Unit 2 Electromagnetic Waves
Worksheet Answers
1. Some neodymium–glass lasers can provide 100 TW of
power in 1.0 ns pulses at a wavelength of 0.26 mm. How much energy
is contained in a single pulse?
If P is the power and Δt is the time interval of one pulse, then the energy in a pulse is
E = PΔt
E = (100 X 10^{12}) (1 X 10^{9})
E = 1.00 X 10^{5} J
2. What is the radiation pressure 1.5 m away from a 500 W
lightbulb? Assume that the surface on which the pressure is exerted
faces the bulb and is perfectly absorbing and that the bulb radiates
uniformly in all directions.
Since the surface is perfectly absorbing, the radiation pressure is given by p_{r} = [I/c] where I is the intensity. Since the bulb radiates uniformly in all directions, the intensity a
distance r from it is given by I = P/4πr^{2}, where P is the power of the bulb.
Thus p_{r} = [P/(4πr^{2}c)
p_{r} = [(500W)/(4π(1.5m)^{2}(3X10^{8})
p_{r} = 5.9X10^{8} Pa.
3. If the magnetic ﬁeld of a light wave oscillates parallel to a y axis
and is given by B_{y} =
B_{m}sin(kz  ωt), (a) in what direction does the
wave travel and (b) parallel to which axis does the associated electric ﬁeld oscillate?
(a) From the equation B_{y} =
B_{m}sin(kz  ωt) the magnetic field vibrates on the y axis as the wave travels along the z direcction.
(b) The associated electric ﬁeld oscillates parallel to the x axis.
Note:
1. The cross product EXB always gives the direction in which the wave travels (you know this as the right hand screw rule (also called righthand grip rule, coffeemug rule, or the corkscrewrule).
2. The electric and magnetic ﬁelds and are always perpendicular to the
direction in which the wave is traveling. Thus, the wave is a transverse wave.
3. The electric ﬁeld is always perpendicular to the magnetic ﬁeld.
4.
The ﬁelds always vary sinusoidally, with the same frequency and in phase
(in step) with each other.
4. If the electric and magnetic ﬁelds of an electromagnetic wave at a certain instant are in the directions shown in the diagram, what is the direction of the wave?.
The wave is traveling into the page.
5. A certain helium – neon laser emits red light in a narrow band
of wavelengths centered at 632.8 nm and with a “wavelength
width” of 0.0100 nm. What is the
corresponding “frequency width” for the emission?
Frequency width = Upper frequency  Lower frequency
Frequency width = c/λ_{u}  c/λ_{L}
Frequency width = 7.49 X 10^{9} Hz
6. Project Seafarer was an ambitious program to construct an
enormous antenna, buried underground on a site about 10 000
km^{2} in area. Its purpose was to transmit signals to submarines
while they were deeply submerged. If the effective wavelengthwere 1.0 X 10^{4} Earth radii, what would be the (a) frequency and
(b) period of the radiations emitted? Ordinarily, electromagnetic
radiations do not penetrate very far into conductors such as seawater,
and so normal signals cannot reach the submarines.
(a) Frequency = c/λ
Frequency = 3.0 X 10^{8}ms^{1}/1.0 X 10^{4} (6.4 X 10^{6}m)
Frequency = 4.7 X 10^{3} Hz
(b) T = 1/f
T = 1/4.7 X 10^{3} Hz
T = 213 s
7. From the diagram, approximate the (a) smaller and (b)
larger wavelength at which the eye of a standard observer has half
the eye’s maximum sensitivity. What are the (c) wavelength, (d)
frequency, and (e) period of the light at which the eye is the most
sensitive?
(a) The smaller wavelength in question to be about 515 nm.
(b) The larger wavelength is approximately 610 nm.
(c) The wavelength at which the eye is most sensitive is about 555 nm.
(d)
Frequency = c/λ
Frequency = 3.0 X 10^{8}ms^{1}/5.55 X 10^{7}
Frequency = 5.41 X 10^{14} Hz
(b) T = 1/f
T = 1/5.41 X 10^{14}Hz
T = 1.85 X 10^{15 }s
8. A plane electromagnetic wave traveling in the positive
direction of an x axis in vacuum has components E_{x} = E_{y} = 0 and E_{z} = (2.0 Vm^{1}) cos[(π X 10^{15} s^{1})(t  x/c)]. (a) What is the amplitude of the magnetic ﬁeld component? (b) Parallel to which axis does the
magnetic ﬁeld oscillate? (c) When the electric ﬁeld component is in
the positive direction of the z axis at a certain point P, what is the
direction of the magnetic ﬁeld component there?
(a) The amplitude of the magnetic field is B_{m} = [E_{m} / c]
B_{m} = [(2.0 Vm^{1}) / (3.0 X 10^{8}ms^{1})]
B_{m} = 6 .7 X 10^{9}T
(b) Since the Ewave oscillates in the z direction and travels in the x direction, we have B_{x} = B _{y} = 0. So, the oscillation of the magnetic field is parallel to the y axis.
(c) The direction (+x) of the electromagnetic wave propagation is determined by
EXB_{x}. If
the electric field points in +z, then the magnetic field must point in the –y direction.
Note:
1. The cross product EXB always gives the direction in which the wave travels (you know this as the right hand screw rule (also called righthand grip rule, coffeemug rule, or the corkscrewrule).
2. The electric and magnetic ﬁelds and are always perpendicular to the
direction in which the wave is traveling. Thus, the wave is a transverse wave.
3. The electric ﬁeld is always perpendicular to the magnetic ﬁeld.
4.
The ﬁelds always vary sinusoidally, with the same frequency and in phase
(in step) with each other.
9. 20. A distant galaxy is moving away from us at approximately 5x10^{7}ms^{1} and we approximate the speed of light as c = 3x10^{8}ms^{1}. (a) What is the resulting wavelength of the Hydrogen spectral line of λ = 434 nm? (b) what would be the shift if the galaxy was moving towards earth?
(a) f_{source} = ^{c}/_{λ}
f_{source} = 6.91x10^{14}Hz
f_{observed} = f_{source}{^{c}/_{[c±Vs]}}
f_{observed} = 6.91x10^{14}Hz{(3x10^{8}ms^{1})/[(3x10^{8}ms^{1})+(5.0x10^{7}ms^{1})]}
f_{observed} =5.92x10^{14}Hz
λ_{observed} = ^{c}/f_{observed}
λ_{observed} = 507 nm
The frequency has shifted from violet to green.
(b) f_{observed} = f_{source}{^{c}/_{[c±Vs]}}
f_{observed} = 6.91x10^{14}Hz{(3x10^{8}ms^{1})/[(3x10^{8}ms^{1})(5.0x10^{7}ms^{1})]}
f_{observed} =8.29x10^{14}Hz
λ_{observed} = ^{c}/f_{observed}
λ_{observed} = 362 nm
The frequency has shifted from violet to ultraviolet.
10. A galaxy is moving away from the Earth at 2.6x10^{7}ms^{1}. Calculate the wavelength and frequency change of a 650 nm line in its spectrum. Take c = 3x10^{8}ms^{1}.
f_{source} = ^{c}/_{λ}
f_{source} = 4.6x10^{14}Hz
f_{observed} = f_{source}{^{c}/_{[c±Vs]}}
f_{observed} = 4.6x10^{14}Hz{(3x10^{8}ms^{1})/[(3x10^{8}ms^{1})+(2.6x10^{7}ms^{1})]}
f_{observed} =4.2x10^{14}Hz
λ_{observed} = ^{c}/f_{observed}
λ_{observed} = 709 nm
The frequency has shifted from red to infrared.
