HMWK 5

HMWK 5
Ch 17: P 6, 11, 30, 31, 34, 42, 50, 56, 58, 60
Ch 18: P 7, 16, 22, 27, 28, 30, 51, 52, 59, 61
Ch. 17
P17.6. Prepare: The laser beam is an electromagnetic wave that travels with the speed of light. We will first
find the speed of light in the unknown liquid and then use Equation 17.1 to find its index of refraction. Finally,
we will use Equation 17.2 to find the wavelength of light in the unknown liquid.
Solve: The speed of light in the liquid is
−2
3 0 × 10 m
8
−9 = 2.174 ×1 0 m/s
1 .38 ×10 s
vliquid =
The liquid’s index of refraction is
n=
c
=
vliquid
3 .0 × 108
8 = 1.38
2.174 × 10
to be reported as 1.4 to two significant figures.
Thus the wavelength of the laser beam in the liquid is
λ liquid =
λvac
n
=
633 nm
= 460 nm
1.38
Assess: The wavelength falls in the visible region.
P17.11. Prepare: Two closely spaced slits produce a double-slit interference pattern. The interference
pattern looks like the photograph of Figure 17.9. Note that there are 11 gaps between a span of 12 fringes or a
distance of 52 mm.
Solve:
The formula for the fringe spacing is
∆y =
λL
52 × 10 m
d
11
−3
−9
=
(633 × 10 m )(3.0 m )
d
d = 0.40 mm
Assess: This is a reasonable distance between the slits, ensuring d/L = 1.34 × 10−4 << 1.
P17.30. Prepare: A narrow slit produces a single-slit diffraction pattern. The intensity pattern for single-slit
diffraction will look like Figure 17.27 with minima given by Equation 17.24. The width of the central maximum
is given by Equation 17.25.
Solve: The width of the central maximum for a slit of width a = 200λ is
−9
w=
2λ L 2(500 × 10 m)(2.0 m)
=
= 0.0040 m = 4.0 mm
a
0.0005 m
P17.31. Prepare: A narrow slit produces a single-slit diffraction pattern. The intensity pattern for single-slit
diffraction will look like Figure 17.27. Angle θ = 0.70° = 0.0122 rad is a small angle (<< 1 rad). Thus we use
Equation 17.23 to find the wavelength of light.
Solve:
The angles of the minima of intensity are
θp = p
λ=
a θp
p
λ
a
p = 1, 2 , 3, . . .
−3
=
( 0.10 × 10 m)(0.0122 rad)
= 610 nm
2
Assess: The wavelength is in the visible region, so its value is reasonable.
P17.34. Prepare: Light passing through a circular aperture leads to a diffraction pattern that has a circular
central maximum surrounded by a series of secondary bright fringes. The intensity pattern will look like Figure
17.31. The diameter D can be found from Equation 17.27.
Solve: From Equation 17.27, the diameter of the circular aperture is
−9
D=
2.44λ L 2.44(633× 10 m)(4.0 m)
=
= 0.25 mm
−2
w
2.5 ×10 m
Assess: The diameter obtained above is typical for observing diffraction.
P17.42. Prepare: For a double slit, the location of the mth interference maxima is determined
by ym = mλ L /d. The wavelength of light in a medium with an index of refraction of n is related to the
wavelength in air by λ n = λ air/n.
Solve: (a) The spacing between the second and third bright fringes under water is
−7
−3
(∆y)water = (y3 )water − (y2 ) water = (3− 2) λwaterL /d = (4.5× 10 m)(0.9 m)/0.34 × 10 m = 1.35 mm
(b) When the water is drained, the wavelength will increase by a factor of n (λ air = nλ water ). When this expression
is inserted into the expression for the location of the interference fringe maxima we have (ym ) air = mλ air L/d =
m(nλ water )L/d.
Using this expression the spacing between the second and third bright fringes in air is
(∆y)air = nλw ater L/d = n(∆y)water = (1.33)(1.35 mm) = 1.80 mm
Assess: Once you get into the lab and use a single slit, you will see that this is a
reasonable fringe spacing.
P17.50. Prepare: A diffraction grating produces an interference pattern that is determined by both the slit
spacing and the wavelength used. The angle of diffraction is determined by the constructive-interference
condition d sinθ m = mλ , where m = 0, 1, 2, 3, . . . If blue light (the shortest wavelengths) is diffracted at angle θ,
then red light (the longest wavelengths) is diffracted at angle θ + 30°.
Solve: (a) In the first order, the equations for the blue and red wavelengths are
sinθ =
λb
d sin(θ + 30°) = λ r
d
Combining the two equations we get for the red wavelength,
λ r = d (sin θ c os30° + c os θ sin 30°) = d( 0.86 60sin θ + 0.50 cos θ ) = d
(0.50)d 1 −
d=
λ2b
d
2
= λr − 0.8660 λb
λ r − 0.866 0 λb
2
2
λb
d
0.8660 + d(0.50)
2
(0.50) ( d − λb ) = ( λr − 0.8660 λ b )
1−
λ2b
d2
2
2
0.50
−9
+ λ 2b
−9
−7
Using λ b = 400 × 10 m and λ r = 700 × 10 m, we get d = 8.125 × 10 m. This value of d corresponds to
1 mm
1.0 × 10−3 m
=
= 1230 lines/mm
d
8.125 × 10−7 m
−9
(b) Using the value of d from part (a) and λ = 58 9 × 10 m, we can calculate the angle of diffraction as follows:
d sinθ1 = (1) λ
−7
−9
(8.125 × 10 m)sinθ1 = 589 ×10 m
θ1 = 4 6.5°
Assess: 1230 lines/mm for the diffraction grating is reasonable.
P17.56. Prepare: The relationship between the diffraction grating spacing (d ), the angle at which a
particular order of constructive interference occurs (θm ), the wavelength of the light, and the order of the
constructive interference (m) is d sinθ m = mλ and N = 1/d.
Solve: The first order diffraction angle for green light is
θ1 = sin −1 (λ /d ) = sin−1(5.5 × 10−7 m/ 2.0 × 10−6 m) = sin−1 (0.275) = 0.278 radian = 16°
Assess: This is a reasonable angle for the first order maximum.
P17.58. Prepare: A diffraction grating produces an interference pattern like the diagram of Figure 17.12.
The angle of diffraction is determined by the constructive-interference condition, Equation
17.12, d sinθ m = mλ , where m = 0, 1, 2, 3, . . . Furthermore, according to Equation 17.13, the distance ym from the
center to the mth maximum is ym = L tan θm . A key statement is that the lines are seen on the screen. This means
that the light is visible light, in the range 400 nm–700 nm. We can determine where the entire visible spectrum
falls on the screen for different values of m. We do this by finding the angles θm at which 400 nm light and 700
nm light are diffracted. We then use ym = L tanθm to find their positions on the screen, which is at a distance L =
75 cm.
−7
Solve: (a) The slit spacing is d = 1 mm/120 0 = 8.333 ×10 m.
For m = 1,
λ = 400 nm: θ1 = sin−1(400 nm/d ) = 28.7°
y1 = 7 5 cm tan 28.7° = 41 cm
λ = 700 nm: θ1 = sin−1 (700 nm/d) = 57 .1°
y1 = 75 cm tan 57.1° = 116 cm
For m = 2,
λ = 400 nm: θ1 = sin−1(2 ⋅ 400 nm/d ) = 73.8°
−1
y2 = 75 cm tan 73.8° = 257 cm
−1
For the 700 nm wavelength at m = 2, θ2 = sin (2 ⋅ 700 nm/d ) = sin (1.68) is not defined, so y2 → ∞. We see that
visible light diffracted at m = 1 will fall in the range 41 cm ≤ y ≤ 116 and that visible light diffracted at m = 2 will
fall in the range y ≥ 257 cm. These ranges do not overlap, so we can conclude with certainty that the observed
diffraction lines are all m = 1.
(b) To determine the wavelengths, we first find the diffraction angle from the observed position by using
θ = tan
−1
y
L
= tan
−1
y
75 cm
This angle is then used in the diffraction grating equation for the wavelength with m = 1, λ = d sin θ1.
Line
56.2 cm
65.9 cm
93.5 cm
θ
36.85
41.30
51.27
λ
500 nm
550 nm
650 nm
P17.60. Prepare: Reflection is maximized for constructive interference of the two reflected waves, but
minimized for destructive interference. The first reflection occurs at the front surface of the oil film. Here, the
index of refraction increases from that of air (n = 1) to that of the film (n = 1.42), so there will be a reflective
phase change. The light then reflects from the rear surface of the film. The index of refraction again increases
from 1.42 to 1.50 for the glass, and there is a reflective phase change. With two phase changes, Tactics Box 17.1
tells us that we should use Equation 17.15 for constructive interference and Equation 17.16 for destructive
interference.
Solve: (a) Constructive interference of the reflected waves occurs for wavelengths given by Equation 17.15:
λm =
2nt 2(1.42)(500 nm) (1420 nm)
=
=
m
m
m
Thus, λ1 = 14 20 nm, λ 2 = 12 (142 0 nm) = 710 nm, λ 3 = 473 nm, λ 4 = 355 nm, . . . Only the wavelength of 473
nm is in the visible range.(b) For destructive interference of the reflected waves,
λ=
2nt
2(1 .42)(500 nm) 1420 nm
=
1 =
m+ 2
m + 12
m+ 12
Thus, λ 0 = 2 × 1420 nm = 2840 nm, λ1 =
2
3
(1420 nm) = 947 nm), λ 2 = 568 nm, λ 3 = 406 nm, . . . The
wavelengths of 406 nm and 568 nm are in the visible range.
(c) Beyond the limits 430 nm and 690 nm the eye’s sensitivity drops to about 1 percent of
its maximum value. The reflected light is enhanced in blue (473 nm). The transmitted light
at mostly 568 nm will be yellowish green.
Ch. 18
P18.7. Prepare: We’ll use the law of reflection and see where rays of light coming from object O and
reflecting in the mirror can go.
Solve:
The figure shows rays coming from O and hitting each end of the mirror. It is clear that object O can be seen in
the mirror from any location between the outer rays, which includes locations B and C. However, at location A
object O can’t be seen in the mirror because there is no way for a ray to obey the law of reflection from O and get
to A. If the mirror extended further to the left, then A could see O. A similar argument can be made that object O
can’t be seen in the mirror from position D.
In summary, the object’s image is visible from B and C.
Assess: Your intuition in looking at the original diagram confirms what the ray tracing
tells us.
P18.16. Prepare: Use the ray model of light. The figure incorporates the first four steps of Tactics Box 18.1.
Solve: Using Snell’s law at the water-diamond boundary,
ndia sin θdia = nwater sinθ water
Assess:
θdia .
sin θwater =
ndia
2.41
sin θdia =
sin3 0°= 0.9060
nwater
1.33
θwater = 65°
Because ndia is much larger than nwater, we expectedθwater to be much larger than
P18.22. Prepare: We know that violet light is refracted more than red (this is reflected in the given values
for n), so we draw a diagram with the violet ray in the glass closer to the normal then the red ray. But they
emerge together at the same point and in the same direction.
We are given nred = 1.5 4, nviolet = 1.5 9, and θair = 22.5°. We also know nair = 1.00.
The strategy will be to solve Snell’s law for θviolet and θred in the glass and then find ∆θ as the difference of the
two angles.
nair sin θair = nviolet sinθ violet = nred sinθ red
Solve: Solve Snell’s law for θviolet and θred in the glass.
θ v i ol et = sin −1
θ red = sin −1
n ai r sin θ ai r
nv i o le t
n ai r sin θa ir
n re d
= sin −1
1.00 sin 22.5°
= 13.93°
1 .59
= sin −1
1 .00 sin 22.5°
= 1 4.39°
1.54
The difference between these angles (i.e., the angle between the two rays in the glass) is ∆θ = 14.39° −
13.93° = 0.462°.
Assess: The angle between the two rays is very small, but that is to be expected since
the index of refraction isn’t very different for the two rays.
P18.27. Prepare: Use ray tracing to locate the image. The figure below shows the ray-tracing diagram using
the steps of Tactics Box 18.2.
Solve:
You can see from the diagram that the image is in the plane where the three special rays converge. The image is
located at s′ = 15 cm to the right of the converging lens, and is inverted and real.
Assess: Ray tracing must be done to scale to obtain useful answers.
P18.28. Prepare: Use ray tracing to locate the image. The figure shows the ray-tracing diagram using the
steps of Tactics Box 18.2.
Solve:
You can see that the rays after refraction do not converge at a point on the refraction side of the lens. On the
other hand, the three special rays, when extrapolated backward toward the incidence side of the lens, meet at P′
which is 15 cm from the lens. That is, s′ = −15 cm. The image is upright and virtual.
Assess: Ray tracing must be done to scale to obtain useful answers.
P18.30. Prepare: Use ray-tracing to locate the image. The figure shows the ray tracing diagram using the
steps of Tactics Box 18.2.
Solve:
The three rays after refraction do not converge at a point, but they appear to come from P′. P′ is 6 cm from the
diverging lens, so s′ = −6 cm. The image is upright and virtual.
Assess: Draw ray tracings accurately to scale.
P18.51. Prepare: This is both a refraction (Snell’s law) problem and a critical angle for total internal reflection problem. Label the normal lines on your sketch carefully.
Solve: We first compute the critical angle inside the fiber optic core using Equation 18.3. The critical angle is
measured from the normal, which is vertical in this drawing.
θc = sin−1
n2
−1 1.45
= sin
= 75°
n1
1.50
Now look at Figure P18.51 carefully. The 75° from the normal on the side with the cladding means the angle for
refraction from the air-glass interface (measured from the new normal at the left) is 90° − 75° = 15°.
θ = θ a = sin−1
ng sin θg
−1 1.50 sin 15°
= sin
= 2 3°
na
1.00
Assess: The numbers all seem reasonable, but the diagram is clearly not drawn to scale
as θ is depicted as much greater than 23°, but that’s OK.
P18.52. Prepare: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of
light. The laser beam enters the water 2.0 m from the edge, undergoes refraction, and illuminates the goggles.
The ray of light from the goggles then retraces its path and enters your eyes. We will use Snell’s law at the airwater boundary and the geometry of the following diagram.
Solve:
From the geometry of the diagram,
1.0 m
tan φ =
2.0 m
φ = tan−1 (0.50) = 26.57°
θair = 90°− 26.57° = 63.43°
Snell’s law at the air-water boundary is nair sin θair = nwater sinθwater. Using the above result,
(1.0)sin 63.43° = 1.33 sinθ water
θwater = sin−1
sin 63.43°
= 42.26°
1.33
Taking advantage of the geometry in the diagram again,
x
= tan θwater
3.0 m
x = (3.0 m) tan 42.26° = 2.73 m
The distance of the goggles from the edge of the pool is 2.73 m + 2.0 m = 4.73 m or 4.7 to two significant
figures.
Assess: The given distances and dimension measurements for the pool certainly
indicate that the result obtained above is reasonable.
P18.59. Prepare: We will use the ray model of light and apply Snell’s law. Please refer to Figure P18.59.
Solve: (a) Using Snell’s law at the air-glass boundary, with φ being the angle of refraction inside the prism,
nair sin β = n sinφ
sin β = n sin φ
Considering the triangle made by the apex angle and the refracted ray,
(90°− φ ) + (90° − φ) + α = 180°
1
2
φ= α
Thus
sin β = n sin
(b) Using the above expression, we obtain
n=
sin β
sin 52.2°
=
= 1.58
1
sin ( 2 α )
sin 30°
1
α
2
β = sin−1 n sin
1
α
2
P18.61. Prepare: Represent the aquarium’s wall as a point source, and use the ray model of light. Paraxial
rays from the outer edge (O) are refracted into the water and then enter into the fish’s eye. When extended into
the wall, these rays will appear to be coming from O′ rather from O. The point on the inside edge (I) of the wall
will not change its apparent location. We are given that sO − sI = 4.00 mm and sO′ − s′I = 3.50 mm.
Solve: Using Equation 18.7,
sO′ =
sO′ − s′I =
nwater
(s − s )
nwall O I
nwater
s
nwall O
3.50 mm =
sI′ =
nwater
s
nwall I
1.33
(4.0 0 mm)
nwall
n wa ll = (1.33)
4.00 mm
3.5 0 mm
Assess: The value obtained above is closer to that of glass, so it is reasonable.
= 1.52