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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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 Spring '08
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 Physics

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