Unformatted text preview: the laser and the detector. Chapter 25 Problems b. Turning the surface normal too far
from due south increases both the angle
of incidence and the angle of reflection,
in this case by 0.004° each. Thus, the
angle α between the incident and
reflected rays increases by twice that
amount. We will use the new angle α,
with tan α = D L (Equation 1.3), to
determine the distance D between the
laser and the reflected beam (see the
drawing). The distance x by which the
reflected beam misses the detector is the
distance D minus the distance d
between the laser and the detector: 1289 Mirror θ θ α L
Detector
Laser d x
D x= D−d (2) We note that the distance d is given in meters, while the distance L is given in kilometers,
where 1 km = 103 m.
SOLUTION
a. Substituting α = tan −1 ( d L ) (Equation 1.6) into Equation (1) yields
d 117 m o = 0.0670
3 50.0 × 10 m θ = 1 α = 1 tan −1 = 1 tan −1 2
2
2
L b. When the mirror is misaligned, the angle of incidence is larger than the result found in
part (a) by 0.004°. From Equation (1), then, the angle α between the incident and reflected
rays becomes
α = 2θ = 2 0.0670o + 0.004o = 0.142o ( ) Solving tan α = D L (Equation 1.3) for the distance D between the laser and the reflected
beam, we obtain
D = L tan α
(3)
Substituting Equation (3) into Equation (2), and using α = 0.142o , we find that ( ) x = D − d = L tan α − d = 50.0 ×103 m tan 0.142o − 117 m = 7 m 1290 THE REFLECTION OF LIGHT: MIRRORS 11. REASONING According to the discussion about relative velocity in Section 3.4, the
velocity vIY of your image relative to you is the vector sum of the velocity vIM of your
image relative to the mirror and the velocity vMY of the mirror relative to you:
vIY = vIM + vMY
As you walk toward the stationary mirror, you perceive the mirror moving toward you in the
opposite direction. Thus, vMY = –vYM. The velocity vYM has the components vYM,x and
vYM,y, while the velocity vIM has the components vIM,x and vIM,y. The x direction is
perpendicular to the mirror, and the two x components have the same magnitude. This is
because the image in a plane mirror is always just as far behind the mirror as the object is in
front of it. For instance, if an object moves 1 meter perpendicularly toward the mirror in 1
second, the magnitude of its velocity relative to the mirror is 1 m/s. But the image also
moves 1 meter toward the mirror in the same time interval, so that the magnitude of its
velocity relative to the mirror is also 1 m/s. The two x components, however, have opposite
directions. The two y components have the same magnitude and the same direction. This is
because an object moving parallel to a plane mirror has an image that remains at the same
distance behind the mirror as the object is in front of it and moves in the same direction as
the object.
SOLUTION According to the discussion in the REASONING,
vIY = vIM + vMY = vIM – vYM
This vector equation is equivalent to two equations, one for the x components and one for
the y components. For the x direction, we note that vYM,x = –vIM,x. vIY,x = vIM,x − vYM,x
= − ( 0.90 m/s ) cos 50.0° − ( 0.90 m/s ) cos 50.0° = −1.2 m/s
For the y direction, we note that vYM,y = vIM,y. vIY,y = vIM,y − vYM,y
= ( 0.90 m/s ) sin 50.0° − ( 0.90 m/s ) sin 50.0° = 0 m/s
Since the y component of the velocity vIY is zero, the velocity of your image relative to you
points in the –x direction and has a magnitude of 1.2 m/s . Chapter 25 Problems 12. REASONING The time
proportional to the total
each path. Therefore,
distances identified in the
have
t reflected
t direct = of travel is
distance for
using the
drawing, we 1291 lamp, L
2s
mirror, M
d LM d LM + d MP d LP (1) d LP 30.0°
30.0° We know that the angle of incidence is
equal to the angle of reflection. We also
know that the lamp L is a distance 2s
from the mirror, while the person P is a
distance s from the mirror. d MP
s
person, P Therefore, it follows that if d MP = d , then d LM = 2 d . We may use the law of cosines (see
Appendix E) to express the distance d LP as 2
bd g+ d d LP = 2 2 b gg b g
b – 2 2 d d cos 2 30.0 ° SOLUTION Substituting the expressions for d MP , d LM , and d LP into Equation (1), we
find that the ratio of the travel times is t reflected
t direct = 2d + d
2
bd g+ d
2 2 = bg b g
bg – 2 2 d d cos 2 30.0 ° 3
5 – 4 cos 60.0 ° = 1.73 13. REASONING When an object is located very far away from a spherical mirror (concave or
convex), the image is located at the mirror’s focal point. Here, the image of the distant
object is located 18 cm behind the convex mirror, so that the focal length f of the mirror is
f = −18 cm and is negative since the mirror is convex.
SOLUTION In constructing a ray diagram, we will need to know the radius R of the mirror.
1
2 The focal length of a convex mirror is related to the radius by f = − R (Equation 25.2).
We can use this expression to determine the radius:
1
2 f =− R or R = −2 f = −2 ( −18 cm ) = 36 cm In the ray diagram that follows, we denote the focal point by F and the center of curvature
by C. Note t...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details