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Figure 49: Figure for Problem 35.23
We now need to use some physical reasoning to ﬁgure out when the two will interfere destruc
tively at the largest
x
P
. Notice that if
x
P
goes oﬀ to inﬁnity, then the diﬀerence between the two
goes to zero. Hence, the maximum
x
P
which leads to destructive interference is when the diﬀerence
is
1
2
.
p
x
2
P
+
d
2

x
P
λ
=
1
2
This is a quadratic equation in
x
P
that we can easily solve.
x
2
P
+
d
2
=
±
x
P
+
λ
2
²
2
⇒
x
2
P
+
d
2
=
x
2
P
+
λx
P
+
λ
2
4
⇒
λx
P
=
d
2

λ
2
4
⇒
x
P
=
d
2
λ

λ
4
Hence, our solution is
x
P
= 7
.
87
μm
.
Problem 35.75:
Figure 50a shows a lens with radius of curvature
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Unformatted text preview: R lying on a ﬂat glass plate and illuminated from above by light with wavelength λ . Figure 50b a photograph taken from above the lens) shows that circular interference fringes (called Newton’s rings) appear, associated with the variable thickness d of the air ﬁlm between the lens and the plate. Find the radii r of the interference maxima assuming r/R ± 1. Figure 50: Figure for Problem 35.75 67...
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This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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