CourseNotes.68 - Our first goal is to set out finding d...

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Unformatted text preview: Our first goal is to set out finding d in terms of r and R since this will control the interference upon reflection. Using a little trigonometry we can see that the hypotenuse of the triangle formed √ by r and R in the figure is h = R2 − r2 . Hence, d is given by d=R− R2 − r 2 ⇒ r= 2dR − d2 Since the problem tells us that r/R 1 we can treat the light as if it were all coming straight into the lens/plate setup and reflecting straight back (i.e. we will neglect the fact that the camera is a single point). The other approximation that we will make is that since the radius of curvature of the lens is very small we can neglect any refraction effects and treat the rays as if they were reflected straight back and not at an angle. With these approximations, the interference upon reflection is simply a function of d just like in thin films. We must be careful in this case because one of the reflections will pick up a phase shift relative to the other. The first reflection off of the air has 0 phase shift while the second reflection will pick up half of a wavelength. Hence, the condition for constructive interference is that 2d be a half integer multiple of λ. 2d = m+ 1 2 λ ⇒ r= 68 m+ 1 2 λR − 1 4 m+ 1 2 2 λ2 ...
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