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Unformatted text preview: the glass and the (unknown) angle θ2 of refraction for the light entering the slab: (1.00 ) sin 35.0° = n2 sin θ 2 . When the incident light is in the liquid, we can again use Snell’s law to express the relation between the index of refraction n1 of the liquid, the angle
of incidence (20.3°), the index of refraction n2 of the glass, and the (unknown) angle of
refraction θ2: n1 sin 20.3° = n2 sin θ 2 . By equating these two equations, we can determine the
index of refraction of the liquid.
SOLUTION Setting the two equations above equal to each other and solving for the index
of refraction of the liquid gives n1 sin 20.3° = (1.00 ) sin 35.0° n1 = (1.00 ) sin 35.0° = 1.65
sin 20.3°
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and 18. REASONING According to Equation 2.1, when an object moves with a constant speed v
d
and travels a distance d in an elapsed time t, these three quantities are related by v = . In
t
this situation, the stone sinks a distance d from the surface to the bottom, and the elapsed
time is t. When viewed from above, the stone still sinks to the bottom in an elapsed time t,
but the apparent depth d ′ to which it sinks is not equal to the actual depth d. Therefore, the
apparent speed v′ of the stone is not the same as its actual speed v, and we have that Chapter 26 Problems v= d
t and v′ = d′
t 1359 (1) n The apparent depth d ′ is given by d ′ = d 2 (Equation 26.3), where n2 = 1.000 is the
n 1
index of refraction of air and n1 = 1.333 (See Table 26.1) is the index of refraction of water.
SOLUTION Solving the first of Equations (1) for t yields t = d
. Substituting this
v expression into the second of Equations (1), we obtain
v′ = d′
d′ d′ =
= v t d d v (2) n Substituting d ′ = d 2 (Equation 26.3) into Equation (2), we find that
n 1
n d 2
n
n d′ 1.000 v′ = v = v 1 = v 2 = ( 0.48 m/s ) = 0.36 m/s
n d
d 1.333 1
______________________________________________________________________________ 19. REASONING AND SOLUTION The horizontal distance of the chest from the normal is
found from Figure 26.4b to be x = d tan θ1 and x = d ' tan θ2, where θ1 is the angle from the dashed normal to the solid rays and θ2 is the angle from the dashed normal to the dashed
rays. Hence, d ' = d (tan θ1/tan θ2) Snell's law applied at the interface gives
n1 sin θ1 = n2 sin θ2
For small angles, sin θ1 ≈ tan θ1 and sin θ2 ≈ tan θ2, so
tan θ1/ tan θ2 ≈ sin θ1/sin θ2 = (n2/n1)
Now d ' = d (tan θ1/tan θ2). Therefore,
n d′ ≈ d 2 n 1 1360 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS ______________________________________________________________________________
20. REASONING Snell’s law will allow us to calculate the angle of refraction θ2, B with which
the ray leaves the glass at point B, provided that we have a value for the angle of incidence
θ1, B at this point (see the drawing). This angle of incidence is not given, but we can obtain it
by considering what happens to the incident ray at point A. This ray is incident at an angle
θ1, A and refracted at an angle θ2, A. Snell’s law can be used to obtain θ2, A, the value for
which can be combined with the geometry at points A and B to provide the needed value for
θ1, B. Since the light ray travels from a material (carbon disulfide) with a higher refractive
index toward a material (glass) with a lower refractive index, it is bent away from the normal
at point A, as the drawing shows. θ2, A B θ2, B
Glass θ1, A = 30.0º A θ1, B
θ2, A Carbon disulfide SOLUTION Using Snell’s law at point B, we have (1.52 ) sin θ1, B = (1.63) sin θ 2, B
14 244
4
3 14 244
4
3 Glass or 1.52 sin θ 2, B = sin θ1, B 1.63 (1) or θ1, B = 90.0° − θ 2, A (2) Carbon disulfide To find θ1, B we note from the drawing that θ1, B + θ 2, A = 90.0° We can find θ2, A, which is the angle of refraction at point A, by again using Snell’s law: (1.63) sin θ1, A = (1.52 ) sin θ 2, A
14 244
4
3 14 244
4
3 Carbon disulfide Glass or 1.63 sin θ 2, A = sin θ1, A 1.52 Chapter 26 Problems 1361 Thus, we have 1.63 1.63 sin θ 2, A = sin θ1, A = sin 30.0° = 0.536 1.52 1.52 or θ 2, A = sin −1 ( 0.536 ) = 32.4° Using Equation (2), we find that θ1, B = 90.0° − θ 2, A = 90.0° − 32.4° = 57.6°
With this value for θ1, B in Equation (1) we obtain 1.52 1.52 sin θ 2, B = sin θ1, B = sin 57.6° = 0.787 1.63 1.63 21. SSM WWW The drawing at the right
shows the geometry of the situation using
the same notation as that in Figure 26.6.
In addition to the text's notation, let t
represent the thickness of the pane, let L
represent the length of the ray in the pane,
let x (shown twice in the figure) equal the
displacement of the ray, and let the
difference in angles θ1 – θ2 be given by φ.
We wish to find the amount x by which
the emergent ray is displaced relative to
the incident ray. This can be done by
applying Snell's law at each interface, and
then making use of the geometric and
trigonometric relations in the drawing. θ 2, B = sin −1 ( 0.787 ) = 51.9° or x
θ3 Air ( n 3 =...
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 Spring '13
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 Physics, The Lottery

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