Φ21 Fall 2006
1
HW27 Solutions
Constructive and Destructive Interference
Parts AD:
Is the interference of the two radio sources in Figure 1 at
each of the labeled points constructive or destructive?
Solution:
The important quantity is the dierence in distances from
the transmitters to the observation point. Point A is equidastant, so the
interference there is
constructive.
wavelengths. The interference at point B is
∆` = 3 m,
destructive.
At point C, the distances are 3.5 m and 2.5 m, so
a half wavelength. The interference at point C is
The distances to point D are 5 m and 1 m, so
2
ent radio waves broadcasting in
phase are located as shown above.
At Point B, the distances are 1.5 m and 4.5 m, so
ference is
Figure 1: Two sources of coher-
constructive.
which is 1.5
and the radio sources broadcast
∆` = 1.0 m,
destructive.
∆` = 4 m,
Each grid square is 0.5 m square,
which is
at
λ = 2.0 m.
and the inter-
Problem 22.41
A diraction grating is illuminated simultaneously with red light of wavelength 650 nm and light of an
unknown wavelength. The fth-order maximum of the unknown wavelength exactly overlaps the third-order
maximum of the red light. What is the unknown wavelength?
Solution:
The key here is that the two maxima overlap and hence have the same
The formula
mλ = d sin θ
applies to each wave, but with dierent values of
λ2 =
3
m.
θand the same d sin θ.
m1 λ1 = m2 λ2 to get:
So, set
3 (650 nm)
= 390 nm
5
Problem 22.58
Scientists use laser range-nding to measure the distance to the moon with great accuracy. A brief laser
pulse is red at the moon, then the time interval is measured until the "echo" is seen by a telescope.
A
laser beam spreads out as it travels because it diracts through a circular exit as it leaves the laser. In order
for the reected light to be bright enough to detect, the laser spot on the moon must be no more than 1
km in diameter.
λ = 537
Staying within this diameter is accomplished by using a special large-diameter laser.
If
nm, what is the minimum diameter of the circular opening from which the laser beam emerges?
The earth-moon distance is 384,000 km.
Solution:
The formula for a round aperture is
θ ≈ sin θ =
Where
θ
1.22λ
D
is the angle between the central bright part of the pattern and the dark edge of the pattern. The
distance between these points is 500 m, and we can get the angle by
tan θ =
If we solve for
D,
we get
y
500 m
=
= 1.302 × 10−6 ≈ θ
L
3.84 × 108 m
(
)
1.22 537 × 10−9 m
1.22λ
=
D=
= 0.503 m = 50.3 cm
θ
1.302 × 10−6
1
4
Problem 23.37
Two light bulbs are 0.800 m apart.
From what distance can these light bulbs be marginally resolved by
a small telescope with a 4.10 cm-diameter objective lens? Assume that the lens is diraction limited and
λ = 600
nm.
Solution:
Diraction limited means to use the Rayleigh Criterion, i.e.
(
)
1.22 600 × 10−9 m
1.22λ
θ ≈ sin θ =
=
= 1.785 × 10−5
D
0.041 m
This minimum angle corresponds to the maximum
θ ≈ tan θ =
5
y
L
→
L
because
L≈
y
0.8 m
=
= 44.8 km
θ
1.785 × 105
Problem 36.24
An interference pattern is produced by two identical parallel slits of width
d = 3a.
a and separation (between centers)
Give the number m of the rst interference maxima that will be missing in the pattern.
Solution:
The angles of the maxima due to the two-slit pattern are given by
mλ = d sin θ
The angle of the minima due to the single-slit diraction are given by
m0 λ = a sin θ
(Except that
m0 = 0
is the central maximum.) Here I'm using
m0
to indicate that this is a dierent set of
counting numbers than those for the two-slit pattern.
The missing fringe is when the single-slit minimum corresponds to a two-slit maximum. So, we set the
sin θ
terms equal.
mλ
m0 λ
=
d
a
We solve for
m
and set
m0 = 1
because we want the rst single-slit minimum.
m=
6
d
m0 d
= =3
a
a
Problem K 22.19
The opening to a cave is a tall, 33.0 cm-wide crack. A bat that is preparing to leave the cave emits a 39.0
kHz ultrasonic chirp. How wide is the "sound beam" 120 m outside the cave opening? Use exact formulas;
don't make small angle approximations. Use
Solution:
vsound = 340 m/s.
We need the wavelength of the sound, which comes from
λ=
v = f λ:
v
340 m/s
=
= 8.718 mm
f
39.0 × 103 Hz
2
Since this is a crock, the opening is long and narrow and we use the single-slit diraction formula
Setting
m=1
gives us the angle of the rst minimum from the center of the beam.
θ = sin−1
mλ
8.718 × 10−3 m
= sin−1
= 1.514◦
a
0.33 m
The actual width of the beam comes from
tan θ =
w/2
y
=
L
L
⇒
w = 2L tan θ = 6.34 m
3
mλ = a sin θ.