Chapter 23
Problems: 5, 11, 12, 20, 25, 39, 48, 49, 59, 62, 64, 66
5. Strategy With respect to the normal at the point of incidence, the angle of incidence equals the angle of
reflection and the reflected ray lies in the same plane, the plane of incidence, as the incident ray and the
normal. Every point on a wavefront is considered a source of spherical wavelets. A surface tangent to the
wavelets at a later time is the wavefront at that time.
11. Strategy Use Snell’s law, Eq. (23-4).
Solution Find the angle the Sun’s rays in air make with the vertical.
12. Strategy Use Snell’s law, Eq. (23-4). Let the index of refraction for air be n, and let
1.20, 1.40, 1.32, and 1.28, respectively. Let
be
be the angle of emergence.
Solution Find the angle the beam makes with the normal when it emerges into the air after passing through
the entire stack of four flat transparent materials.
20. Strategy The speed of light in matter is given by
Use Snell’s law, Eq. (23-4).
Solution
(a) Find the indices of refraction.
For red light:
For blue light:
(b) To get the ratio of the speeds, express the speeds using the indices of refraction.
(c) With no dispersion, all colors would undergo the same refraction. The diamond would be clear, with no
color.
25. Strategy Total internal reflection occurs when the angle of incidence is greater than or equal to the critical
angle. Use Eqs. (23-5).
Solution
(a) Find the index of refraction of the glass.
(b) No; rays from the defect could reach all points above the glass since
39.
Strategy For a plane mirror, a point source and its image are at the same distance from the mirror
(on opposite sides) and both lie on the same normal line. Draw a ray diagram. Use geometry and the laws of
reflection. (See last page for sketch.)
48. Strategy The object distance is
and the magnification is
mirror equations. The focal length is half of the radius of curvature.
Use the magnification and
Solution Use the magnification equation to find the image distance.
Find the focal length.
Compute the radius of curvature.
49. Strategy Since the image is real, the image distance is positive and the image is inverted (negative). The
image is twice the size of the object, so
Use the magnification and mirror equations.
Solution Relate the object and image locations.
Find the distance of the object from the mirror.
The object is 18.8 cm in front of the mirror.
59.Strategy Sketch a ray diagram for a converging lens using the three principal rays. Place the object twice the
focal length from the lens.
Solution As can be seen in a ray diagram (see
last page), when an object is placed twice the
focal length away from a converging lens, an
inverted, real, and same-sized image is formed.
62. Strategy The object distance is
thin lens equation.
Since the image is virtual, the image distance is
Use the
Solution Find the focal length of the lens.
64. Strategy All images are virtual since a diverging lens is used. The magnification determines the size and
orientation of the image. Use the thin lens and magnification equations.
Solution
(a) Solve the thin lens equation for q.
Substitute the given values.
For p-values of 14.0 cm, 16.0 cm, and 20.0 cm, the corresponding q values are
and
respectively. The results are summarized in the table.
p (cm)
q (cm)
Real or virtual
Orientation
Relative size
5.00
–3.08
0.615
virtual
upright
diminished
8.00
– 4.00
0.500
virtual
upright
diminished
14.0
–5.09
0.364
virtual
upright
diminished
16.0
–5.33
0.333
virtual
upright
diminished
20.0
–5.71
0.286
virtual
upright
diminished
(b) Solving the magnification equation for the image height yields
For
,
, and
For
,
, and
, we have
, we have
66. Strategy Sketch a ray diagram for a converging lens using two principal rays. Place the object a distance less
than the focal length from the lens.
Solution As can be seen in a ray diagram (see last page), the
image is upright. It is virtual since the rays only seem to
come from the image.