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Unformatted text preview: 7.7, sin θ = mλ / d , the
angle for the m = 1 bright fringe of red light is greater than that for yellow light, since red
has a longer wavelength. 19. (b) As the number of lines per centimeter increases, the separation d between adjacent slits
becomes smaller. The angle θ that specifies a principal maxima of a diffraction grating is
given by sin θ = mλ / d , where λ is the wavelength and m = 0, 1, 2, 3, … (Equation 27.7).
For fixed values of m and the wavelength, this relation shows that as d decreases, θ
increases. For a fixed screen location, this means that the distance between two adjacent
principal maxima also increases.
20. θ = 9.4º Chapter 27 Problems 1431 CHAPTER 27 INTERFERENCE AND THE WAVE NATURE OF LIGHT
PROBLEMS 1. SSM REASONING AND SOLUTION To decide whether constructive or destructive
interference occurs, we need to determine the wavelength of the wave. For electromagnetic
waves, Equation 16.1 can be written f λ = c , so that c 3.00 × 10 8 m
λ= =
= 5.60 × 10 2 m
f 536 × 10 3 Hz
Since the two wave sources are in phase, constructive interference occurs when the path
difference is an integer number of wavelengths, and destructive interference occurs when
the path difference is an odd number of half wavelengths. We find that the path difference is
8.12 km – 7.00 km = 1.12 km. The number of wavelengths in this path difference is (1.12 × 103 m ) / (5.60 × 102 m ) = 2.00 . Therefore, 2. constructive interference occurs . REASONING In order for the two rays to interfere constructively and thereby form a
bright interference fringe, the difference ∆l between their path lengths must be an integral
multiple m of the wavelength λ of the light: ∆l = mλ (1) In Equation (1), m can take on any integral value (m = 0, 1, 2, 3, …). In this case, the rays
meet at the eightorder bright fringe, so we have that m = 8.
SOLUTION Solving Equation (1) for λ, and substituting m = 8, we obtain λ= ∆l 4.57 × 10−6 m
=
= 5.71× 10−7 m
m
8 Using the equivalence 1 nm = 10−9 m, we convert this result to nanometers: ( ) 101−nmm = 571 nm 9 λ = 5.71× 10−7 m 1432 INTERFERENCE AND THE WAVE NATURE OF LIGHT 3. REASONING Let l1 and l2 be the distances from source 1 and source 2, respectively. For
constructive interference the condition is l2 – l1 = mλ, where λ is the wavelength and m = 0,
1
1, 2, 3, …. For destructive interference the condition is l2 – l1 = (m + 2 )λ, where m = 0, 1,
2, 3, …. The fact that the wavelength is greater than the separation between the sources will
allow us to decide which type of interference we have and to locate the two places in
question. SOLUTION a. Since the places we seek lie on the line between the two sources and since
the separation between the sources is 4.00 m, we know that l2 – l1 cannot have a value
greater than 4.00 m. Therefore, with a wavelength of 5.00 m, the only way constructive
interference can occur is at a place where m = 0. But this would mean that l2 = l1, and there
is only one place where this is true, namely, at the midpoint between the sources. But we
know that there are two places. As a result, we conclude that
destructive interference is occurring .
b. For destructive interference, the possible values for the integer m are m = 0, 1, 2, 3, …,
and we must decide which to use. The condition for destructive interference is
1
l2 – l1 = (m + 2 )λ. Since λ = 5.00 m, all values of m greater than or equal to one result in
l2 – l1 being greater than 4.00 m, which we already know is not possible. Therefore, we
conclude that m = 0, and the condition for destructive interference becomes l2 – l1 = 1
2 λ. We also know that l2 + l1 = 4.00 m, since the separation between the sources is 4.00 m. In
other words, we have
1
l 2 − l1 = 2 λ = 1
2 5
=
b.00 m g 2.50 m and l 2 + l 1 = 4 .00 m Adding these two equations gives
2l2 = 6.50 m or l2 = 3.25 m Since l2 + l1 = 4.00m, it follows that
l1 = 4.00 m – l2 = 4.00 m – 3.25 m = 0.75 m
These results indicate that the two places are 3.25 m and 0.75 m from one of the sources. 4. REASONING The loudspeakers are inphase sources of identical sound waves. The waves
from one speaker travel a distance l1 in reaching point A and the waves from the second
speaker travel a distance l2. The condition that leads to constructive interference is
l2 – l1 = mλ, where λ is the wavelength of the waves and m = 0, 1, 2, 3 …. In other words, Chapter 27 Problems 1433 the two distances are the same or differ by an integer number of wavelengths. Point A is the
midpoint of a side of the square, so that the distances l1 and l2 are the same, and
constructive interference occurs.
As you walk toward the corner, the waves from one of the sources travels a greater distance
in reaching you than does the other wave. This difference in the distances increases until it
reaches one half of a wavelength, at which spot destructive interference occurs and you hear
no sound. As you walk on, the difference in distances continues to increase, and you
gradually hear a louder and louder sound. Ultimately, at the corner, the difference in
distances becomes one wavelength, constructive interference occurs, and you hear a
maximally loud sound.
The general condition that leads to constructive interference is l 2 − l 1 = m λ , where
m = 0, 1, 2, 3 … There are many...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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