Unformatted text preview: uals the critical angle, we have from Equation 26.4 that
sin α = sin θ c = n
nq (2) According to the geometry in the drawing above, α = 90° – θ2. As a result, Equation (2)
becomes sin ( 90° – θ 2 ) = cos θ 2 = n
nq (3) Squaring Equation (3), using the fact that sin2θ2 + cos2θ2 = 1, and substituting from
Equation (1), we obtain cos 2 θ 2 = 1– sin 2 θ 2 = 1– n2
2
nq sin 2 θ1 = n2
2
nq (4) Solving Equation (4) for n and using the value given in Table 26.1 for the refractive index
of crystalline quartz, we find
n= nq
1+ sin θ1
2 = 1.544
1+ sin 2 34° = 1.35 ______________________________________________________________________________ Chapter 26 Problems 1371 36. REASONING Using the value given for the critical angle in Equation 26.4 (sin θc = n2/n1),
we can obtain the ratio of the refractive indices. Then, using this ratio in Equation 26.5
(Brewster’s law), we can obtain Brewster’s angle θB.
SOLUTION From Equation 26.4, with n2 = nair = 1 and n2 = nliquid, we have
sin θ c = sin 39° = 1
nliquid (1) According to Brewster’s law, tan θ B = n2
1
=
n1 nliquid (2) Substituting Equation (2) into Equation (1), we find
tan θ B = 1
nliquid = sin 39° = 0.63 θ B = tan –1 ( 0.63) = 32° or ______________________________________________________________________________
37. REASONING Since the light reflected from the coffee table is completely polarized
parallel to the surface of the glass, the angle of incidence must be the Brewster angle
(θB = 56.7°) for the airglass interface. We can use Brewster's law (Equation 26.5: tan θ B = n2 / n1 ) to find the index of refraction n2 of the glass. SOLUTION Solving Brewster's law for n2, we find that the refractive index of the glass is n2 = n1 tan θ B = (1.000)(tan 56.7°) = 1.52
______________________________________________________________________________
38. REASONING Given the height h of
the laser above the glass pane, the
distance d between the edge of the pane
and the point where the laser reflects
depends upon the angle θ between the
incident ray and the vertical (see the
drawing). The height h is adjacent to the
angle θ, and the distance d is opposite,
d
so from tan θ =
(Equation 1.3) we
h
have that Laser θ h
θ d Glass pane 1372 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS d = h tan θ (1) The glass pane is horizontal, so the normal to the surface is vertical, and we see that θ is also
the angle of incidence when the incident ray strikes the glass pane. The reflected beam is
100% polarized, so we conclude that the angle θ of incidence is equal to Brewster’s angle θB
n
for the glassair interface: θ = θB. Brewster’s angle is given by tan θB = 2 (Equation 26.5),
n1
where n2 = 1.523 and n1 = 1.000 are the indices of refraction for crown glass and air,
respectively (see Table 26.1). Therefore, tan θ = tan θB = n2
n1 (2) SOLUTION Substituting Equation (2) into Equation (1), we obtain
n 1.523 d = h tan θ = h 2 = ( 0.476 m ) = 0.725 m
n 1.000 1
______________________________________________________________________________ 39. SSM WWW REASONING Brewster's law (Equation 26.5: tan θ B = n2 / n1 ) relates the angle of incidence θB at which the reflected ray is completely polarized parallel to the surface to the indices of refraction n1 and n2 of the two media forming the interface. We
can use Brewster's law for light incident from above to find the ratio of the refractive indices
n2/n1. This ratio can then be used to find the Brewster angle for light incident from below
on the same interface.
SOLUTION The index of refraction for the medium in which the incident ray occurs is
designated by n1. For the light striking from above n2 / n1 = tan θ B = tan 65.0° = 2.14 .
The same equation can be used when the light strikes from below if the indices of refraction
are interchanged n1 –1 1 –1 1 = tan n /n = tan 2.14 = 25.0° n2 2 1
______________________________________________________________________________ θ B = tan –1 40. REASONING The reflected light is 100% polarized when the angle of incidence is equal to
the Brewster angle θB. The Brewster angle is given by tan θ B = nliquid / nair (Equation 26.5),
where nliquid and nair are the refractive indices of the liquid and air (neither of which is
known). However, nliquid and nair are related by Snell’s law (Equation 26.2), Chapter 26 Problems 1373 nair sin θ1 = nliquid sin θ 2 , where θ1 and θ2 are, respectively, the angles of incidence and refraction. These two relations will allow us to determine the Brewster angle.
SOLUTION The Brewster angle is given by
tan θ B =
Snell’s law is nliquid
nair nair sin θ1 = nliquid sin θ 2 (26.5) (26.2) from which we obtain nliquid / nair = sin θ1 / sin θ 2 . Substituting this result into Equation 26.5
yields tan θ B = nliquid
nair = sin θ1
sin θ 2 Thus, the Brewster angle is sin θ1 −1 sin 53.0° = 55.0° = tan sin 34.0° sin θ 2 θ B = tan −1 ______________________________________________________________________________
41. REASONING AND SOLUTION From Snell’s law we have n sin θ 2 sin θ B =...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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