Phys 197 Homework Solution 33A
Q7.
Sometimes when looking at a window, you see two reflected images slightly displaced from
each other. What causes this?
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One reflection from each of the front and back surfaces. For double-pane glass, you can get 4 –
and use then to determine the glass thickness and spacing. This shows best in a well-lit room
at night, and holding the object so that the angles of incidence and reflection are about 45◦ . A
good object is an index card with one vertical line.
Q12.
Does it make sense to talk about the polarization of a longitudinal wave, such as a sound
wave? Why or why not?
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No. Polarization is a property of transverse waves.
P2.
Light Inside the Eye. The vitreous humor, a transparent, gelatinous fluid that fills
most of the eyeball, has an index of refraction of 1.34. Visible light ranges in wavelength from 380 nm
(violet) to 750 nm (red), as measured in air. This light travels through the vitreous humor and strikes
the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the
frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?
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(a) Wavelength is directly proportional to speed (v = f λ), so inversely proportional to n
(n = c/v):
λred = (750 nm)/1.34 = 560 nm λviolet = 284 nm
(b) Frequencies are the same as in air :
fred = (3.00 × 108 m/s)/(750 × 10−9 m) = 4.00 × 1014 Hz fviolet = 7.89 × 1014 Hz.
(c) Speeds are inversely proportional to n: v = (3 × 108 m/s)/1.34 = 2.24 × 108 mps for all
colors.
P5.
A light beam travels at in quartz. The wavelength of the light in quartz is 355 nm. (a)
What is the index of refraction of quartz at this wavelength? (b) If this same light travels through
air, what is its wavelength there?
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Referring to Fig 33.19, we see that it plots index of refraction as a function of wavelength in
vacuum. So we will need to guess and iterate. We note that the n for quartz is around 1.55 for
visible wavelengths, so start with:
λair ≈ (355 nm)(1.55) = 550.3 nm.
Rechecking the figure, n(550.3 nm) = 1.545, so light of vacuum wavelength 550.3 nm would
have a wavelength in quartz of 550.3/1.545 = 356.1 nm, so we are close. The index doesn’t
change significantly over a few nanometers, so:
(a) n = 1.545; and re-do the estimate with this value for n.
(b) λair = (355 nm)(1.545) = 548.5 nm.
P8.
A laser beam shines along the surface of a block of transparent material (see Fig. E33.8.).
(Figure shows the length to be 2.5 m). Half of the beam goes straight to a detector, while the other
half travels through the block and then hits the detector. The time delay between the arrival of the
two light beams at the detector is 6.25 ns. What is the index of refraction of this material?
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Let t1 be the time through vacuum (or air in this context) and t2 the time through the material.
Use L for the length of 2.5 m. Then:
t1 = L/c and t2 = L/(c/n) = nL/c.
Subtract the first from the second to get the given ∆t on the LHS:
t2 − t1 = (n − 1)L/c. Solve for n and plug:
n = 1 + c(t2 − t1 )/L = 1 + (3 × 108 m/s)(6.25 × 10−9 s)/(2.5 m) = 1.75.
P11.
As shown in Fig. E33.11, a layer of water covers a slab of material X in a beaker. A ray
of light traveling upward follows the path indicated. Using the information on the figure, find (a) the
index of refraction of material X and (b) the angle the light makes with the normal in the air.
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Recognize a Snell’s Law problem; note that the 65◦ angle is NOT the angle of incidence, so is
not the number to plug.
(a) nX sin 25◦ = nW sin 48◦ ⇒ nX = (1.333)(0.7431)/0.4226 = 2.34.
(b) nair sin φ = 1.333 sin 48◦ ⇒ sin φ = (1.333)(0.7431)/1 = 0.9906.
φ = 82.1◦ .
P20.
At the very end of Wagner’s series of operas Ring of the Nibelung, Brünnhilde takes the
golden ring from the finger of the dead Siegfried and throws it into the Rhine, where it sinks to the
bottom of the river. Assuming that the ring is small enough compared to the depth of the river to be
treated as a point and that the Rhine is 10.0 m deep where the ring goes in, what is the area of the
largest circle at the surface of the water over which light from the ring could escape from the water?
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Sketch a ray leaving the ring at an angle θcrit from vertical. Calculate θcrit :
θcrit = Sin−1 (1/n) = 48.61◦ .
Calculate b, the horizontal distance from the axis, with trig:
b = (10 m)(tan 48.61◦ ) = 11.35 m. This is the radius of the circle in question.
A = πb2 = 404.4 m2 .
P25.
A narrow beam of white light strikes one face of a slab of silicate flint glass. The light is
traveling parallel to the two adjoining faces, as shown in Fig. E33.25. For the transmitted light inside
the glass, through what angle is the portion of the visible spectrum between 400 nm and 700 nm
dispersed? (Consult the graph in Fig. 33.18.)
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The incoming ray makes an angle of 55◦ with the face of the glass, so its angle with the normal,
θa , is the complement of this: θa = 35.0◦ .
Snell’s Law: nb sin θb = na sin θa = θa ⇒
sin θb = sin(35◦ )/n. Plugging n = {1.66, 1.61},
sin θb = {0.3455, 0.3563} ⇒ θb = {20.21◦ , 20.87◦ }.
∆θ = 0.66◦ . Back-of-book says 0.6◦ , which is close enough, given the need to read the values
of n off the graph.