32-15 We can reasonably model a 60 W incandescent light-bulb as a sphere 5.2 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. a) What is the visible light intensity at the surface of the bulb? b) What is the am-plitude of the electric field at this surface, for a sinusoidal wave with this intensity? c) What is the amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity?
(a) The intensity of light is given by the power per unit area, so the visible light intensity Ivis at the surface of the light bulb is the power due to visible light divided by the surface area of the light bulb:
Ivis =Pvis
A = 0.05Ptot
4π(d/2)2 = (0.05)(60.0)
4π(0.026)2 = 353 W/m2. (b) The intensity at the surface of the bulb calculated above is just the magnitude ¯S of the time averaged Poynting vector, which is related to the electric field amplitude E0 by
S =¯ 120cE02.
Using this expression we can solve for the amplitude of the electric field:
(c) The amplitudes of the electric and magnetic fields are related by
B0= E0
c = 1.7 µT.
We could have found the magnetic field amplitude directly from ¯S using an alternate expression for ¯S:
S =¯ 12 c
µ0B02 =⇒ B0=
√ 2µ0S¯
c .
32-41 A small helium-neon laser emits red visible light with a power of 3.40 mW in a beam that has a diameter of 2.00 mm. (a) What is the amplitude of the electric field of the light? (b) What is the amplitude of the magnetic field of the light? (c) What is the average energy density associated with the electric field? (d) What is the average energy density associated with the magnetic field? (e) What is the total energy contained in a 1.00 m length of the beam?
(a) The amplitude of the electric field of the light is related to the time average of the Poynting vector ¯S by
S =¯ 1 20cE02.
The time average of the Poynting vector is in units of power per unit area; it can be calculated by dividing the power given by the area of the beam:
S =¯ .0034
π0.0012W/m2= 1082 W/m2.
Solving the above equation for electric field yields E0=
(b) One can find the amplitude of the magnetic field in a similar manner, using another relation from the equation sheet:
(c) The energy density associated with the electric field is uelec =120E2.
However, this expression gives the instantaneous energy den-sity at the time when the electric field magnitude is E. To find the average energy density associated with the elec-tric field, you must use the time-averaged value of E2. Be-cause the electric field is sinusoidal, E(t) = E0cos(ωt) and E(t)2= E20cos2(ωt). The time average of cos2ωt =12, so
hueleci = 120hE2i =140E02= 1.80× 10−6J/m3
(d) Similarily, the average density associated with the mag-netic field is
humagi = 1
2µ0hB2i = 1 4µ0
B02= 1.80× 10−6J/m3. Note that the average energy in the electric field is the same as the average energy in the magnetic field.
(e) To find the total energy contained in a 1.00 m length of the beam, use the total (average) energy density multiplied by the volume of that length. The total energy density is
hutoti = hueleci + humagi = 3.60 × 10−6J/m3 The volume of a 1.00 m length of beam is
V = πr2L = π× .0012× 1.0 = 3.14 × 10−6m3, and the total energy is
hutotiV = 1.13 × 10−11J.
December 2, 2011
32-18 A sinusoidal electromagnetic wave from a radio sta-tion passes perpendicularly through an open window that has area of 0.500 m2. At the window, the electric field of the wave has an rms value 2.60× 10−2 V/m. How much energy does this wave carry through the window during a 30.0 s commercial?
The Poynting vector gives the energy per unit area per unit time carried by the electromagnetic wave. To get the energy carried through the window in 30 s, we just multiply the magnitude ¯S of the time-averaged Poynting vector by the area of the window and then by the length of the time inter-val. (Since the wave propagates perpendicular to the win-dow, the surface integral over the window is∫
S· dA = ¯SA.) S is given on the equation sheet in terms of the electric field¯ amplitude by
S =¯ 120cE02.
Since we are given Erms, it is convenient to replace the peak amplitude E0by Erms, using Erms= E0/√
2. The result is S = ¯ 0cErms2 .
The energy is then
E = ¯SA∆t = 0cErms2 A∆t
= (8.85× 10−12)(3.00× 108)(0.026)2(0.500)(30.0)
= 2.69× 10−5J.
34-5 An object 0.550 cm tall is placed 17.0 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Determine the position of the image. (b) Determine the size of the image. (c) Determine the orientation of the image. (d) Determine the nature (real or virtual) of the image. (e) Make a ray diagram and bring it to recitation.
Since we are dealing with mirrors, the equations we will use are
1 do + 1
di = 1
f and m = hi
ho =−di
do
The focal length f is positive since the mirror is convex, and f = R
2 = 22.0 cm
2 = 11.0 cm
We take do = 17.0 cm to be positive since it is in front of the mirror. We take ho= 0.550 cm to be positive since the object is always assumed to be upright.
(a) 1 di
= 1 f − 1
do
=⇒ di= f do
do− f = 11.0 cm· 17.0 cm
17.0 cm− 11.0 cm = 31.2 cm (b)
hi=−ho
di do
=−0.550 cm31.2 cm
17.0 cm =−1.01 cm
The fact that we get a negative result means that the image is inverted. Mastering Physics only asks for the size of the image in this part, so enter the absolute value of the result.
(c) The image is inverted.
(d) The fact that we got a positive result in part (a) means that the image is in front of the mirror and it is a real image.
(This image could be focussed on a screen.) (e) Here is the ray diagram for this situation:
YF 34-8 An object is a distance of 25.0 cm from the CEN-TER of a silvered spherical glass Christmas tree ornament which has a diameter of 5.70 cm. (a) What is the position of its image? Use the mirror equation to answer this question, but draw a ray diagram and bring it with you to recitation.
Hint: be careful to determine do (the distance to the SUR-FACE of the mirror) correctly. (b) What is the magnification of its image?
YF 34-14 A spherical, concave, shaving mirror has a ra-dius of curvature of 32.5 cm. (a) What is the magnification of a person’s face when it is a distance do = 11.6 cm from the center of the mirror? (b) What is the distance di to the image? (c) Is the image real or virtual? Use the mirror equation to answer this question, but draw a ray diagram and bring it with you to recitation.
Work part (b) first:
(b) The focal length is half the radius of curvature, or
−16.25 cm. (distances in cm.) The sign is positive for a concave mirror.
1 di
= 1 f − 1
do
= 1
16.25− 1
11.6= 1
−40.5 The image is 40.5 cm behind the mirror.
(a) The magnification is m =−di
do
=40.5
11.6 = 3.49.
(c) The image is virtual.