Minimum thickness of thin film constructive interference

AI Thread Summary
The discussion centers on calculating the minimum thickness of a benzene thin film for constructive interference when illuminated with orange light. The correct formula for this scenario involves using the refractive index of benzene (n = 1.501) and accounts for two reflective phase reversals. The derived thickness is 204 nm, achieved by applying the equation t = (lambda) / (2*n). Participants clarify the distinction between constructive and destructive interference based on phase shifts and the context of the problem. Understanding the wording of physics problems is emphasized as crucial for determining the type of interference required.
crazyog
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Homework Statement


What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when viewed at normal incidence and illuminated with orange light (lamba = 615 nm)? A glass slide (ng = 1.620) supports the thin film
answer= 204 nm

Homework Equations


I rearranged an equation to get
t= (m=.5)(lambda) / 2*n

but I'm thrown off by how they give me two n values?


The Attempt at a Solution


I tried using my equation plugging in each n, but did not get the answer.

please help, thanks in advance :)
 
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Hi crazyog,

crazyog said:

Homework Statement


What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when viewed at normal incidence and illuminated with orange light (lamba = 615 nm)? A glass slide (ng = 1.620) supports the thin film
answer= 204 nm

Homework Equations


I rearranged an equation to get
t= (m=.5)(lambda) / 2*n

I don't believe that is the correct formula to use for this case.

but I'm thrown off by how they give me two n values?

They are really giving you three n values because there are three materials: the thin film of benzene is between air and glass. The question is, which material does the n in the equation refer to?
 
well they are referring to the benzene so would it be the 1.501 for n.
would the equation be
t= (lambda)/4*n

hm...but I am still not getting the answer?
 
crazyog said:
well they are referring to the benzene so would it be the 1.501 for n.

Exactly; in the thin film constuctive/destructive condition equations, the thin film index of refraction is used.

would the equation be
t= (lambda)/4*n

hm...but I am still not getting the answer?

No, that's the same equation as in your original post. The equation in your original post would find the smallest thickness of destructive interference. (Remember that whether the think film equations are constructive or destructive depends on how many reflective phase shifts there are.)

So what equation give constructive interference for this case? (I think there should only be two alternatives.)
 
oh ok so (615*10^-9)/(2*1.501) = 2.04*10^-7

Got it! thanks!
so the key was that it is constructive interference, which they stated in the problem...
is there a way to tell if they don't state it?
like if they say they would the find the maximum or minimum, does that refer to constructive and deconstructive?
 
crazyog said:
oh ok so (615*10^-9)/(2*1.501) = 2.04*10^-7

Got it! thanks!

That's right; the important point was that there are two reflective phase reversals in this problem.

If instead, for example, the benzene was on top of water (with an index n=1.333), then the equation in your original post would have been the correct one for constructive interference, because that case would only have one phase reversal.


so the key was that it is constructive interference, which they stated in the problem...
is there a way to tell if they don't state it?
like if they say they would the find the maximum or minimum, does that refer to constructive and deconstructive?

They'll often use the words bright or dark, for constructive and destructive. Or perhaps they'll say they want to reduce reflection (off of an aircraft coated with a thin film, for example), which would be destructive interference.

I'm sure you've noticed by now in your studies that there are many non-standard way of wording physics problems, so no matter how many keywords you keep in mind, you'll still run across problems that are worded in a new way, and you just have to think about it to figure out what they want.
 
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