number of maxima increases and the distance between the maxima will decrease

now

in statement II says that if you change the air (n=1) to carbon disulfide (n=1.6) the distance between maxima will decrease....heres where i have an issue with TBRs answer they claim that if the difference in refractive index is smaller it will result in less bending. so i drew it out and applied their logic to the equation above air will result in a smaller value for cosϴ which means the number constructive points will decrease which means you'll have a greater distance between maxima HOWEVER
carbon disulfide cosϴ is larger relative to air's cosϴ thus carbon disulfide must have more maxima and smaller distances between the maxima making choice 2 correct

even when you plug in the refractive indices you get the same result
n carbon disulfide > n air
more m
smaller distance between maxima

so the answer should be all of the above

Summary for those that don't want to read what i wrote:
Why choice II should be right...i think

2 n2 d cosϴ2 = (m+0.5) λ

Carbon disulfide (CS2) vs Water

n CS2 > n H2O ====> m CS2 > m H2O ======> d between maxima CS2 < d between H2O

cos&#1012; CS2 > cos&#1012; H2O ====> m CS2 > m H2O ======> d between maxima CS2 < d between H2O


please correct me if im wrong this question is annoying me

 

[MedPR]

Reading the explanation worked well for me. Throw out all the equations you know and think about it in terms of extremes.

 

[MrNeuro]

Reading the explanation worked well for me. Throw out all the equations you know and think about it in terms of extremes.

thats what i did but then i got the answer wrong...then i read their explanation thought about it drew it out and it doesn't match up

if you're going from a n of 1.5 (glass) to 1.0 (air) light will refract away from the normal/away from the path......right?
if you now go from a n of 1.5 (glass) to 1.6 (carbon disulfide) light will refract toward the normal line

if light refracts away from the normal the path length will increase
if light refracts towards the normal the path length will decrease

a shorter path length will result in more maxima and a decreased distance between the maxima.....making choice 2 correct...

thats what i did the first time through....is that thought process incorrect?

 

[ilovemedi]

What equation do we use for Example 10.12a (above @ Mr.Neuro's post)?

 

[roody]

So I know its been a long time since you guys have been discussing this question but I thought it should be helpful to add some insight incase someone is looking for further explanation for this question (Like myself yesterday and today I hated this question but thanks to it I have a better understanding of the topic now) anyway so here is my solution and correct me if I'm wrong

For Statement I : First of all we need to acknowledge what is constant and what isn't.... so since the distance between the 2 glasses is the same, therefore, "L" (as in Y= L*Wavelength / Distance between slits) is the same. is also the same. So, if we use the equation 2d=(m+0.5)* Wavelength/n, only the wavelength is change as proposed by the choice therefore we use Y= L*Wavelength / Distance between slits,,,, so decreasing wavelength will decrease Y (Which is the distance between the peaks which means more bright spots)

[[[[Just an fyi, the 2d here is not the distance between the 2 slits as in Y=L* Wavelength/Distance between Slits; it is actually the (L) the thickness of the medium or in this case the distance between the glasses due to the Hair Diameter]]]]


For Statement II (The naughty one) .... Same approach as above, determine what is constant and what isn't.... Since 2d (L) is still the same (because we didnt move the hair or increased the diameter of the hair so the distance between the glasses is still the same.. L is the same) and Wavelength is still the same, therefore, we will have to think about what will change when changes ....well, according to snells law the angle of refraction will change ..... now you need to draw it because I dont know how to explain it here but I will try my best.... if you use Air as the medium the light will be deflected a lot away from the normal line.... then if we use Carbon Disulfide (which has higher n than glass) then the light will be deflected towered the normal line.... if you can imagine what I just wrote then you are almost done.... now think of the point where the light reflects as the slits (the first slit is when B/W air and glass and the second Slit is between Glass and the medium which in this case is either Air or Carbon Disulfide) so the distance between the slits will change when we change because [[[[fyi, this distance is the distance between the slits it has nothing to do with L]]]] so we use the old equation again Y= L* Wavelength / Distance Between Slits since with Carbon disulfide will have light originating from points closer to each other so the distance between slits is small therefore Y will be large and we said that "Y" is the distance between bright spots

Statement III : this one is weird and I think the statement should be more informative as to say how far the hair will be moved left because if the hair moves the maximum distance to the left while the left edges are still in contact then then we should get less bright light (because "L" the distance between the glasses will increase which will increase Y the distance between the bright spots)
However, if you move it just a little bit to the left then the increase in "L" is not that much but light will have more suface to reflect off (because this small increase in L caused less touching between the 2 glasses so more surface is now available for light to reflect off) so presumably we should get more bright spots .....

 

[snowys435]

I still dont understand how this is a slit problem and not a thin film problem?

 

[brood910]

Sorry for bumping this, but I want to hear more on this from diff people..

 

[manohman]

I still dont understand how this is a slit problem and not a thin film problem?

Bringing this back up as well because im having trouble understanding why this is a slit problem and not a thin film problem.

Am i visualizing this the wrong way?

 

[whistle47]

hey so I just spent a while puzzling through this problem (anyone else cramming for january test date?) and I think I figured it out. the posted BR explanation just doesn't do it for me so I wanted to come up with an answer based on only intuition.

two very important things to understand:

1. why is there interference in the first place? this is a thin film problem. ray 1 reflects off the boundary between the top glass slide and the air wedge, while ray 2 refracts through the air, reflects off the bottom glass slide, then refracts through the glass slide. where do you get a path length difference? ray 2 travels through the air wedge but ray 1 doesn't.

2. why do you get an interference pattern? the space between the glass slides changes as you scan along the length of the slide due to the hair at one end (from 0 at the very left to the diameter of the hair at the very right, as drawn), which means the path length difference increases from left to right.

now the answers:

I - suppose you have wavelength = lambda_1, and you are on a part of the slide with a bright spot. the next maxima will be found when you have moved down the slide enough so that the space between the two glass slides has increased by some quantity directly proportional to lambda_1. this means when you decrease the wavelength to lambda_2, you have to move down the slide a smaller distance to find the next maxima. so maxima become more frequent.

II - if the air is replaced by carbon disulfide, ray 2 will refract towards the normal when it encounters the wedge instead of away from the normal (1 < 1.5 but 1.6 > 1.5). refracting towards the normal makes ray 2 take a more direct path between the two glass slides, while refracting away from the normal makes the ray take a wider and longer path. therefore, replacing air with carbon disulfide lowers the path length difference at a given slide separation distance; the slide separation distance is directly related to your position on the slide, so you need to move further down the slide to get enough of a path length difference to match lambda_1. this means this answer choice will increase the distance between adjacent maxima.

III - this should be obvious by now. moving the hair to the left increases the amount of slide separation you get (and hence path length difference) for a set movement down the slide. to achieve the same amount of path length difference, you don't have to move as far down the slide. this means maxima are more frequent.

what a total ***** of a problem. expecting all mcat problems to look like this.

 

[DrknoSDN]

If it helps I discussed this problem in another thread also. You can find it Here:

There is also a good website that explains it here. (Just ignore that it this is considered highschool level physics, hah)

 

[manohman]

ah thank both of you that helped immensely.

Just to clarify, regarding the pattern you see in a wedge fringe, the reason is that as you go from left to right, the path length difference changes, so that you will get either constructive or destructive interference at different points.

depending on how far away you are from the hair, your path length will be different, and so the wavelength will get shifted by different amounts as you go across. This should result in varying amounts of destructive or constructive interference right? Or is it all or nothing, so if it's not perfectly constructive (path length *2 = one half wavelength), then its automatically destructive. I think it's not all or nothing (based on basic interference concepts).

If you had two hairs so there was no angle, so the space in between was constant, you'd basically just get one "result" (im having trouble imagining what that would look like). Would it just be one big light, that's dimmer or brighter depending on the degree of interference that occured? Since normal constructive or destructive interference only affects amplitude? (unless they are two different waves completely with different wavelengths that are out of phase - Lord feels like im going down the rabit hole here). but it should be just one resulting wave right? So one signal on a board (like one line?)

 

[amy_k]

@Manoman you've got the reason for the interference pattern right. There are two reflected rays, one from the top and one from the bottom slide. The difference in path length between the two waves causes interference, and the difference in path length is 2r, where r is the distance between the slides. R decreases moving from right to left.

We can have either total constructive interference, total destructive interference, or partial interference. So to be classified as constructive or destructive it must be "all or nothing," but there can be partial interference which is anything between those two extremes.

If we had two hairs we would treat this as a thin film problem, and we would get either constructive or destructive interference depending on film thickness and wavelength. Constructive interference would produce a light band along the length of the slide and destructive interference would produce a dark band.

 

[manohman]

Awesome man that clears up so much. I can finally connect this problem to the interference from the double slit interference haha.

 

[amy_k]

Great! I would look at this initial problem as a double-slit problem rather than single-slit or thin-film