This preview shows page 1. Sign up to view the full content.
Unformatted text preview: hich can then be
used to find the angle that locates the dark fringe for m = 1.
SOLUTION Applying Equation 27.2 to the dark fringes for m = 0 and m = 1, we have ( )λ
d sin θ 0 = 0 + 1
2 and ( )λ
d sin θ1 = 1 + 1
2 Dividing the expression for sin θ1 by the expression for sin θ 0 gives Chapter 27 Problems sin θ1 sin θ 0 1469 (1 + ) λ = 3.0
d
=
(0 + ) λ
d
1
2 1
2 The angle that locates the dark fringe for m = 1 can now be found:
sin θ1 = 3.0 sin θ 0 θ1 = sin −1 ( 3.0 sin θ 0 ) = sin −1 ( 3.0 sin15° ) = 51° or 57. SSM REASONING The slit separation d is given by Equation 27.1 with m = 1; namely
d = λ / sin θ . As shown in Example 1 in the text, the angle θ is given by θ = tan −1 ( y / L ) . SOLUTION The angle θ is θ = tan −1 0
F.037 m I = 0.47 °
G4 .5 m J
HK Therefore, the slit separation d is
d= λ
sin θ = 490 × 10 –9 m
=
sin 0.47 ° 6.0 × 10 –5 m 58. REASONING The drawing shows a top
Third dark
view of the slit and screen, as well as the Slit
fringe
position of the central bright fringe and the
y
third dark fringe. The distance y can be
θ
obtained from the tangent function as y = L
Central bright
fringe
tan θ. Since L is given, we need to find the
L
angle θ before y can be determined.
According to Equation 27.4, the angle θ is related to the wavelength λ of the light and the
width W of the slit by sin θ = mλ / W , where m = 3 since we are interested in the angle for
the third dark fringe.
SOLUTION We will first compute the angle between the central bright fringe and the third
dark fringe using Equation 27.4 (with m = 3):
−9
(
m) mλ −1 3 668 × 10 = 17.3° = sin W 6.73 × 10−6 m θ = sin −1 The vertical distance is y = L tan θ = (1.85 m ) tan17.3° = 0.576 m 1470 INTERFERENCE AND THE WAVE NATURE OF LIGHT 59. REASONING For the hunter’s eye to resolve the squirrels as separate objects, the Rayleigh
criterion must be satisfied. The Rayleigh criterion relates the angle θmin (the angle subtended
at the eye by the objects) to the wavelength λ of the light and the diameter D of the pupil.
Thus, we can calculate the diameter if values are available for the wavelength and the angle.
The wavelength is given. To obtain the angle, we will use the concept of the radian and
express the angle approximately as an arc length (approximately the separation distance s
between the squirrels) divided by a radius (the distance L between the hunter and the
squirrels), as discussed in Section 8.1.
SOLUTION
a. The Rayleigh criterion specifies the angle θmin in radians as θ min ≈ 1.22
Solving for D gives
D≈ λ (27.6) D 1.22λ (1) θ min The drawing at the right shows the hunter’s eye and the
squirrels. According to the discussion in Section 8.1, the
angle θmin can be expressed approximately in radians as θmin ≈ s
L Squirrel 1
Hunter’s
eye L θmin (8.1) s
Squirrel 2 This expression for the angle is approximate because we are assuming that the distance s
between the squirrels is nearly the same as the arc length on a circle of radius L centered on
the hunter’s eye. Substituting Equation 8.1 into Equation (1), we find that
D≈ D≈ ( 1.22λ )( θ min ≈ 1.22λ 1.22λ L
=
s
s
L 1.22 498 × 10−9 m 1.6 × 103 m
0.10 m ) = 9.7 ×10−3 m or 9.7 mm b. A diameter of 9.7 mm is beyond the normal range for the human eye. Moreover, the
human eye would increase the diameter of its pupil to a maximum only under dark
conditions. Hunting in the dark is difficult. The hunter’s claim is not reasonable. Chapter 27 Problems 1471 60. REASONING The drawing shows
th
m bright fringe
Double
a top view of the double slit and the
slit
screen, as well as the position of the
central bright fringe (m = 0) and the
θ
th
Central bright
m bright fringe. The angle θ to the
th
fringe (m = 0)
m
bright fringe is given by
Equation 27.1 as sin θ = mλ / d ,
where λ is the wavelength of the
light and d is the separation between the slits. The largest that the angle can be is θ = 90.0°.
This condition will tell us how many bright fringes can be formed on either side of the
central bright fringe.
SOLUTION Substituting θ = 90.0° into Equation 27.1 and solving for m gives
m= d sin 90.0° λ = ( 3.76 ×10−6 m ) sin 90.0° = 6.02
625 × 10−9 m Therefore, the greatest number of bright fringes is 6 . 61. SSM REASONING The angle θ that locates the firstorder maximum produced by a
grating with 3300 lines/cm is given by Equation 27.7, sin θ = mλ / d , with the order of the
fringes given by m = 0 , 1, 2, 3, . . . Any two of the diffraction patterns will overlap when
their angular positions are the same.
SOLUTION Since the grating has 3300 lines/cm, we have d= 1
= 3.0 × 10 –4 cm = 3.0 × 10 –6 m
3300 lines / cm a. In first order, m = 1; therefore, for violet light, θ = sin –1 –9 –1 –6 Similarly for red light, θ = sin 4
F λ I = sin L1) F10 × 10
m
(
Gd J MG × 10
H K NH
3.0 –1 6
F λ I = sin L1) F60 × 10
m
(
Gd J MG × 10
H K NH
3.0 –9 –1 –6 m
m IO 7.9 °
J=
P
K
Q m
m IO 13°
J=
P
K
Q 1472 INTERFERENCE AND THE WAVE NATURE OF LIGHT b. Repeating the calculation for the second order maximum ( m = 2 ), we find that
( m = 2)
for violet
for red θ = 16 °
θ = 26 ° c. Repea...
View
Full
Document
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

Click to edit the document details