Monochromatic Light Through Two Microscope Slides

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Homework Help Overview

The problem involves monochromatic light of wavelength 500 nm interacting with a pair of identical glass microscope slides arranged to form a narrow wedge. The setup includes reflective inner surfaces and non-reflective outer surfaces, leading to a fringe pattern observed from above.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the distance "d" using a specific equation related to dark fringes but arrives at an incorrect value. Participants question the validity of the equation used and explore modifications to account for the observed fringe pattern.

Discussion Status

Participants are actively discussing the discrepancies in the calculations and the underlying physics of the dark fringes. Some guidance has been offered regarding the phase shift associated with reflections, but no consensus has been reached on the correct approach to the problem.

Contextual Notes

There is an ongoing discussion about the assumptions made regarding path length differences and the conditions for dark versus bright fringes. The original poster's confusion about the application of the interference equations is noted.

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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


[tex]\lambda_{D}[/tex]=[tex]\frac{2nd}{m-0.5}[/tex]


The Attempt at a Solution


Since the "P" points to the 4th dark fringe,
m=4
[tex]\lambda_{D}[/tex]=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.
[tex]\lambda[/tex]=[tex]\frac{2nd}{m}[/tex]
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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