This preview shows pages 1–2. Sign up to view the full content.
© 2011 E. Szarmes
PHYS 274
DAILY ASSIGNMENT 4 — DUE WEDNESDAY, JAN 26/11
PROF. SZARMES
Q2S.2
D = 5.0 km
f = 100 MHz
λ
= c/f
= 3(10
8
)m/s/100(10
6
) Hz
= 3.0 m
Note that radio waves travel at the speed of light, not the speed of sound! Now, for two antennas
emitting in phase, the angle
ϑ
n
of the n
th
maximum in the interference pattern is
where y
n
is the transverse position of the n
th
maximum at distance D. Thus, the separation between
two adjacent maxima (in the region near the axis where
ϑ
is small), is
For
λ
= 3 m (i.e. f = 100 MHz), D = 5.0 km, and d = 60 m, we calculate
Δ
y = 250 m.
Q2S.8
The criterion for resolving the headlights
(separated by d = 1.4 m at distance D) is
where
α
~
d
/
D
and
ϑ
1
(not shown in the
diagram) is the half angle of the diffraction
patterns. We assume that the diameter of
the eye’s pupil is a = 8 mm and that the
average wavelength of the light is
λ
= 550 nm.
Then
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '11
 EricSzarmes
 Physics, Light

Click to edit the document details