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Unformatted text preview: Chapter 37 Q37.14 383 The metal body of the airplane is reflecting radio waves broadcast by the television station. The
reflected wave that your antenna receives has traveled an extra distance compared to the stronger
signal that came straight from the transmitter tower. You receive it with a short time delay. On the
television screen you see a faint image offset to the side. SOLUTIONS TO PROBLEMS
Section 37.1 Conditions for Interference Section 37.2 Young’s DoubleSlit Experiment P37.1 ∆y bright = P37.2 y bright = ja f e −9
5.00
λL 632.8 × 10
=
m = 1.58 cm
−4
d
2.00 × 10 λL
m
d
λ= For m = 1 ,
P37.3 e je j 3.40 × 10 −3 m 5.00 × 10 −4 m
yd
=
= 515 nm
3.30 m
L Note, with the conditions given, the small angle approximation does not
work well. That is, sin θ , tan θ , and θ are significantly different. We treat
the interference as a Fraunhofer pattern.
(a) At the m = 2 maximum, tan θ = 400 m
= 0.400
1 000 m 400 m θ = 21.8°
λ= so
(b) 300 m a f 300 m sin 21.8°
d sin θ
=
= 55.7 m .
2
m The next minimum encountered is the m = 2 minimum; FG
H IJ
K 1
λ
2 and at that point, d sin θ = m + which becomes d sin θ = or sin θ = and θ = 27.7° so y = 1 000 m tan 27.7° = 524 m . b 5
λ
2 FG
H IJ
K 5 λ 5 55.7 m
=
= 0.464
2 d 2 300 m g Therefore, the car must travel an additional 124 m .
If we considered Fresnel interference, we would more precisely find
1
550 2 + 1 000 2 − 250 2 + 1 000 2 = 55.2 m and (b) 123 m.
(a) λ =
2 FH IK 1 000 m FIG. P37.3 ...
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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