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Unformatted text preview: milar to that outlined in Example 6 in the
text. According to Equation 27.4, the dark fringes for single slit diffraction are located by
sin θ = mλ/W. For firstorder fringes, m = 1, and sin θ 1 = λ / W . For secondorder fringes,
m = 2, and sin θ 2 = 2 λ / W . Therefore, we find that θ 1 = sin –1 480
λ
F I = sin F × 10 m I = 1.4 °
G J G.0 × 10 m J
HK H
W
2
K
O
2
2
Fλ I = sin Lc × 10 mh= 2.8°
G J M2480× 10 m P
H K M .0
W
P
N
Q
–9 –1 –5 –9 θ 2 = sin –1 –1 –5 From Example 6 we know that the distance y is given by y = L tan θ, where L is the distance
between the slit and the screen. Therefore, it follows that c y 2 – y 1 = L tan θ 2 – tan θ 1 0
b
h= b.50 mgtan 2.8° – tan 1.4° g= 0 . 012 m 30. REASONING The distance y between the center of the diffraction pattern and a particular
dark fringe depends upon the distance L between the slit and the screen and the angle θ at
which the dark fringe appears
y = L tan θ (1) For light with a wavelength λ passing through a slit of width W, the angles of the dark
fringes are given by sin θ = m λ (Equation 27.4), where m is an integer (m = 1, 2, 3, …).
W
The screen will only be completely dark when both diffraction patterns have a dark fringe at
the same angle θ . We will use Equation 27.4 to determine the integers (m1 and m2) for
which the two wavelengths (λ1 = 632 nm and λ2 = 474 nm) have dark fringes at the same
angle. Then, Equation (1) will determine the distance y. SOLUTION For an angle θ at which both diffraction patterns have dark fringes, Equation
27.4 gives sin θ = m1 λ1 W and sin θ = m2 Setting the right sides of Equations (2) equal, we see that λ2 W (2) Chapter 27 Problems m1 λ1
W = m2 λ2 m1 λ2 474 nm
==
= 0.75
m2 λ1 632 nm or W 1453 (3) The first dark fringes of the two diffraction patterns do not coincide, because setting
m1 = m2 = 1 yields a ratio of m1/m2 = 1/1 = 1, which does not satisfy Equation (3). But we
can see that other dark fringes do coincide, because Equation (3) is satisfied when m1 = 3
and m2 = 4 (m1/m2 = 3/4 = 0.75), or when m1 = 6 and m2 = 8 (m1/m2 = 6/8 = 0.75), and so
forth. The first time the dark fringes overlap occurs when m1 = 3 and m2 = 4. Solving the
first of Equations (2) for θ , and taking m1 = 3 yields θ = sin −1 m1 632 ×10−9 m o
= sin −1 3 7.15 ×10−5 m = 1.52 W λ1 Then, from Equation (1), we have that y = L tan θ = (1.20 m ) tan1.52o = 3.18 ×10−2 m = 3.18 cm 31. SSM REASONING The width of the
central bright fringe is defined by the
location of the first dark fringe on
either side of it. According to
Equation 27.4, the angle θ locating the
first dark fringe can be obtained from
sin θ = λ/W, where λ is the wavelength
of the light and W is the width of the
slit. According to the drawing,
tan θ = y/L, where y is half the width of
the central bright fringe and L is the
distance between the slit and the screen. First dark fringe
y θ W Midpoint of central
bright fringe
L SOLUTION Since the angle θ is small, we can use the fact that sin θ ≈ tan θ. Since
sin θ = λ/W and tan θ = y/L, we have λ
W = y
L or W= λL
y 1454 INTERFERENCE AND THE WAVE NATURE OF LIGHT Applying this result to both slits gives
W2
W1
W2 = W1 y1
y2 c = = 3.2 × 10 −5 λ L / y 2 y1
=
λ L / y1 y 2
m 1 cm g
hb.2 cmg
1.9
b=
1
2
1
2 2 .0 × 10 −5 m 32. REASONING AND SOLUTION It is given that 2y = 450W and L = 18 000W. We know
λ/W = sin θ. Now sin θ ≈ tan θ = y/L, so λ
W = 225 W
y
=
=
L
18 000 W 0.013 33. REASONING AND SOLUTION The minimum angular separation of the cars must be
θmin = 1.22 λ/D, and the separation of the cars is y = Lθmin = 1.22 Lλ /D.
a. For red light, λ = 665 nm, and
y = (1.22)(8690 m)(665 × 10 m)/(2.00 × 10 m) = 3.53 m
–9 –3 b. For violet light, λ = 405 nm, and
y = (1.22)(8690 m)(405 × 10–9 m)/(2.00 × 10–3 m) = 2 .15 m 34. REASONING The drawing shows the two stars, separated by a distance s,
that are a distance r from the earth. The angle θ (in radians) is defined as
the arc length divided by the radius r (see Equation 8.1). Assuming that
r >> s, the arc length is very nearly equal to the distance s, so we can write
θ = s/r. Since s is known, this relation will allow us to find the distance r if
we can determine the angle. If the stars can just be seen as separate objects,
we know that the minimum angle θmin between them is given by Equation 27.6 as θ min ≈ 1.22λ / D , where λ is the wavelength of the light being
observed and D is the diameter of the telescope’s objective. By combining
these two relations, we will be able to find the distance r. s r θ Earth Chapter 27 Problems SOLUTION Solving the relation θ
θ = θmin ≈ 1.22λ / D , we obtain r= s θ s = 1.22 λ
D = 1455 = s/r for r, and substituting in the relation 3.7 ×1011 m
= 5.6 × 1017 m 550 ×10−9 m 1.22 1.02 m 35. SSM REASONING According to Rayleigh's criterion, the two taillights must be
separated by a distance s sufficient to subtend an angle θ min ≈ 1.22 λ / D at the pupil of the
observer's eye. Recalling that this angle must be expressed in ra...
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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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