Φ21 Fall 2006
1
HW23 Solutions
Problem 34.46
Assume that a 120 W light bulb radiates all its energy as a single wavelength of visible light. Part A.
Estimate the electric eld strength at the surface of the bulb. Assume that the bulb is a sphere and its
radius is 2.8 cm.
The electric eld strength can be calculated directly from the intensity of the light, measured in
W/m2 . Just from the units, a uniform intensity is the power divided by the area through which the power
ows, in this case the surface of the light bulb. The intensity is equal to the average of the Poynting Vector.
Solution:
P
120 W
1
=
= 12180 W/m2 = ²0 cE02
2
A
2
4π (0.028 m)
S=
The electric eld strength is then
√
E0 =
Part B.
√
2S
=
²0 c
2 (12180 W/m2 )
= 3029 N/C
(8.85 × 10−12 C 2 / (N · m2 )) (3.00 × 108 m/s)
Estimate the magnetic eld strength at the surface of the bulb.
From the two formulas for S , the relationship between the electric and magnetic elds in an
electromagnetic wave is
Solution:
B0 =
2
E0
3029 N/C
=
= 1.01 × 10−5 T
c
3 × 108 m/s
Problem 23.6
What is the angle φ in Figure 1?
Draw the normal line perpendicular to the mirror. By using complimentary
(add to 90◦ ) and supplimentary (add to 180◦ ) angles and opposite interior angles, it is
Figure 1:
A
easy to show that φ = 30◦ .
mirror tilted at
an angle diverts
light by 60◦ .
Solution:
1
3
Raytracing for a Concave Mirror
An object O is placed at the location shown in front of a concave spherical mirror. Use ray tracing to
determine the location and size of the reected image. Solutions are Shown.
Figure 2: Ray 2 passes through (or proceeds toward) the focal point F and is reected parallel to the axis.
Figure 3: Ray 1 is parallel to the axis; after reection, it passes through the focal point F of a concave mirror
or appears to come from the (virtual) focal point of a convex mirror.
Figure 4: The image is formed at the intersection of the two reected rays. In some cases, the image may be
behind the mirror, in which case the reected rays only appear to be coming from the image because there
is no light behind the mirror.
2
4
Raytracing for a Convex Mirror
An object O is placed at the location shown in front of a convex spherical mirror. Use ray tracing to
determine the location and size of the image in the mirror. Solutions are Shown.
Figure 5: Ray 2 passes through (or proceedes toward) the focal point F and is reected parallel to the axis.
(Note that the website requires the length of Fr to be about what is pictured.)
Figure 6: Ray 1 is parallel to the axis; after reection, it passes through the focal point F of a concave
mirror or appears to come from the (virtual) focal point of a convex mirror. (When you construct Pr , rst
place the tail on the focal point, then move the arrowhead so that the vector crosses the tip of Pi , then x
the tail so that there is no light behind the mirror.)
Figure 7: To nd the image, extend the reected rays until they cross. This crossing is the location of the
(virtual) image.
3