Monochromatic Light Through Two Microscope Slides

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Monochromatic light of 500nm wavelength is used to analyze the interference pattern created by two glass microscope slides forming a wedge. The user initially calculated the distance "d" for the fourth dark fringe as 875nm, but the correct value is 750nm. The confusion arises from the application of the interference equation, which is typically used for constructive interference, leading to a misunderstanding of the dark fringe formation. A key point is that a 180° phase shift occurs during reflection, effectively altering the path length difference and resulting in dark fringes. Understanding this phase shift clarifies why the first dark fringe appears despite the expected path length difference being zero.
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Homework Statement


Monochromatic light of wavelength 500nm shines on a pair of identical glass microscope slides that form a very narrow wedge, as shown in the diagram. The top surface of the upper slide and the bottom surface of the lower slide have special coatings on them so that they do not reflect light. The inner surfaces are reflective. The indices of refraction of air and glass are 1.00 and 1.50 respectively. Looking down on the slide from above, you see the fringe pattern shown.
http://img519.imageshack.us/img519/1517/picture2dt0.png


Homework Equations


\lambda_{D}=\frac{2nd}{m-0.5}


The Attempt at a Solution


Since the "P" points to the 4th dark fringe,
m=4
\lambda_{D}=500nm
n=1 (as the light is traveling the distance "d" in air)
d=?

Isolating for d and solving,
My answer is 875nm.
However, the correct answer is 750nm.
If anyone could help me out I'd really appreciate it.
Thanks in advanced.
 
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Your equation disagrees with the case d=0 at the edge of the slides, where we can say n=0 for the dark fringe that appears there.

How would you modify your equation in order to explain the dark fringe at the left edge?
 
i see.
\lambda=\frac{2nd}{m}
if m=0, then d would equal zero

also, setting P as the 3rd dark fringe, you would get the answer 750nm.

however, the equation i stated above happens to be the same equation on my formula sheet as for constructive interference (whereas the dark fringes are obviously destructive).

im assuming for the first dark fringe there is no path length difference, but why does it produce a dark fringe and not a bright spot as my formula predicts?

thank you
 
MightyMan11 said:
im assuming for the first dark fringe there is no path length difference, but why does it produce a dark fringe and not a bright spot as my formula predicts?

There is a 180° phase shift for one of the two reflections. This has the same effect as changing the path length difference by ½λ.
http://theory.uwinnipeg.ca/physics/light/node10.html
 
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