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Unformatted text preview: n 1) x
L θ2
θ2 t φ Glass
Glass ( n2) ( n2 )
Air ( n1) θ1 SOLUTION If we apply Snell's law (see Equation 26.2) to the bottom interface we obtain
n1 sin θ1 = n2 sin θ 2 . Similarly, if we apply Snell's law at the top interface where the ray
emerges, we have n2 sin θ 2 = n3 sin θ 3 = n1 sin θ 3 . Comparing this with Snell's law at the bottom face, we see that n1 sin θ1 = n1 sin θ 3 , from which we can conclude that θ3 = θ1.
Therefore, the emerging ray is parallel to the incident ray. From the geometry of the ray and thickness of the pane, we see that L cos θ 2 = t , from which it follows that L = t /cos θ 2 . Furthermore, we see that x = L sin φ = L sin (θ1 – θ 2 ) .
Substituting for L, we find 1362 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS x = L sin(θ1 – θ 2 ) = t sin(θ1 – θ 2 )
cos θ 2 Before we can use this expression to determine a numerical value for x, we must find the
value of θ2. Solving the expression for Snell's law at the bottom interface for θ2, we have sin θ 2 = n1 sin θ1 (1.000) (sin 30.0°)
=
= 0.329
n2
1.52 θ 2 = sin –1 0.329 = 19.2° or Therefore, the amount by which the emergent ray is displaced relative to the incident ray is t sin (θ1 – θ 2 ) (6.00 mm) sin (30.0° –19.2°)
=
= 1.19 mm
cos θ 2
cos 19.2°
______________________________________________________________________________
x= 22. REASONING The fish appears to be a distance less than 1 L from the front wall of the
2
aquarium, where L is the distance between the back and front walls. The phenomenon of
apparent depth is at play here. According to Equation 26.3, the apparent depth or distance d ′
n is related to the actual depth 1 L by d ′ = 1 L air , where nair and nwater are the refractive
2
2 n water indices of air and water. Referring to Table 26.1, we see that nair < nwater, so that d ′ < 1 L .
2
Since the fish is a distance
1
2 1
2 L in front of the plane mirror, the image of the fish is a distance L behind the plane mirror. Thus, the image is a distance 3
2 L from the front wall. The image of the fish appears to be at a distance less than 3 L from the front wall, because of
2
the phenomenon of apparent depth. The explanation given above applies here also, except
that the actual depth or distance is 3 L instead of 1 L .
2
2
SOLUTION
a. Using Equation 26.3, we find that the apparent distance d ′ between the fish and the front
wall is
n 1.000 d ′ = 1 L air = 1 ( 40.0 cm ) = 15.0 cm
2 n
2 1.333 water b. Again using Equation 26.3, we find that the apparent distance between the image of the
fish and the front wall is
n
d ′ = 3 L air
2 n water = 3
2 ( 40.0 cm ) 1 = 45.0 cm
1.333 Chapter 26 Problems 23. SSM REASONING Following the discussion in
Conceptual Example 4, we have the drawing at the
right to use as a guide. In this drawing the symbol d
refers to depths in the water, while the symbol h refers
to heights in the air above the water. Moreover,
symbols with a prime denote apparent distances, and
unprimed symbols denote actual distances. We will use
Equation 26.3 to relate apparent distances to actual
distances. In so doing, we will use the fact that the
refractive index of air is essentially nair = 1 and denote
the refractive index of water by nw = 1.333 (see Table
26.1). 1363 h′
h d′ Air
Water d SOLUTION To the fish, the man appears to be a
distance above the airwater interface that is given by
Equation 26.3 as h′ = h ( nw /1) . Thus, measured above
the eyes of the fish, the man appears to be located at a
distance of nw h ′ + d = h +d
1 (1) To the man, the fish appears to be a distance below the airwater interface that is given by
Equation 26.3 as d ′ = d (1/ nw ) . Thus, measured below the man’s eyes, the fish appears to
be located at a distance of 1
h + d ′ = h + d
(2) nw Dividing Equation (1) by Equation (2) and using the fact that h = d, we find h′+ d
=
h +d′ nw h +d
1 1
h + d nw = nw + 1
= nw
1
1+
nw (3) In Equation (3), h ′ + d is the distance we seek, and h + d ′ is given as 2.0 m. Thus, we find h′ + d = nw ( h + d ′ ) = (1.333) ( 2.0 m ) = 2.7 m
______________________________________________________________________________ 1364 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 24. REASONING The drawing at
the right shows the situation. As
discussed in the text, when the
observer is directly above, the
apparent depth d ′ of the object is
related to the actual depth by
Equation 26.3: dw = 1.50 cm water
(nw = 1.333) n2 d′ = d n1 dg = 3.20 cm glass
(ng = 1.52) In this case, we must apply
logo
Equation 26.3 twice; once for the
rays in the glass, and once again for the rays in the water. SOLUTION We refer to the drawing for our notation and begin at the logo. To an observer ( ) depth is, ′
in the water directly above the logo, the apparent depth of the logo is d g = d g nw / ng .
When ( viewed ′
′
dw = dw + d g directly from above in air, the logo’s apparent ) ( nair / nw ) , where we have used the fact that wh...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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