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Unformatted text preview: these values into Equation (1), we obtain
vIY = (–0.90 m/s) – (+0.90 m/s) = –1.80 m/s 5. SSM REASONING The geometry is shown below. According to the law of reflection,
the incident ray, the reflected ray, and the normal to the surface all lie in the same plane, and
the angle of reflection θ r equals the angle of incidence θ i . We can use the law of reflection
and the properties of triangles to determine the angle θ at which the ray leaves M2. 1286 THE REFLECTION OF LIGHT: MIRRORS θ
γ
β
M2
65° φ α 120° M1 SOLUTION From the law of reflection, we know that φ = 65° . We see from the figure
that φ + α = 90° , or α = 90° – φ = 90° – 65° = 25° . From the figure and the fact that the
sum of the interior angles in any triangle is 180°, we have α + β + 120° = 180° . Solving for
β , we find that β = 180° – (120° + 25°) = 35° . Therefore, since β + γ = 90° , we find that
the angle γ is given by γ = 90° – β = 90° – 35° = 55° . Since γ is the angle of incidence of
the ray on mirror M2, we know from the law of reflection that θ = 55° . 6. REASONING AND SOLUTION
a. The height of the shortest mirror would be onehalf the height of the person. Therefore,
h = H/2 = (1.70 m + 0.12 m)/2 = 0.91 m
b. The bottom edge of the mirror should be above the floor by
h' = (1.70 m)/2 = 0.85 m 7. REASONING AND SOLUTION The two arrows, A and B are located in front of a plane
mirror, and a person at point P is viewing the image of each arrow. As discussed in
Conceptual Example 1, light emanating from the arrow is reflected from the mirror and is
reflected toward the observer at P. In order for the observer to see the arrow in its entirety,
both rays, the one from the top of the arrow and the one from the bottom of the arrow, must
pass through the point P. Chapter 25 Problems Plane
mirror P 1287 Plane
mirror P A
B According to the law of reflection, all rays will be reflected so that the angle of reflection is
equal to the angle of incidence. The ray from the top of arrow A strikes the mirror and
reflects so that it passes through point P. Likewise, the ray from the bottom of the arrow is
reflected such that it too passes through point P. Therefore, the observer at P
sees the arrow at A in its entirety .
Similar reasoning shows that the ray from the top of arrow B passes through point P.
However, as the drawing shows, the ray from the bottom of the arrow does not pass through
P. This conclusion is true no matter where the bottom ray strikes the mirror. The observer
does not see the arrow at B in its entirety . 8. REASONING The drawing shows the ray
of light reflecting from the mirror and
striking the floor. The angle of reflection θ
is the same as the angle of incidence. The
angle θ is also the angle that the light ray
makes with the floor (see the drawing).
Therefore, we can use the inverse tangent
function to find θ as a function of y and x; y
θ = tan −1 . Since the mirror is facing
x
east and the sun is rising, the angle of
incidence θ becomes larger (see the
drawing). As the angle θ becomes larger,
the distance x becomes smaller. θ
θ
y θ
x SOLUTION When the horizontal distance is x1 = 3.86 m, the angle of incidence θ1 is
y
−1 1.80 m = tan 3.86 m = 25.0° x1 θ1 = tan −1 When the horizontal distance is x2 = 1.26 m, the angle of incidence θ2 is 1288 THE REFLECTION OF LIGHT: MIRRORS y
−1 1.80 m = tan 1.26 m = 55.0° x2 θ 2 = tan −1 As the sun rises, the change in the angle of incidence is 55.0° − 25.0° = 30.0°. Since the
earth rotates at a rate of 15.0° per hour, the elapsed time between the two observations is 1h Elapsed time = ( 30.0° ) = 2.00 h 15.0° 9. REASONING AND SOLUTION
a. After the mirror has been rotated, the new angle of incidence is θi = 45° + 15° = 60°. The
angle of reflection, then, is also equal to 60°. The reflected ray which was originally 90°
(45° + 45°) from the original angle of incidence, is now 120° (60° + 60°) from the incident
ray's direction. Therefore, the reflected ray has been rotated through β = 120° − 90° = 30°
b. The angle through which the reflected ray is rotated depends only on the angle through
which the mirror is rotated, and is independent of the angle of incidence. Therefore,
β ′ = 30° . 10. REASONING
a. The incident ray from the laser
travels a distance of L = 50.0 km due
north to the mirror (see the drawing,
which is not to scale). The angle θ we
seek, that between the normal to the
surface of the mirror and due south, is
the angle of incidence. The angle of
incidence must equal the angle of
reflection, so the angle α between the
incident and reflected rays is bisected
by the normal to the surface of the
mirror. Therefore, we have that Mirror North θ L θ East α
South Laser Detector d θ = 1α
2 (1) Because the laser, the mirror, and the detector sit at the corners of a right triangle, we will
use α = tan −1 ( d L ) (Equation 1.6) to determine the angle α, where d = 117 m is the
distance between...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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