2. Picture the Problem: The image shows two mirrors oriented at 120° with respect to each other. A light ray strikes the first mirror with an incident angle of 55°. The reflected light then reflects off the second mirror.
. Strategy: Set the reflected angle from the first mirror equal
to the incident angle. The two mirrors and the ray form a triangle. The sum of the interior angles of a triangle is 180°. Two of the interior angles are the complementary angles
(
90° −θ)
for the incident and reflected rays. Use this relation to calculate the angle of incidence for mirror 2, and set angle of reflection for mirror 2 equal to the angle of incidence for mirror 2.Solution: 1. Write the angle of reflection for mirror
1: θ1r =θ1i = °55
2. Use the interior angles of the triangle to calculate the incident angle of the second mirror:
(
1r)
(
2i)
2i 1r 90 120 90 180 120 120 55 65 θ θ θ θ ° − + ° + ° − = ° = ° − = ° − ° = °3. Set the reflected angle equal to the incident angle: θ2r =θ2i= 65°
Insight: For any incident angle, the sum of the incident angle and the reflected angle will equal the angle between the two mirrors.
12. Picture the Problem: The image shows a person looking at a mirror of length dmirror and observing a
full image of the rear window of length dwindow.
Strategy: Since the incident angle and the reflected angle are equal, the light rays from the ends of the window will intersect at the same distance behind the mirror as the eye is in front of the mirror. Using this relationship, create two similar triangles to relate the ratio of the length of the mirror to the mirror-eye distance and the ratio of the length of the window to the sum of the eye and mirror-window distances. Set these two ratios equal and solve for the mirror length. The same method will solve for the mirror height, using the window height instead of the window length.
Solution: 1. Set the ratios of the lengths to distances equal:
mirror window m-e m-e m-w
d d
x =x +x
2. Solve for the mirror length: mirror window m-e
(
)
m-e m-w 1.3 m 0.50 m 33 cm 0.50 m 1.50 m d x d x x = = = + +
3. Replace length with height of the mirror and window:
(
)
window m-e mirror m-e m-w 0.3 m 0.50 m 7.5 cm 0.50 m 1.50 m h x h x x = = = + +28. Picture the Problem: The figure shows a lightbulb placed in front of a concave mirror and its image that is projected on a distant wall.
Strategy: Use equation 26-8 to calculate the magnification of the image. Then use equation 26-6 to calculate the focal length of the mirror. Solution: 1. (a) Calculate the magnification: i o 3.5 m 1.9 1.8 m d m d = − = − = −
2. (b) Since m < 0, the image is inverted 3. (c) Calculate the focal length:
1 1 o i 1 1 1 1 1.2 m 1.8 m 3.5 m f d d − − ⎛ ⎞ ⎛ ⎞ =⎜ + ⎟ =⎜ + ⎟ = ⎝ ⎠ ⎝ ⎠
Insight: If the lightbulb has a diameter of 12 cm, its image on the wall will be inverted and have a diameter of 23 cm.
32. Picture the Problem: A convex mirror produces a virtual, upright, and reduced image of an object. Strategy: Use the magnification equation (equation 26-8) to calculate the image distance from the mirror, and then solve equation 26-6 for the focal length.
Solution: 1. (a) Calculate the image
distance: i o
(
)
1
32 cm 8.0 cm
4
d = −md = − = −
2. (b) Calculate the focal length from equation 26-6: 1 1 o 1 1 1 1 11 cm 32 cm 8.0 cm i f d d − − ⎛ ⎞ ⎛ ⎞ =⎜ + ⎟ =⎜⎝ +− ⎟⎠ = − ⎝ ⎠
Insight: As expected, the image distance is negative, indicating a virtual image, and the focal length is negative, indicating a convex mirror.
46. Picture the Problem: The image shows a light beam incident on a semicircular disk with index of refraction 1.52. After passing through the disk, the light beam hits a wall 5.00 cm above the center of the disk and at a distance 20.0 cm behind the disk.
Strategy: Since the light enters the semicircular disk at the center, it will exit perpendicular to the radius of curvature. It will therefore not refract upon exiting. Calculate the angle of refraction θ2 for the light entering
Solution: 1. Calculate the refracted angle: 2 1
(
)
1(
)
2 tan tan tan 5.00 cm 20.0 cm 14.0 y x y x θ θ − − = = = = °2. Solve Snell’s Law for the angle of incidence:
(
)
air g 2 g 1 1 2 air sin sin 1.52sin sin sin sin 14.0 21.6
1.000 n n n n θ θ θ − θ − = ⎛ ⎞ ⎡ ⎤ = ⎜ ⎟= ⎢ ⎥= ° ⎣ ⎦ ⎝ ⎠
Insight: Increasing the disk’s index of refraction, while keeping the angle of incidence constant, will decrease the height that the beam hits the wall. For example, if the index of refraction of the glass in this problem were 1.65, the beam would hit the wall at a height of 4.58 cm.
50. Picture the Problem: The image shows a light beam entering a 45°-90°-45° prism perpendicular to the long side. It undergoes total internal reflection on both short sides before exiting vertically downward through the long sides.
Strategy: The angles of incidence for both internal reflections are exactly 45°. Solve equation 26-12 for the index of refraction of the prism. Since total internal reflection still occurs when the external index is 1.21, use this index to determine the prism’s minimum index of refraction. Since total internal reflection does not occur when the external index is 1.43, use this index to determine the prism’s maximum index of refraction.
Solution: 1. Calculate the minimum index of refraction: min c 1.21 1.71 sin sin 45 n n θ = = = °
2. Calculate the maximum index of refraction: max c 1.43 2.02 sin sin 45 n n θ = = = °
3. List the possible range for the index of refraction: 1.71≤nprism<2.02
Insight: Because the light is not totally internally reflected inside the prism when it is submerged in liquid with an index of refraction of 1.43, the angle at which the beam refracts into the liquid could be used to determine the index of refraction of the prism by using Snell’s Law.
54. Picture the Problem: Light from a lamp reflects off and refracts into the varnish on a tabletop. The angle between the incident and reflected rays is 110°.
Strategy: Since the reflected light is completely polarized, the incident angle is the Brewster’s angle. Set the reflected and incident angles equal to each other (from equation 26-1) and solve for the incident angle. Then use the incident angle as the Brewster’s angle in equation 26-13 to solve for the index of refraction.
2. (b) Calculate the index of refraction:
(
)
2 B 1 2 1 B tan tan 1.00 tan 55 1.4 n n n n θ θ = = = ° =Insight: Verify for yourself that the angle of refraction is 35° by using the index of refraction and Snell’s Law . Note that the reflected and refracted angles sum to 90° when the incident angle is the Brewster’s angle.
68. Picture the Problem: An object placed to the left of a convex lens produces an upright image with a
magnification of 3.0.
Strategy: Solve equation 26-16 for the image distance and insert it into equation 26-18 to write an equation for the object distance as a function of magnification. Insert the two magnifications (3.0 and 4.0) to determine the distances and calculate the difference.
Solution: 1. (a) Solve equation 26-16 for di:
1 o i o o 1 1 f d d f d d f − ⎛ ⎞ =⎜ − ⎟ = − ⎝ ⎠
2. Eliminate di from the magnification
equation: o i o o o o 1 f d d f m d d d f f d ⎛ ⎞ = − = − ⎜ ⎟= − − ⎝ ⎠
3. In order to increase the magnification, the object should be moved farther away from the lens. 4. (b) Solve for the object distance: do = f
(
1 1− m)
5. Calculate the change in distance to increase the magnification:
(
)
(
)
(
)
(
)
o4 o3 1 1 4 1 1 3 1 3 1 4 36 cm 1 3 1 4 3.0 cm d −d = f − m − f − m = f m − m = − =6. The object should be moved 3.0 cm away from the lens.
Insight: The magnification increases to infinity as the object approaches the focal point.
72. Picture the Problem: The figure shows an object in front of a concave lens with focal length −36 cm. The magnification of the image is 1
3.
m= The object is to be moved so that the magnification
becomes 1
4.
m=
Strategy: Solve equation 26-16 for the image distance and insert it into equation 26-18 to write an equation for the object distance as a function of magnification. Insert the two magnifications (1 3
and 1 4) to determine the distances and calculate the difference.
Solution: 1. (a) Solve equation 26-16 for the image distance:
2. Eliminate di from the magnification equation: o i o o o o 1 fd d f m d d d f f d ⎛ ⎞ = − = − ⎜ ⎟= − − ⎝ ⎠
3. Explain how the object must be
moved to decrease the magnification: Since f is negative, increasing dcation. The object should be moved farther away from the o will decrease the magnifi-lens.
4. Solve for the object distance: do = f
(
1 1− m)
5. (b) Calculate the change in distance to increase the magnification:
(
)
(
)
(
)
(
)
o4 o3 1 1 4 1 1 3 1 3 1 4 36 cm 3 4 36 cm d −d = f − m − f − m = f m − m = − − =The object should be moved 36 cm further away from the lens.
