Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
24
The Wave Nature of Light
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24-3: Double-Slit Interference
24-4: Visible Spectrum; Dispersion
24-5: Single-Slit Diffraction
24-6 and 24-7: Diffraction Gratings
24-8: Thin-Film Interference
24-9: Michelson Interferometer
24-10: Polarization

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 53
Q

How thick (minimum) should the air layer be between two flat glass surfaces if the glass is to appear bright when 450-nm light is incident normally? What if the glass is to appear dark?

A
110 nm
Giancoli 7th Edition, Chapter 24, Problem 53 solution video poster
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COMMENTS
By spanda2u on Sun, 11/29/2015 - 11:25 PM

Hello,
there is a second question in this problem: "What if the glass is to appear dark?"
I am curious to see that solution.
thanks,
Spanda

By Mr. Dychko on Mon, 11/30/2015 - 4:40 AM

Hi spanda2u, well, in a pinch I can give a written solution: the minimum separation would be zero, or if you want a non-zero answer then the first separation for destructive interference would be 225 nm225 \textrm{ nm} (230 nm230 \textrm{ nm} with sig. figs. considered). It's important to know why "zero" creates destructive interference ("zero" isn't really zero, but rather a microscopic separation in which a very tiny air gap still exists despite the two pieces of glass touching). At "zero" separation, the ray reflecting from the bottom piece of glass undergoes a 12λ\dfrac{1}{2}\lambda phase shift since it's initially in a medium of low index of refraction (a microscopic layer of air) and then reflecting off a medium of high index of refraction (bottom piece of glass). This business about a ray undergoing 12λ\dfrac{1}{2}\lambda phase shift when reflecting from a medium of higher index of refraction that the medium in which it is initially is just a rule that has to be memorized. Since the ray reflected from the bottom piece of glass is shifted 12λ\dfrac{1}{2}\lambda, whereas the ray reflected from the bottom of the top piece of glass is not phase shifted at all (since it's initially in a high index medium, glass, reflecting off a low index medium of air), and there is no additional path length travelled by the bottom ray (or a negligible path length to be precise), these two rays are 12λ\dfrac{1}{2}\lambda out of phase with respect to each other and so they destructively interfere (dark!). The next dark separation would occur when the separation between the pieces of glass results in an additional path length travelled by the bottom ray to be λ\lambda. The bottom ray travels the separation twice (once down, then again back up), so the separation that causes an additional path length of λ\lambda is 12λ\dfrac{1}{2}\lambda. This makes the first non-zero dark separation equal to 12450 nm=225 nm230 nm\dfrac{1}{2}450 \textrm{ nm} = 225 \textrm{ nm} \simeq 230 \textrm{ nm}.

Hope that helps,
Mr. Dychko

By spanda2u on Thu, 12/3/2015 - 4:47 AM

Thank you.

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