Unit 1 Worksheet
Unit 1 Worksheet Answers
Unit 1.3.5 Doppler Effect Worksheet
Unit 1.3.5 Doppler Effect Worksheet Answers
Unit 2 Worksheet
Unit 2 Worksheet Answers
Final Worksheet
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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 1012) (1 X 10-9)
E = 1.00 X 105 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 pr = [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/4πr2, where P is the power of the bulb.
Thus pr = [P/(4πr2c)
pr = [(500W)/(4π(1.5m)2(3X108)

pr = 5.9X10-8 Pa.

3. If the magnetic field of a light wave oscillates parallel to a y axis and is given by By = Bmsin(kz - ωt), (a) in what direction does the wave travel and (b) parallel to which axis does the associated electric field oscillate?
em(a) From the equation By = Bmsin(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 right-hand grip rulecoffee-mug rule, or the corkscrew-rule).
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.

em4. 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/λu - c/λL
Frequency width = 7.49 X 109 Hz

6. Project Seafarer was an ambitious program to construct an enormous antenna, buried underground on a site about 10 000 km2 in area. Its purpose was to transmit signals to submarines
while they were deeply submerged. If the effective wavelengthwere 1.0 X 104 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 108ms-1/1.0 X 104 (6.4 X 106m)
Frequency = 4.7 X 10-3 Hz
(b) T = 1/f
T = 1/4.7 X 10-3 Hz
T = 213 s

em7. 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?

em(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 108ms-1/5.55 X 107
Frequency = 5.41 X 1014 Hz
(b) T = 1/f
T = 1/5.41 X 1014Hz
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 Ex = Ey = 0 and Ez = (2.0 Vm-1) cos[(π X 1015 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 Bm = [Em / c]
Bm = [(2.0 Vm-1) / (3.0 X 108ms-1)]
Bm = 6 .7 X 10-9T
(b) Since the E-wave oscillates in the z direction and travels in the x direction, we have Bx = 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 EXBx. 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 right-hand grip rulecoffee-mug rule, or the corkscrew-rule).
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 5x107ms-1 and we approximate the speed of light as c = 3x108ms-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) fsource = c/λ

fsource = 6.91x1014Hz

fobserved = fsource{c/[c±Vs]}

fobserved = 6.91x1014Hz{(3x108ms-1)/[(3x108ms-1)+(5.0x107ms-1)]}

fobserved =5.92x1014Hz

λobserved = c/fobserved

λobserved = 507 nm

The frequency has shifted from violet to green.

(b) fobserved = fsource{c/[c±Vs]}

fobserved = 6.91x1014Hz{(3x108ms-1)/[(3x108ms-1)-(5.0x107ms-1)]}

fobserved =8.29x1014Hz

λobserved = c/fobserved

λobserved = 362 nm

The frequency has shifted from violet to ultraviolet.

10. A galaxy is moving away from the Earth at 2.6x107ms-1. Calculate the wavelength and frequency change of a 650 nm line in its spectrum. Take c = 3x108ms-1.

fsource = c/λ

fsource = 4.6x1014Hz

fobserved = fsource{c/[c±Vs]}

fobserved = 4.6x1014Hz{(3x108ms-1)/[(3x108ms-1)+(2.6x107ms-1)]}

fobserved =4.2x1014Hz

λobserved = c/fobserved

λobserved = 709 nm

The frequency has shifted from red to infrared.



     
Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.