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Phase difference b/w air-water interface and water-air interface

  • Thread starter c-murda
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Homework Statement


Light of wavelength 418 nm is incident normally on a film of
water 1.0 μm thick. The index of refraction of water is 1.33.
(a) What is the wavelength of the light in the water?
(b) How many wavelengths are contained in the distance 2t,
where t is the thickness of the film? (Do not round your answer
to a whole number.)
(c) What is the phase difference between the wave reflected from
the top of the air-water interface and the one reflected from the
bottom of the water-air interface in the region where the two
reflected waves superpose?


Homework Equations





The Attempt at a Solution


my attemp at a solution:
(a)lambda"water" = lambda/n = 445nm
(b)N = 2t/ lambda"water" = 4.5
(c) Light is going from a medium of smaller index of refraction
(air n = 1) to a medium with higher one (water n = 1.33) so the
reflected wave is out of phase by π respect to the transmitted
wave right?

So the phase diff = π + 2π(2t/lambda"water") = -π +2π(4.5) = ???

My answers i have tried:
25.1 rad, 3.14rad, 31.4 rad, 1 rad, 4.72 rad, 0 rad 31.42rad 1 rad

Im really stumped...pls help!!!!!!!!!!!!!!!!!!!!!!!
 

Answers and Replies

  • #2
Doc Al
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(a)lambda"water" = lambda/n = 445nm
Redo this calculation. This effects parts (b) and (c).
 
  • #3
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Redo this calculation. This effects parts (b) and (c).
Sorry typo...original problem was from the book...

This is from my HW:

Correct Problem Statement:
Light of wavelength 592 nm is incident normally on a film of water 1.0 µm thick. The index of refraction of water is 1.33.
then part b and c follow
 
  • #4
Doc Al
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my attemp at a solution:
(a)lambda"water" = lambda/n = 445nm
(b)N = 2t/ lambda"water" = 4.5
(c) Light is going from a medium of smaller index of refraction
(air n = 1) to a medium with higher one (water n = 1.33) so the
reflected wave is out of phase by π respect to the transmitted
wave right?
All good.

So the phase diff = π + 2π(2t/lambda"water") = -π +2π(4.5) = ???
OK. You're comparing π to 2π(4.5), so the phase difference is that difference mod (2π).

My answers i have tried:
25.1 rad, 3.14rad, 31.4 rad, 1 rad, 4.72 rad, 0 rad 31.42rad 1 rad
One of those should be right.

What book is this from?
 
  • #5
Doc Al
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Correct Problem Statement:
Light of wavelength 592 nm is incident normally on a film of water 1.0 µm thick. The index of refraction of water is 1.33.
What is the film of water on? It sounds like you are treating it as a film of water on air--that doesn't seem likely.
 
  • #6
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All good.


OK. You're comparing π to 2π(4.5), so the phase difference is that difference mod (2π).


One of those should be right.

What book is this from?
I know i thought long and hard about this and only have come up with these solutions repeatedly...is it possible they want a negative answer(which is my only guess left)?

this is web based homework.

The problem is from tipler 6th ed ch32 prob22

What is the film of water on? It sounds like you are treating it as a film of water on air--that doesn't seem likely
your right it doesn't seem likely but the problem doesn't state any other medium is involved
 
Last edited:
  • #7
Doc Al
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I know i thought long and hard about this and only have come up with these solutions repeatedly...is it possible they want a negative answer(which is my only guess left)?
I seriously doubt it.

this is web based homework.
Those web-based systems can be picky. Some insist on 3 significant figures, so it may be worth redoing your calculations without rounding off to two digits in any intermediate step. For example, is 4.5 closer to 4.49?

The problem is from tipler 6th ed ch32 prob22
Unfortunately, I only have an earlier edition.
 
  • #8
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Those web-based systems can be picky. Some insist on 3 significant figures, so it may be worth redoing your calculations without rounding off to two digits in any intermediate step. For example, is 4.5 closer to 4.49?
I got:
-π +2π(4.49) = 8.89π

subtracting 7π leaves 1.98π = 6.22 radians
 
  • #9
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which is correct!
 
  • #10
Doc Al
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which is correct!
Sweet! (Pain in the butt, isn't it? :smile:)
 
  • #11
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Sweet! (Pain in the butt, isn't it? :smile:)
HA youre telling me...This is the 13 assignment ive had to do for this class all with 30+ questions...sig figs start to become a little annoying. Luckily they have improved the software side of the web-based homework(WEBASSIGN). with chem is was 2x as bad. Thank goodness there is a 1% marginal error aloud

Thanks for your help DOC as usual you come through!
 
  • #12
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Quick question: So C-Murda got -n + 2n(4.5) because 4.5 is the value of the 2t/lambda(water), and 2n because it travels through water first, but where did he get the 7n that he uses in his calculations? And which n is air and which is water?
 
  • #13
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Sweet! (Pain in the butt, isn't it? :smile:)
Appreciate if you could help me out on this one.
 
  • #14
Doc Al
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Quick question: So C-Murda got -n + 2n(4.5) because 4.5 is the value of the 2t/lambda(water), and 2n because it travels through water first, but where did he get the 7n that he uses in his calculations? And which n is air and which is water?
First, realize that what you are calling 'n' is actually 'π' = pi.

The phase difference is 2π(4.49) - π = 7.98π. When you express that mod 2π, you end up subtracting 6π, for a phase difference of 1.98π.

(His post #8 was not quite clear and contained a few typos.)
 
  • #15
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Thank you very much - makes WAY more sense now. Regards.
 
  • #16
I don't understand what "Expressing mod 2pi" means. I'm really having a hard time understanding why you subtract that 6pi.
 
  • #17
Doc Al
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I don't understand what "Expressing mod 2pi" means.
That means subtracting out whole multiples of 2pi until you're left with just a fraction of 2pi. For more on this, see: Modular arithmetic
I'm really having a hard time understanding why you subtract that 6pi.
The reason why you subtract the multiples of 2pi is that 2pi is equal to a whole wavelength. If two things have a phase difference of 2pi, they're really in phase. Only the fractions of a wavelength that are left after subtracting the whole wavelengths are relevant to figuring out the net phase difference.
 
  • #18
RIGHT RIGHT. Oh my goodness I'm so dumb haha. Thanks for clearing that up haha.
 

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