Ch. 26 HW Solutions
2.) Using the law of reflection we can calculate how tall the mirror must be for the person
to see the top of his head and the bottom of his feet. Examination of the figure illustrates
that the top two triangles are identical and the bottom two triangles are identical (due to
the law of reflection). Since the mirror only spans one of each triangle, the mirror must be
at least half the height of the person, or 1.7/2 = 0.85 m.
6.) The sun is very far away so its rays at the mirror will be approximately parallel. Parallel rays reflect from a sperical mirror through the focal point, which is half the radius of
curvature, f = R/2 = 3/2 = 1.5 m. The object should be placed here to receive the most
radiation for heating.
11.) With f = 40 cm and given value of p for each part, we calculate the image location using 1/p + 1/q = 1/f or rewritten q = pf /(p − f ), and we calculate the image height
using M = −q/p = hi /ho or hi = ho (−q/p).
(a) p = 120 cm: q = (120)40/(120 − 40) = 60 cm. The image is real since q > 0.
hi = (3)(−60/120) = −1.5 cm. The image is inverted and diminished.
(b) p = 80 cm: q = (80)40/(80 − 40) = 80 cm. The image is real since q > 0.
hi = (3)(−80/80) = −3.0 cm. The image is inverted and unchanged in size, as is the case
when the object is placed at R = 2f .
(c) p = 15 cm: q = (15)40/(15 − 40) = −24 cm. The image is virtual since q < 0 (behind
the mirror).
hi = (3)(−(−24)/15) = 4.8 cm. Upright and enlarged, as is the case when the object is
within the focus of the mirror.
14.) A convex mirror with radius of curvature of 60 cm has f = −30 cm (negative since the
mirror is convex).
(a) p = 120 cm: q = (120)(−30)/(120 − (−30)) = −24 cm. The image is virtual.
hi = (3)(−(−24)/120) = 0.6 cm. Upright and diminished.
(b) p = 80 cm: q = (80)(−30)/(80 − (−30)) = −21.8 cm. The image is virtual.
hi = (3)(−(−21.8)/80) = 0.82 cm. Upright and diminished.
(c) p = 15 cm: q = (15)(−30)/(15 − (−30)) = −10 cm. The image is virtual.
hi = (3)(−(−10)/15) = 2 cm. Upright and diminished.
15.) ho = 0.5 cm, p = 25 cm, q = +200 cm. Thus, M = −q/p = −200/25 = −8. (a)
The size of the image is hi = ho M = (0.5)(−8) = −4 cm. (b) The image is inverted.
16.) ho = 0.4 cm, p = 12 cm. Since it is a convex mirror the image must be virtual, so
q = −5 cm. (a) hi = ho (−q/p) = (0.4)(5/12) = 0.17 cm. (b) Image is virtual and (c) upright.
18.) Concave mirror with R = 60 cm means f = +30 cm. A concave mirror creating
a real image will invert it, so M = −4. Use the magnification equation and the mirror
equation to solve for p and q. M = −4 = −q/p so q = 4p. Plug into the mirror equation: 1/f = 1/30 = 1/p + 1/q = 1/p + 1/4p = 5/4p or p = (30)(5)/4 = 37.5 cm. Then
q = 4p = 150 cm.
20.) For a convex mirror of R = 60 cm, f = −30 cm. q = −p/2. (a) 1/f = 1/(−30) =
1/p + 1/q = 1/p + (−2)/p = −1/(p). So p = 30 cm. (b) M = −q/p = 1/2
25.) λv = 589 nm, µglass = 1.55. vglass = c/µglass = 1.94 × 108 m/s. f = c/λv =
(3 × 108 )/(589 × 10−9 ) = 5.09 × 1014 Hz. λglass = λv /µglass = 380 nm.
29.) Use Snell’s law: µb sin θb = µw sin θw . θb = 48◦ . From the table in the chapter,
µb = 1.50, µw = 1.33. θw = sin−1 (µb sin θb /µw ) = 56.9◦
30.) θc = sin−1 (µw /µb ) = 62.457◦
45.) The critical angle for total internal reflection off the inside edge of the prism is
θc = sin−1 (1/1.4) = 45.5847◦ . The light will be totally internally reflected off the inside
of the prism until the incident angle on a side is less than 45.5847◦ .
The ray first enters at a 90 degree angle so there is no change in angle. The ray then strikes
the right side of the right side of the prism, forming a 90-50-40 triangle with the top part
of the prism. The angle of incidence on the right side is 50◦ (measure with respect to the
normal) and so it is totally internally reflected at an angle of 50◦ (law of reflection).
The ray next strikes the bottom surface of the prism, forming a 40-65-75 triangle with the
bottom right of the prism. The angle of incidence there is 15◦ and so total internal reflection
does not occur. Snell’s law gives the exit angle, θexit = sin−1 (sin(15)/1.4) = 21.24◦ . (Draw
it!)
47.) The ray cannot escape if the incident angle is so large that the incident angle on the glassair interface exceeds the critical angle there. At this interface, θc = sin−1 (1/1.6) = 38.68◦ .
Geometry shows that both angles inside the glass are the same, so the maximum incident
angle occurs when the refracted angle from the water-glass interface is θc . Using Snell’s law,
θ = sin−1 (µg sin(θc )/µw ) = 48.75◦ , the incident angle must be less than this angle.
2