Unformatted text preview: he speed and frequency of the sound. Substituting this
value for the wavelength into the expression for θ, we find that Chapter 27 Problems λ
W θ = sin −1 1449 v
343 m/s = 1.1°
= sin −1 = sin −1 2.0 ×104 Hz ( 0.91 m ) fW ( ) b. For the light waves, Equation 27.4 (with m = 1) reveals that
sin θ = λ
W W= or λ
sin θ = 580 × 10−9 m
= 3.0 × 10−5 m
sin1.1° Thus, in order for the diffraction of the light waves to match that of the sound waves, the
light waves must pass through a very narrow “doorway.” In passing through a doorway that
is 0.91 m wide, the light waves would be diffracted by an amount so small that it would be
unobservable. 24. REASONING AND SOLUTION We have θ = sin –1 a. Thus, θ= 2
−1 sin θ= 2
−1 sin b. and (2λ/W) for the second dark fringe. ( 430 × 10−9 m ) = 24° ( 660 × 10−9 m ) = 39° 2.1 × 10−6 m 2.1 × 10−6 m 25. SSM REASONING This problem can be solved by using Equation 27.4 for the value of
the angle θ when m = 1 (first dark fringe).
SOLUTION
a. When the slit width is W = 1.8 × 10 –4 m and λ = 675 nm = 675 × 10 –9 m , we find,
according to Equation 27.4, F λ I = sin L1) 675 × 10
(
GW J M 1.8 × 10
HK N θ = sin −1 m −1 –9 –4 O
P
Q m
= 0.21°
m b. Similarly, when the slit width is W = 1.8 × 10 –6 m and λ = 675 × 10 –9 m , we find L 675 × 10
M 1.8 × 10
N θ = sin −1 ( 1) –9 –6 O
P
Q m
= 22 °
m 1450 INTERFERENCE AND THE WAVE NATURE OF LIGHT 26. REASONING Since the width of the central bright fringe on the screen is defined by the
locations of the dark fringes on either side, we consider Equation 27.4, which specifies the
angle θ that determines the location of a dark fringe. This equation is sin θ = mλ/W, where
λ is the wavelength, W is the slit width, and m = 1, 2, 3, ….
The fact that the width of the central bright fringe does not change means that the positions
of the dark fringes to either side also do not change. In other words, the angle θ remains
constant. According to Equation 27.4, the condition that must be satisfied for this to happen
is that the ratio λ/W of the wavelength to the slit width remains constant.
SOLUTION Since the width of the central bright fringe on the screen remains constant, the
angular position θ of the dark fringes must also remain constant. Thus, according to
Equation 27.4, we have
mλ 1
mλ 2
sin θ =
and sin θ =
W1
W2
14243
14243
Case 1 Case 2 Since θ is the same for each case, it follows that
mλ 1
W1 = mλ 2
W2 The term m can be eliminated algebraically from this result, so that solving for W2 gives
W2 = W1 λ 2 λ1 2
c.3 × 10
= −6 b
h m 740 nm 510 nm g
= 3. 3 × 10 −6 m 27. SSM WWW REASONING The width W of the slit and the wavelength λ of the light
are related to the angle θ defining the location of a dark fringe in the singleslit diffraction
pattern, so we can determine the width from values for the wavelength and the angle. The
wavelength is given. We can obtain the angle from the width given for the central bright
fringe on the screen and the distance between the screen and the slit. To do this, we will use
trigonometry and the fact that the width of the central bright fringe is defined by the first
dark fringe on either side of the central bright fringe.
SOLUTION The angle that defines the location of a dark fringe in the diffraction pattern
can be determined according to sin θ = m λ
W m = 1, 2, 3,... (27.4) Chapter 27 Problems 1451 Recognizing that we need the case for m = 1 (the first dark fringe on either side of the
central bright fringe determines the width of the central bright fringe) and solving for W give W= λ (1) sinθ Referring to Figure 27.22 in the text, we see that the width of the central bright fringe is 2y,
where trigonometry shows that
2 y = 2 L tan θ or 2y −1 0.050 m = 2.4° = tan 2L 2 ( 0.60 m ) θ = tan −1 Substituting this value for θ into Equation (1), we find that
W= λ
sinθ = 510 × 10−9 m
= 1.2 × 10−5 m
sin 2.4° 28. REASONING The wavelength λ of the light and the width W of the single slit together
determine the angle θ of the first dark fringe, via sin θ = m
sin θ = λ W (Equation 27.4), with m = 1: λ
W (1) Both the angle θ and the position y of the dark fringe on the screen are measured from the
middle of the central bright fringe. Therefore, the angle θ may be found from y
θ = tan −1 (Equation 1.6), where L is the distance between the slit and the screen.
L
SOLUTION Solving Equation (1) for λ, we obtain λ = W sin θ
y
Substituting θ = tan −1 (Equation 1.6) into Equation (2) yields L −1 3.5 × 10 −3 m y (
−4
−7 = 4.9 × 10 m = 5.6 × 10 m ) sin tan 4.0 m L λ = W sin tan −1 (2) 1452 INTERFERENCE AND THE WAVE NATURE OF LIGHT 29. REASONING AND SOLUTION The width of the bright fringe that is next to the central
bright fringe is y2 − y1, where y1 is the distance from the center of the central bright fringe to
the firstorder dark fringe, and y2 is the analogous distance for the secondorder dark fringe.
To find the fringe width, we take an approach si...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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