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Unformatted text preview: ance y is y = L sin θ . But sin θ = mλ/d, according to Equation 27.7, so
that our expression for y becomes y = L sin θ = Lm λ
d The separation between adjacent principal maxima is
∆y = b g − Lmλ = Lλ L m +1 λ
d d d The number of lines per meter N is the reciprocal of the slit separation d, so N = 1/d.
Substituting this result into the equation above gives ∆ y = Lλ N . Applying this expression
to gratings A and B, we have
∆y A = LλN A ∆ y B = L λN B and Forming the ratio of these two expressions, we have
∆y B
∆y A
NB = N A ∆y B
∆y A = LλN B
LλN A = NB
NA 2000
b=
c m h3.2 cmg
=
−1 2.7 cm 2400 m −1 47. REASONING For a diffraction grating the angle θ that locates a bright fringe can be found
using Equation 27.7, sin θ = mλ/d, where λ is the wavelength, d is the separation between
the grating slits, and the order m is m = 0, 1, 2, 3, …. By applying this relation to both cases,
we will be able to determine the unknown wavelength, because the order and the slit
separation are the same for both.
SOLUTION We will use λ1 and λ2 to denote the known and unknown wavelengths,
respectively. The corresponding angles are θ1 and θ2. Applying Equation 27.7 to the two
cases, we have
mλ 1
mλ 2
sin θ 1 =
and sin θ 2 =
d
d Dividing the expression for case 2 by the expression for case 1 gives
sin θ 2
sin θ 1 = mλ 2 / d
mλ 1 / d = λ2
λ1 Chapter 27 Problems Solving for λ2, we find λ 2 = λ1 sin θ 2
sin θ 1 b = 420 nm sin 41
g 26°° =
sin 1463 630 nm 48. REASONING The angle that specifies the thirdorder maximum of a diffraction grating is
sin θ = mλ/d (Equation 27.7), where m = 3, λ is the wavelength of the light, and d is the
separation between the slits of the grating. The separation is equal to the width of the grating
(1.50 cm) divided by the number of lines (2400).
SOLUTION Solving Equation 27.7 for the wavelength, we obtain 1.50 ×10−2 m sin18.0°
d sin θ 2400 =
= 6.44 ×10−7 m = 644 nm
λ=
m
3 th 49. REASONING The angle that specifies the m principal maximum of a diffraction grating
is sin θ = mλ/d (Equation 27.7), where λ is the wavelength of the light, and d is the
separation between the slits of the grating. The angle θ is known, and the separation d (in
cm) between the slits is equal to the reciprocal of the number of lines per centimeter
(2604 cm−1 ).
SOLUTION Solving Equation 27.7 for the wavelength, we obtain
1 sin 30.0° 1.92 ×10−4 cm 1920 nm
d sin θ 2604 cm −1 λ=
=
=
=
m
m
m
m For m = 1, λ = 1920 nm; for m = 2, λ = 960 nm; For m = 3, λ = 640 nm; For m = 4,
λ = 480 nm; for m = 5, λ = 384 nm. For m > 5, the values for the wavelength are smaller
than 384 nm. Thus, the two wavelengths that fall within the range of 410 to 660 nm are
640 nm and 480 nm . 50. REASONING
Diffraction
grating
a. From the following drawing we see
that tan θ = y/L, so that y = L tan θ .
The angle between the central bright
θ
fringe and the 2nd order bright fringe is
L
given by sinθ = mλ / d (Equation
27.7), where m = 2 in this case, λ is the
wavelength of the light, and d is the separation between the slits. 2nd order
bright fringe Screen
y Central bright
fringe (m = 0) 1464 INTERFERENCE AND THE WAVE NATURE OF LIGHT Since y = L tan θ and tan θ ≈ sin θ , we have that y = L sin θ. But sinθ = mλ / d , so
y = Lmλ/d.
b. The wavelength λwater of light in water is related to the wavelength λvacuum in vacuum by λwater = λvacuum/nwater (Equation 27.3), where nwater is the index of refraction of water. Since
nwater = 1.33, the wavelength in water is less than that in a vacuum (or in air). Therefore, the
distance y, which is proportional to the wavelength (see part a above), is smaller when the
apparatus is submerged in water.
SOLUTION
a. From the REASONING, the distance from the central bright fringe to the 2nd order bright
fringe is
Lmλ ( 0.15 m ) ( 2 ) ( 410 × 10−9 m )
y=
=
= 0.010 m
d
1.2 × 10−5 m b. When the apparatus is immersed in water, the wavelength of the light becomes smaller: λwater = λvacuum
nwater = 410 × 10−9 m
= 308 × 10−9 m
1.33 In this case, the distance y becomes ywater Lmλwater ( 0.15 m ) ( 2 ) ( 308 × 10−9 m )
=
=
= 0.0077 m
d
1.2 × 10−5 m 51. SSM REASONING For a diffraction grating, the angular position θ of a principal
maximum on the screen is given by Equation 27.7 as sin θ = mλ / d with m = 0 , 1, 2, 3, . . .
SOLUTION When the fourthorder principal maximum of light A exactly overlaps the
thirdorder principal maximum of light B, we have sin θ A = sin θ B
4λ A
d = 3λ B
d or λA
=
λB 3
4 Chapter 27 Problems 1465 52. REASONING The angle θ of a principal maximum formed by light passing through a
diffraction grating with slit separation d is given by sin θ = m λ (Equation 27.7), where m is
d
an integer (m = 0, 1, 2, 3, …) and λ is the wavelength of the light. We seek the change ∆θ in
the angle of the seventhorder principal maximum, so we will take m = 7 in Equation 27.7.
As the diffraction gr...
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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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