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Unformatted text preview: 2 n 1 But from Brewster’s law, Equation 26.5, n2/n1 = tan θB. Substituting this expression for n2/n1 into Snell’s law, we see that sin θ B sin θ B = tan θ B sin θ 2 = cos θ sin θ 2 B
This result shows that cos θB = sin θ2. Since sin θ2 = cos (90° – θ2), we have that cos θB = sin θ2 = cos (90° – θ2). Thus, θB = 90° – θ2, so θB + θ2 = 90°, and the reflected and refracted rays are perpendicular .
______________________________________________________________________________ 1374 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 42. REASONING When light is incident at the Brewster angle, we know that the angle
between the refracted ray and the reflected ray is 90°. This relation will allow us to
determine the Brewster angle. By applying Snell’s law to the incident and refracted rays,
we can find the index of refraction of the glass.
SOLUTION The drawing shows the incident, reflected, and refracted rays. θB θB Vacuum 90°
Glass θ2 a. We see from the drawing that θB + 90° + θ2 = 180°, so that θB = 90° – 29.9° = 60.1° .
b. Applying Snell’s law at the vacuum/glass interface gives (1.00 ) sin 60.1° nglass = = 1.74 sin 29.9° ______________________________________________________________________________
nvacuum sin θ B = nglass sin θ 2 or 43. REASONING The angle of each refracted ray in the crown glass can be obtained from
Snell’s law (Equation 26.2) as ndiamond sin θ1 = ncrown glass sin θ2, where θ1 is the angle of
incidence and θ2 is the angle of refraction.
SOLUTION The angles of refraction for the red and blue rays are: Blue ray Red ray sin θ1 ( 2.444 ) sin 35.00° = sin −1 = 66.29° ncrown glass 1.531 n θ 2 = sin −1 diamond n sin θ1 2.410 sin 35.00° = sin −1 = 65.43° ncrown glass 1.520 θ 2 = sin −1 diamond The angle between the blue and red rays is θ blue − θ red = 66.29° − 65.43° = 0.86°
______________________________________________________________________________ Chapter 26 Problems 1375 44. REASONING Since the angles of refraction are the same, the angles of incidence must be
different, because the refractive indices of the red and violet light are different. This follows
directly from Snell’s law. We can apply the law for each color, obtaining two equations in
the process. By eliminating the common angle of refraction from these equations, we can
obtain a single expression from which the angle of incidence of the violet light can be
determined.
SOLUTION Applying Snell’s law for each color, we obtain
n1, red sin θ1, red = n2, red sin θ 2, red
14 244
4
3 14 244
4
3
Air n1, violet sin θ1, violet = n2, violet sin θ 2, violet
144 2444 144 2444
4
3
4
3 and Glass Air Glass Dividing the equation on the right by the equation on the left and recognizing that the angles
of refraction θ2, red and θ2, violet are equal, we find
n1, violet sin θ1, violet
n1, red sin θ1, red = n2, violet sin θ 2, violet
n2, red sin θ 2, red = n2, violet
n2, red Since both colors are incident in air, the indices of refraction n1,
equal to 1.000, and this expression simplifies to
sin θ1, violet
sin θ1, red = red and n1, violet are both n2, violet
n2, red Solving for the angle of incidence of the violet light gives n2, violet
sin θ1, violet = sin θ1, red n 2, red 1.538 = ( sin 30.00° ) = 0.5059 1.520 θ1, violet = sin −1 ( 0.5059 ) = 30.39° 45. SSM REASONING Because the refractive index of the glass depends on the wavelength
(i.e., the color) of the light, the rays corresponding to different colors are bent by different
amounts in the glass. We can use Snell’s law (Equation 26.2: n1 sin θ1 = n2 sin θ 2 ) to find
the angle of refraction for the violet ray and the red ray. The angle between these rays can
be found by the subtraction of the two angles of refraction.
SOLUTION In Table 26.2 the index of refraction for violet light in crown glass is 1.538,
while that for red light is 1.520. According to Snell's law, then, the sine of the angle of 1376 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS refraction for the violet ray in the glass is sin θ 2 = (1.000 /1.538) sin 45.00° = 0.4598 , so
that
θ 2 = sin −1 (0.4598) = 27.37° Similarly, for the red ray, sin θ 2 = (1.000 /1.520) sin 45.00° = 0.4652 , from which it follows
that
θ 2 = sin −1 (0.4652) = 27.72°
Therefore, the angle between the violet ray and the red ray in the glass is
27.72° – 27.37° = 0.35°
______________________________________________________________________________ 46. REASONING AND SOLUTION Applying Snell’s law at the gassolid interface gives the
angle of refraction θ2 to be (1.00) sin 35.0° = (1.55) sin θ 2 or θ 2 = 21.7° Since the refractive index of the liquid is the same as that of the solid, light is not refracted
when it enters the liquid. Therefore, the light enters the liquid at an angle of 21.7° .
______________________________________________________________________________
47. SSM REASONING We can use Snell's law (Equation 26.2: n1 sin θ1 =...
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