Teach Yourself Electricity and Electronics, 5th edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 33
1. In a pulsed laser, the ratio of the peak output power to the average output power depends on the duty cycle (the proportion of time that the laser actually produces output), assuming all other factors remain constant. The correct choice is (d).
2. A helium-neon (He-Ne) laser produces output at a wavelength of 633 nm. The correct choice is (c). Some He-Ne lasers also produce output at other wavelengths, but not at 1155 nm (a) or 793 nm (b), and not in the UV range (d).
3. If we reverse-bias a laser diode and keep the voltage below the avalanche point, we get no current through the device, and we observe no radiation from it. In fact, under these conditions, the laser diode behaves exactly as an ordinary diode does. The correct choice is (d).
4. In a pulsed cavity laser (or any other type of cavity laser), the output wavelength depends on the length of the cavity. The correct choice is (a). The wavelength doesn't depend on the duty cycle (b), the peak energy output (c), or the characteristic shape of the output pulses (d).
5. In a gallium-arsenide (GaAs) laser, the high carrier mobility (that is, the speed with which the charge carriers move) makes the task of modulation easy. The correct choice is (b).
6. When an electron moves from a certain shell to a smaller shell within an atom, that electron loses energy. In the process, it radiates a photon having a specific wavelength corresponding to the amount of energy that it loses. The correct choice is (b).
7. If know the peak output power (Ppk-out) and the average output power (Pavg-out) from a pulsed laser with rectangular pulses, then we can calculate the duty cycle as

D = Pavg-out / Ppk-out

We're told that the peak output power from a particular laser having rectangular pulses is 660 GW, or 6.60 x 1011 W. We're also told that the average output power is 33 W. To obtain the duty cycle, we divide the latter quantity by the former, getting

D = 33 / (6.60 x 1011)
= 5.0 x 10-11

The correct choice is (d).
8. We can determine the efficiency eff of any laser device by dividing the average output power Pavg-out by the average input power Pavg-in to get

eff = Pavg-out / Pavg-in

The statement of Question 7 (the previous problem) tells us that Pavg-out = 33 W. The statement of Question 8 (the problem at hand) tells us that Pavg-in = 1.0 kW, which equals 1000 W. We can therefore calculate the efficiency to two significant figures as

eff = 33 / 1000
= 0.033

The correct choice is (b).
9. By definition, coherent radiation occurs in such a way that all the waves have the same frequency, and they all follow each other in phase coincidence. The correct choice is (c).
10. A typical vertical-cavity surface-emitting laser (VCSEL) has a narrower beam, at great distances, than a typical injection laser operating a similar wavelength. In other words, the VCSEL has less beam divergence. The correct choice is (a).
11. We want to find the average output power, Pavg-out. We know that the peak pulse power output equals 1200 W and the device works at a duty cycle of 0.020. Let's set Ppk-out = 1200 and D = 0.020, and then plug these numbers into the formula

D = Pavg-out / Ppk-out

to get

0.020 = Pavg-out / 1200

We can rewrite this equation as

Pavg-out = 1200 x 0.020

which solves to

Pavg-out = 24 W

The correct choice is (d).
12. As we did in the solution to Question 8, we can determine the efficiency eff with the formula

eff = Pavg-out / Pavg-in

In the solution to Question 11, we determined that Pavg-out = 24 W. The statement of this question tells us that Pavg-in = 30 W. We can therefore calculate the efficiency as

eff = 24 / 30
= 0.8

The correct choice is (a).
13. To calculate the power (in watts) that a gas-discharge tube dissipates, we multiply the applied voltage (in volts) by the current that flows through the gas (in amperes). In this case, we apply 5.0 kV DC, which equals 5000 V DC, and get a current of 2.0 mA, which equals 0.0020 A. The dissipated power P is therefore

P = 5000 x 0.0020
= 10 W

The correct choice is (c).
14. This problem constitutes a "repeat" of Question 7 with different numbers. If know the peak output power (Ppk-out) and the average output power (Pavg-out) from a pulsed laser with rectangular pulses, then the duty cycle is

D = Pavg-out / Ppk-out

We're told that the peak output power from a particular laser having rectangular pulses is 500 GW, or 5.00 x 1011 W. The average output power is 30.0 W. We can therefore calculate the duty cycle as

D = 30.0 / (5.00 x 1011)
= 6.00 x 10-11

The correct choice is (b).
15. Let's turn the formula for efficiency "inside-out." Recall that we can calculate the efficiency eff by dividing the average output power Pavg-out by the average input power Pavg-in to obtain

eff = Pavg-out / Pavg-in

Using algebra, we can rewrite this formula as

Pavg-in = Pavg-out / eff

Now we can plug in the values Pavg-out = 30.0 and eff = 0.0300 to get

Pavg-in = 30.0 / 0.0300
= 1000 W
= 1.00 kW

The correct choice is (b).
16. If we reverse-bias a laser diode, the laser diode will behave essentially as an ordinary diode does. Beyond the avalanche voltage, the device will conduct current, but it will not radiate. The correct choice is (c).
17. Energy radiates from ionized hydrogen gas in the form of photons having multiple discrete wavelengths w according to the Rydberg-Ritz formula

w = [RH (n1-2 - n2-2)]-1

where we express w in meters, RH represents the Rydberg constant, n1 represents the principal quantum number of the smaller shell to which a particular electron "falls," and n2 represents the principal quantum number of the larger shell from which that same electron has just "fallen." For our purposes, we can assume that

RH = 1.10 x 107

In this case, we're told that an electron falls from a shell with n2 = 4 to a shell with n1 = 2. Plugging in these values to the Rydberg-Ritz formula, we get

w = [1.10 x 107 (2-2 - 4-2)]-1
= 1 / [1.10 x 107 (1/4 - 1/16)]
= 1 / (3/16 x 1.10 x 107)
= 1 / (2.0625 x 106)

When we work out this quotient and round off to three significant figures, we get

w = 4.85 x 10-7 m
= 485 nm

The correct choice is (a).
18. Look back at the solution to Question 11 for reference. This time, we want to find the peak output power, Ppk-out. We know that the average pulse power output equals 24 W (it hasn't changed) and the device works at a duty cycle of 0.010 (half of what it was before). Now we can set Pavg-out = 24 and D = 0.010, and then plug these numbers into the formula

D = Pavg-out / Ppk-out

to get

0.010 = 24 / Ppk-out

We can rewrite this equation as

Ppk-out = 24 / 0.010

which solves to

Ppk-out = 2400 W

The peak output power has doubled from 1200 W, its value in the scenario described by Question 11. The correct choice is (c).
19. Look back at Question 11 for reference. Now we want to find the efficiency, eff. The average pulse power output, Pavg-out, was 24 W in that situation, and it hasn't changed. The average input power, Pavg-in, was 30 W back then, and it's still the same now. Although the duty cycle has changed, it has no effect on the calculation of efficiency in this particular situation. We have simply

eff = Pavg-out / Pavg-in
= 24 / 30
= 0.8

which is the same as it was in the scenario of Question 11. The correct choice is (b).
20. At first, the P-N junction does not conduct, and the device doesn't radiate at all. When forward breakover occurs, current begins to flow, and the diode emits incoherent infrared (IR). When the voltage rises to a certain critical point, the IR emission becomes coherent and we observe lasing. The correct choice is (d).