Unformatted text preview: dians, we relate θ min to the
distances s and L. θmin
Taillight
Observer s
Taillight L SOLUTION The wavelength λ is 660 nm. Therefore, we have from Equation 27.6 θ min ≈ 1.22 λ
D = 1.22 F × 10
660
G × 10
H
7.0 –9 –3 m
m I = 1.2 × 10
J
K −4 rad According to Equation 8.1, the distance L between the observer and the taillights is
L= s θ min = 1.2 m
= 1.0 × 10 4 m
1.2 × 10 –4 rad 36. REASONING The maximum allowable dot separation should be chosen so that neither the
red, the green, nor the blue dots can be resolved separately. The Rayleigh criterion is
θmin ≈ 1.22 λ/D (Equation 27.6), where λ is the wavelength of the light and D is the
diameter of the pupil in this case. In the Rayleigh criterion the angle θmin is expressed in
radians. According to Equation 8.1, the angle in radians is θ min ≈ s / L, where s is the dot
separation and L is the distance from the dots to the eye (the viewing distance). The
maximum allowable dot separation should be chosen so that neither the red, the green, nor
the blue dots can be resolved separately. This means that the distance should be determined 1456 INTERFERENCE AND THE WAVE NATURE OF LIGHT by the blue dots. A dot separation that is slightly smaller than sblue will automatically be
smaller than sred and sgreen. This is because the min values for red and green light are
larger than for blue light. In other words, if the blue dots can’t be resolved separately, then
neither can the red dots nor the green dots.
SOLUTION Using the Rayleigh criterion and Equation 8.1, we have s
λ
≈ 1.22
L
D
{ s≈ or 1.22 λ L
D θ min The maximum allowable dot separation, then, is s blue b
h c g −9
1.22 λ L 1.22 470 × 10 m 0.40 m
≈
=
= 1.1 × 10 −4 m
D
2 .0 × 10 −3 m 37. REASONING The diameter D of the spectator’s pupils and the wavelength λ of the light
being viewed determine the minimum angular separation θmin (in radians) at which adjacent
cards can be separately resolved, according to θmin ≈ 1.22 λ (Equation 27.6). If the
D
spectator’s pupils are narrower than the diameter D given by Equation 27.6, the cards will
not appear separated, and their colors will blur together. Thus D is the upper limit on the
diameter of the spectator’s pupils. The angular separation (in radians) of the cards is found
s
from Equation 8.1: θmin ≈ , where s is the distance between adjacent cards, and L is the
L
distance between the cards and the spectator.
SOLUTION Setting the right sides of θmin ≈ 1.22 (Equation 8.1) equal, we obtain θmin ≈ λ
D (Equation 27.6) and θmin ≈ s
λ
≈ 1.22
L
D Solving Equation (1) for D yields
D ≈ 1.22 λL
s ( 480 ×10−9 m) (160 m) = 1.9 ×10−3 m = 1.9 mm
= 1.22
5.0 ×10−2 m s
L (1) Chapter 27 Problems 1457 38. REASONING AND SOLUTION
a. Equation 27.6 θ min = 1.22 λ / D gives the minimum angle θ min that two point objects
can subtend at an aperture of diameter D and still be resolved. The angle must be measured
in radians. For a dot separation s and a distance L between the painting and the eye,
Equation 8.1 gives the angle in radians as θ min = s / L . Therefore, we find that c h 1.22 λ
s
=
D
L L= or sD
1.22 λ With λ = 550 nm and a pupil diameter of D = 2.5 mm, the distance L is
L= c h
c h 1.5 × 10 –3 m 2 .5 × 10 –3 m
sD
=
= 5.6 m
1.22 λ
1.22 550 × 10 −9 m bg
c h b. The calculation here is similar to that in part a, except that n = 1.00 and D = 25 mm for
the camera. Therefore, the distance L for the camera is c h
c h 1.5 × 10 –3 m 25 × 10 –3 m
sD
L=
=
= 56 m
1.22 λ
1.22 550 × 10 –9 m bg
c h 39. REASONING At the aperture the star and its planet must subtend an angle at least as large
as that given by the Rayleigh criterion, which is θmin ≈ 1.22 λ/D, where λ is the wavelength
of the light and D is the diameter of the aperture. The angle θmin is given by this criterion in
radians. We can obtain the angle subtended at the telescope aperture by using the separation
s between the planet and the star and the distance L of the star from the earth. According to
Equation 8.1, the angle in radians is θmin = s/L.
SOLUTION Using the Rayleigh criterion and Equation 8.1, we have
s
λ
≈ 1.22
L
D or D≈ c h
c h −9
17
1.22 λ L 1.22 550 × 10 m 4 .2 × 10 m
=
= 2 .3 m
s
1.2 × 10 11 m 40. REASONING The beam diffracts
outward when it leaves the spotlight,
so that the circular spot on the moon
has a diameter d that is greater than
the diameter D of the spotlight. The
diffraction angle θ , measured from
the middle of the beam to the first θmin
θ θspot Earth Circular d spot on
the moon r 1458 INTERFERENCE AND THE WAVE NATURE OF LIGHT circular dark fringe, which is the edge of the central bright spot on the moon, is expected to
be small. Therefore, the diameter d of the central bright spot on the moon is approximately
equal to the arc length s = rθspot (Equation 8.1), where r is the distance between the Earth
and the moon and θspot is the angle subtended by the central bright spot (see the drawing): d = rθspot (1) Note that θ is measu...
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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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