NYCSampleFinalExam_v2

Dawson College

Physics 203-NYC-05 Waves, Optics & Modern Physics

Sample Final ExaminationThis exam is divided into two parts:

Part I: Problems (10 marks each) Solve all six problems. Show all of your work, clearly andin order, to receive full marks. If you use a formula not given on the formula sheet, a derivationmust be shown.

Part II: Multiple Choice Questions (2 marks each) Answer all twenty questions. Circlethe best response from the choices given. If your final selection is unclear you will notbe given the marks. No marks will be awarded for diagrams, calculations, or reasoning.

Additional instructions:

1. The time allotted for this examination is three hours.

2. Answer directly on the question sheet. If you need extra room, you may use the back ofanother page.

3. Use vsound = 343 m/s.

4. When finished, return this entire package and the formula sheet to your instructor.

Good luck! ,

P1 P2 P3 P4 P5 P6 MC Total /100

203-NYC Sample Final Page 1 of 22

Part I: Problems (10 marks each)

Solve all six problems. Show all of your work, clearly and in order, to receive full marks. If youuse a formula not given on the formula sheet, a derivation must be shown.

1. A transverse wave with an amplitude of 2.00cm travels down a string under 50.0N of tension.The distance between one wave crest and the next is 70.0 cm. A fixed point on the stringoscillates 200 cycles in 12.0 s.

(a) What are the wave speed and the maximum transverse speed of a point on the string?

(b) Explain the difference between the speeds found in part (a).

(c) What is the mass-per-unit-length of the string?

(d) What is the power transmitted along the string by these oscillations?


2. A speaker emitting sound of frequency 440 Hz is attached to a block that oscillates in simpleharmonic motion with a maximum speed of 6.00 m/s and an amplitude of 0.500 m. Thespeaker moves towards and away from a listener who is standing in front of it. The minimumseparation between the speaker and the listener as it oscillates is 1.50 m.

listener

1.50 m

(a) What are the highest and lowest frequencies heard by the listener?

(b) If the maximum sound intensity level heard by the listener is 60.0 dB when the speakeris closest to him, what is the sound intensity level he hears when the speaker is furthestaway from him?


3. A source, S, emits an electromagnetic wave of wavelength 250 m. The source is located3.00 km from a flat surface that reflects the wave. A receiver, R, is 8.00 km away from thesource along a line parallel to the reflecting surface. The reciever detects the superpositionof two waves: the one that travels directly from S to R and the one that leaves S, reflects offof the surface, and arrives at R. The reflected wave has a phase shift of π radians.

reflecting surfaceS′

3.00 km

8.00 kmS R

(a) What is the phase difference between the two waves arriving at the receiver?

(b) At what distance from S along the line SS′ is the phase difference the same as in (a)?

(c) At how many points along the line SS′ is there constructive interference?

(d) At how many points along the line SR is there constructive interference?


4. Shown below is the path of a ray of orange light (λ = 620 nm) passing through a transparentplastic cylinder with an index of refraction of 1.63. The radius of the cylinder is 0.20 mm.

R

R

θ1

θ2d

n = 1.63

(a) What is the angle of incidence θ1? [Hint : The identity sin(2θ) = 2 sin θ cos θ is useful]

(b) What is the distance d between the incident ray and a line parallel to it and passingthrough the centre of the cylinder?

(c) What are the frequency and the wavelength of the light inside the cylinder?


5. Hydrogen emits visible light at specific wavelengths. The wavelengths are traditionally namedas follows: Hα = 656.3 nm (red), Hβ = 486.1 nm (cyan), Hγ = 434.1 nm (blue), Hδ =410.2 nm (violet). Light from hydrogen passes through a diffraction grating creating anemission spectrum of fringes.

(a) If the second-order Hα line is observed at 41.018◦, what is the number of lines permillimetre of the grating?

(b) What is the angular separation of the Hγ and Hδ lines in the 1st order? [Keep at least3 decimal places in your work.]

(c) How many complete orders are visible on one side of the centre of the spectrum?

(d) Are there any overlapping orders? If yes, which ones?


6. A by-product of some nuclear power reactors is the isotope 239Pu. This isotope has a half-lifeof 24120 years and decays by alpha emission: 239Pu → 235U + α. Suppose a sample of thisplutonium isotope has a mass of 0.7939109 kg at t = 0.

(a) How many atoms are initially in the sample?

(b) What is the initial activity of the sample in Becquerels?

(c) How many years does it take for the activity to decrease to 0.100Bq? [Use the conversionfactor: 1 year = 365.25 days]

(d) What the total energy released between t = 0 and the time found in part (c)? [Use themasses: mPu = 239.052157 u, mU = 235.043924 u, mα = 4.002602 u]


Part II: Multiple Choice Questions (2 marks each)

Answer all twenty questions. Circle the best response from the choices given. If your final selectionis unclear you will not be given the marks. No marks will be awarded for diagrams, calculations,or reasoning.

1. A mass at the end of an ideal spring vibrates with period T . If you double the mass, howmust you change the force constant of the spring to achieve a period of 2T?

(a) Decrease it by a factor of 4.

(b) Increase it by a factor of√

2.

(c) Decrease it by a factor of√

2.

(d) Decrease it by a factor of 2.

(e) Increase it by a factor of 2.

2. Which of following graphs represents simple harmonic motion with an amplitude of 4 cm andan angular frequency of 2 rad/s?

x (cm)

-4

-2

2

4

t2 4 6 8 10

(a)

x (cm)

-4

-2

2

4

t2 4 6 8 10

(b)

x (cm)

-4

-2

2

4

t (s)2 4 6 8 10

(c)

x (cm)

-4

-2

2

4

t2 4 6 8 10

(d)

x (cm)

-4

-2

2

4

t2 4 6 8 10

(e)


3. Which of the following is a false statement?

(a) The speed of a wave and the speed of the vibrating particles that constitute the waveare different entities.

(b) For a transverse wave the particle motion is perpendicular to the direction of propagationof the wave.

(c) A wave in which the particle motion is parallel to the direction of propagation of thewave is called a longitudinal wave.

(d) Waves transport energy and matter from one region to another.

(e) Not all waves are mechanical waves.

4. Four separate travelling waves are described by the following equations, where y(x, t) repre-sents the displacement:

I: y(x, t) = (0.12 m) cos [(3.0 rad/m)x+ (2.0 rad/s)t]

II: y(x, t) = (0.15 m) sin [(6.0 rad/m)x− (3.0 rad/s)t]

III: y(x, t) = (0.23 m) cos [(3.0 rad/m)x+ (6.0 rad/s)t]

IV: y(x, t) = (0.29 m) sin [(1.5 rad/m)x+ (1.0 rad/s)t]

Which two waves have the same wave speed?

(a) I and III

(b) I and II

(c) I and IV

(d) III and IV

(e) II and III

5. An open-open tube 0.46 m long supports a standing wave of frequency 1114 Hz. What is thedistance from the centre of the tube to the nearest antinode?

(a) 0.15 m

(b) 0.12 m

(c) 0.077 m

(d) 0.038 m

(e) There is an antinode exactly at the centre of the tube.


6. A 137 cm long wire that is fixed at both ends has a fundamental frequency of 87.5 Hz. Thetension in the wire is then changed so that the fundamental frequency becomes 175.0 Hz.What is the ratio of the new fundamental wavelength to the old fundamental wavelength,λnew/λold?

(a) 1

(b)√

2

(c) 2

(d) 1/√

2

(e) 1/2

7. A car is on a road parallel to a railroad track and right beside it. The car is travelling eastat 30.0 m/s while a train travelling west at 50.0 m/s is approaching it. If the car honks itshorn at a frequency of 1.00 kHz, what is the wavelength of the sound perceived by a personriding in the train?

(a) 0.235 m

(b) 0.275 m

(c) 0.294 m

(d) 0.313 m

(e) 0.344 m

8. If the distance from a point source of sound is tripled, the intensity will be what multiple ofits original value?

(a) 3

(b) 1/3

(c) 1

(d) 9

(e) 1/9

9. At a distance of 2.00 m from a point source of sound, the sound intensity level is 80.0 dB. Ata distance of 4.00 m from the source, what is the sound intensity level?

(a) 74.0 dB

(b) 77.0 dB

(c) 60.0 dB

(d) 20.0 dB

(e) 40.0 dB


10. Which one of the following is a correct ranking of electromagnetic waves from longest wave-length to shortest wavelength?

(a) radio waves, infrared, microwaves, UV, visible, X-rays, gamma rays

(b) radio waves, microwaves, visible, X-rays, infrared, UV, gamma rays

(c) radio waves, infrared, X-rays, microwaves, UV, visible, gamma rays

(d) radio waves, UV, X-rays, microwaves, infrared, visible, gamma rays

(e) radio waves, microwaves, infrared, visible, UV, X-rays, gamma rays

11. A ray of light strikes a boundary between two transparent materials and there is no trans-mitted ray, as shown in the figure. What can you conclude about the indices of refraction ofthese two materials?

n1 n2

(a) n1 ≥ n2(b) n1 > n2

(c) n1 = n2

(d) n1 < n2

(e) n1 ≤ n2


12. A tank holds a layer of oil, 1.88 m thick, which floats on a layer of syrup that is 0.69 m thick.Both liquids are transparent and do not intermix. A ray of light originates from the bottom ofthe tank and crosses the oil-syrup interface a horizontal distance of 0.90 m its starting point.The ray continues and arrives at the oil-air interface a horizontal distance of 2.00 m from thestarting point. The angle of incidence at the oil-air interface is the critical angle. What isthe index of refraction of the oil?

Syrup

Oil

Air

θc

0.69 m

1.88 m

0.90 m

2.00 m

(a) noil = 2.02

(b) noil = 1.94

(c) noil = 2.00

(d) noil = 1.96

(e) noil = 1.98

13. In a double-slit experiment, if the slit separation is increased, which of the following happensto the interference pattern shown on the screen?

(a) The minima and maxima stay at the same positions.

(b) The maxima stay at the same positions.

(c) The minima stay at the same positions.

(d) The minima get closer together.

(e) The maxima get further apart.

14. Light passes through a pair of very thin parallel slits. The resulting interference pattern isviewed far from the slits at various angles θ relative to the centre-line coming outward fromthe midpoint between the slits. The central bright fringe is at θ = 0◦, and the next brightfringes on either side of the centre are at θ = ±15◦. If the central bright fringe has an intensityI0, what is the intensity of the next bright fringe on either side of it?

(a) I0

(b) I0 cos(15◦)

(c) I0 cos2(15◦)

(d) I0/√

2

(e) I0/2


15. A diffraction grating has 450 lines per mm. What is the highest order that contains the entirevisible spectrum from 400 nm to 700 nm?

(a) m = 2

(b) m = 3

(c) m = 4

(d) m = 5

(e) m = 6

16. If the diameter of a radar dish is doubled, what happens to its resolving power assuming thatall other factors remain unchanged? [Use the small angle approximation]

(a) The resolving power is reduced to 1/4 of its original value.

(b) The resolving power is reduced to 1/2 of its original value.

(c) The resolving power doubles.

(d) The resolving power quadruples.

(e) The resolving power does not change unless the focal length changes.

17. A laser pulse of duration 25 ms has a total energy of 1.2 J. If the wavelength of the light is463 nm, how many photons are emitted in that one pulse?

(a) 3.4× 1019 photons

(b) 1.1× 1017 photons

(c) 2.2× 1017 photons

(d) 6.9× 1019 photons

(e) 2.8× 1018 photons

18. Monochromatic light strikes a metal surface and electrons are ejected from the metal. If theintensity of the light is increased, what will happen to the ejection rate and maximum energyof the electrons?

(a) Same ejection rate; same maximum energy

(b) Greater ejection rate; same maximum energy

(c) Same ejection rate; greater maximum energy

(d) Greater ejection rate; greater maximum energy

(e) Electrons would no longer be ejected


19. When a blackbody has a temperature T , the peak intensity of its radiation is at wavelengthλ. If the blackbody has a temperature 2T , what is the wavelength of the peak intensity?

(a) 16λ

(b) λ/2

(c) λ

(d) λ/16

(e) 2λ

20. The hydrogen emission wavelength 434.0nm is produced by the transition to the n = 2 energylevel from which energy level?

(a) n = 3

(b) n = 4

(c) n = 5

(d) n = 6

(e) n =∞


Answers

Problems

1. (a) v = 11.7 m/s, vy,max = 2.10 m/s

(b) The wave speed is a property of the medium while the maximum transverse speed isdetermined by the source. These speeds are independent of each other.

(c) µ = 0.365 kg/m

(d) P = 9.40 W

2. (a) fhigh = 448 Hz, flow = 432 Hz

(b) β2 = 55.6 dB

3. (a) ∆φ = 17π rad

(b) d = 2.00 km

(c) 24 points of constructive interference

(d) 16 points of constructive interference

4. (a) θ1 = 70.8◦

(b) d = 0.19 mm

(c) f = 4.84× 1014 Hz, λ = 380 nm

5. (a) 500 lines/mm

(b) ∆θ = 0.701◦

(c) 3 complete orders

(d) m = 2 and m = 3

6. (a) N0 = 2.00× 1024 atoms

(b) R0 = 1.82× 1012 Bq

(c) t = 1.06× 106 years

(d) E = 1.68× 1012 J

Multiple Choice

1. (d)

2. (a)

3. (d)

4. (c)

5. (c)

6. (a)

7. (d)

8. (e)

9. (a)

10. (e)

11. (b)

12. (e)

13. (d)

14. (a)

15. (b)

16. (c)

17. (e)

18. (b)

19. (b)

20. (c)


Solutions

Problems

1. (a) The wave speed is v = λf . We’re given that λ = 0.700 m, and determine that

f =200 cycles

12.0 s= 16.7 Hz

Thus,v = λf = (0.700 m)(16.7 Hz) = 11.7 m/s.

The maximum transverse speed of a piece of the string is vy,max = ωA. This is becauseeach piece is performing simple harmonic motion. From the given data:

vy,max = ωA = (2πf)A = 2π(16.7 Hz)(0.0200 m) = 2.10 m/s

(b) The wave speed is a property of the medium while the maximum transverse speed isdetermined by the source. These speeds are independent of each other.

(c) Since v =√F/µ we find that

µ = Fv2 = (50.0 N)(11.7 m/s)2 = 0.365 kg/m

(d) The average power transmitted is

P =1

2µvω2A2

P =1

2µv(2πf)2A2

P =1

2(0.365 kg/m)(11.7 m/s)4π2(16.7 Hz)2(0.0200 m)2

P = 9.40 W

2. (a) The source reaches its maximum speed at the equilibrium position. As the source moveswith its maximum speed towards the listener

fhigh =v + vLv + vS

fS =343 m/s + 0

343 m/s + (−6.00 m/s)(440 Hz) = 448 Hz

Moving with the same speed away from the listener

flow =v + vLv + vS

fS =343 m/s + 0

343 m/s + 6.00 m/s(440 Hz) = 432 Hz

(b) When the source is closest to the listener

β1 = (10 dB) log

(I1I0

)−→ I1 = I010β1/10 dB

I1 = (10−12 W/m2)1060 dB/10 dB

I1 = 10−6 W/m2


When the source is furthest from the listener (i.e. further away by twice the amplitude)

I2 =r21r22I1 =

(1.50 m)2

(2.50 m)2(10−6 W/m2) = 3.6× 10−7 W/m2

At this intensity, the corresponding sound intensity level is

β2 = (10 dB) log

(I2I0

)= (10 dB) log

(3.6× 10−7 W/m2

10−12 W/m2

)= 55.6 dB

3. (a) The path length from S directly to R is r1 = 8.00 km. The path length of the reflectedwave is r2 = 10.0km, because it goes from S to the midpoint of the reflecting surface thenback up to R. The reflected light also gets a phase shift of π radians, which is analogousto having out of phase sources (∆φ0 = π). Thus, the phase difference between the twowaves that arrive at the listener is

∆φ =2π

λ∆r + ∆φ0

∆φ =2π

0.250 km(10.0 km− 8.00 km) + π

∆φ = 17π

(b) If the point in question is a distance d from the source, then the path length from Sdirectly to the point is r1 = d. The path length of the reflected wave is r2 = 3.00 km +(3.00 km − d) = 6.00 km − d, because it travels from from S to S′ then back up again.We can solve for d using the phase difference

∆φ =2π

λ∆r + ∆φ0

17π =2π

0.250 km((6.00 km− d)− d) + π

d = 2.00 km

(c) Consider the point S′. The path length from S directly to S′ is r1 = 3.00 km. The pathlength of the reflected wave is r2 = 3.00 km, because it also travels from S to S′ but nofurther. The phase difference is

∆φ =2π

λ∆r + ∆φ0

∆φ =2π

0.250 km(3.00 km− 3.00 km) + π

∆φ = π =

(0 +

1

2

)2π

Therefore m = 0 at point S′.

Now Consider the point S. The path length from S to itself r1 = 0 km. The path lengthof the reflected wave is r2 = 6.00 km, because it also travels from S to S′ then all the


way back again. The phase difference is

∆φ =2π

λ∆r + ∆φ0

∆φ =2π

0.250 km(6.00 km− 0 km) + π

∆φ = 49π =

(24 +

1

2

)2π

Therefore m = 24 at point S.

This means that along the line SS′ there are 24− 0 = 24 points of constructive interfer-ence.

(d) At R the phase difference (found in part (a)) is ∆φ = 17π, which corresponds to m = 8.At the source m = 24, which means that along the line SR there are 24− 8 = 16 pointsof constructive interference.

4. (a) The geometry inside the cylinder shows

(180◦ − θ1) + 2θ2 = 180◦ −→ θ1 = 2θ2

Applying Snell’s law at the point of incidence gives

n1 sin θ1 = n2 sin θ2

n1 sin(2θ2) = n2 sin θ2

2n1 sin θ2 cos θ2 = n2 sin θ2

2n1 cos θ2 = n2

θ2 = cos−1(n22n1

)θ2 = cos−1

(1.63

2(1)

)θ2 = 35.4◦

Returning back to the first line, we can now solve for the angle of incidence

n1 sin θ1 = n2 sin θ2 −→ θ1 = sin−1(n2 sin θ2n1

)θ2 = sin−1

(1.63 sin 35.4◦

1

)θ2 = 70.8◦

(b) From the geometry in the diagram it’s plain to see

d = R sin θ1 = (0.20 mm) sin 70.8◦ = 0.19 mm

(c) Due to conservation of energy the frequency of an electromagnetic wave remains constantregardless of what medium it is in. Thus the frequency in the plastic is the same asoutside

f =c

λvac=

3× 108 m/s

620× 10−9 m= 4.84× 1014 Hz


The wavelength, however, does change on entering the plastic

λmat =λvacn

=620 nm

1.63= 380 nm

5. (a) Using the angle given we can find the grating spacing

d sin θ = mλ −→ d =mλ

sin θ=

(2)656.3 nm

sin 41.018◦= 2000 nm = 2.00× 10−3 mm

The inverse of the grating spacing is the number of lines per millimetre

1

d=

1

2.00× 10−3 mm= 500 lines/mm

(b) For the blue wavelength

d sin θ = mλ −→ θ = sin−1(mλ

d

)= sin−1

((1)434.1 nm

2000 nm

)= 12.536◦

For the violet wavelength

d sin θ = mλ −→ θ = sin−1(mλ

d

)= sin−1

((1)410.2 nm

2000 nm

)= 11.835◦

And so the angular separation

∆θ = θblue − θviolet = 12.536◦ − 11.835◦ = 0.701◦

(c) The longest wavelength, and therefore the most deviated in each order, is Hα (red). Allcomplete orders have the Hα line at an angle θ ≤ 90◦. Solving for the order in whichthe red line appears at 90◦ we get

d sin θ = mλ −→ m =d sin θ

λ=

(2000 nm) sin 90◦

656.3 nm= 3.047

Since m must be an integer we round down, so there are three (3) complete orders.

(d) Overlapping orders means that the most deviated wavelength from order m appearsbefore the least deviated wavelength from order m+ 1:

θred,m > θviolet,m+1

mλredd

>(m+ 1)λviolet

d

m >λviolet

λred − λvioletm >

410.1 nm

656.3 nm− 410.1 nm

m > 1.666

Since m must be an integer this tells us that the m = 2 and m = 3 orders overlap. Ifany higher orders are even partially visible, they will overlap.


6. (a) The mass of a single atom of plutonium is

mPu = (239.052157 u)(1.660539× 10−27 kg/u) = 3.696554× 10−25 kg

If we divide the total mass of the sample by the mass of one atom we get the totalnumber of atoms

N0 =msample

mPu=

0.7939109 kg

3.696554× 10−25 kg= 2.00× 1024 atoms

(b) Using the conversion factor given we find t1/2 = 24120 years = 7.6117× 1011 s. From theequation for activity we find

R0 = rN0 =ln 2

t1/2N0 =

ln 2

7.6117× 1011 s(2.00× 1024) = 1.82× 1012 Bq

(c) Using the decay equation we can solve for time

R = R0e−t/τ −→ t = −τ ln

(R

R0

)t = −

t1/2

ln 2ln

(R

R0

)t = −24120 years

ln 2ln

(0.100 Bq

1.82× 1012 Bq

)t = 1.06× 106 years

(d) The energy released in each decay is

Edecay = (mPu − (mU +mα))c2

Edecay = (239.052157 u− (235.043924 u + 4.002602 u))(931.494061 MeV/u)

Edecay = 5.245243 MeV

At the time found in part (c), the number of atoms remaining is

N =R

r=t1/2R

ln 2=

(7.6117× 1011 s)(0.100 Bq)

ln 2= 1.098× 1011 atoms

The total number of decays is ∆N = N0 −N . The number above is thirteen orders ofmagnitude smaller than the value of N0 we found in part (a), meaning that ∆N ≈ N0.Thus, the total energy released is

E = Edecay∆N

E = (5.245243 MeV/decay)(2.00× 1024 decays)

E = 1.05× 1025 MeV

E = 1.68× 1012 J


Multiple Choice

1. Since T = 2π√m/k for a mass on a spring, a doubling of the mass must be joined by a

halving of the spring constant. The answer is (d).

2. Only graphs (a), (c), and (d) have the 4 cm amplitude. An angular frequency of 2 rad/scorresponds to a period T = 2π/ω = π s. Only graph (a) has the proper period. The answeris (a).

3. Waves transport energy, but they do not transport material. The answer is (d).

4. In each equation the argument of the function is kx ± ωt. The wave speed v = λf = ω/k.The values of wave speed are vI = 0.67 m/s, vII = 0.5 m/s, vIII = 2 m/s, and vIV = 0.67 m/s.The answer is (c).

5. The wavelength of the sound λ = v/f = 0.308 m. Using the wavelength and length of thetube we can find the mode number m = 2L/λ = 3. For the m = 3 mode there is a node atthe centre of the tube. The nearest antinode is λ/4 = 0.077 m away. The answer is (c).

6. For a fixed-fixed string the fundamental wavelength λ1 = 2L. If we change the tension, butnot the length, the fundamental wavelength will stay the same. The answer is (a).

7. The wavelength of the sound is affected by the motion of the source only, not by the motionof the listener. All listeners, whether moving or not, should measure the same wavelength.The frequency measured by a person standing at rest as the car approaches is

fL =

(v

v + vS

)fS =

(343 m/s

343 m/s + (−30 m/s)

)1.00 kHz = 1.096 kHz

Which corresponds to a wavelength λ = v/f = 0.313 m. The answer is (d).

8. Intensity decreases as the inverse-square of the distance. The answer is (e).

9. Doubling the distance will decrease the intensity by a factor of four. The change in thesound intensity level ∆β = (10 dB) log(I2/I1) = (10 dB) log(1/4) = −6.0 dB. The new soundintensity level is β2 = 80.0 dB− 6.0 dB = 74.0 dB. The answer is (a).

10. Recalling the electromagnetic spectrum, radio waves have the largest wavelength (lowestfrequency), followed by microwaves, infrared, visible, UV, X-rays, and finally gamma rays.The answer is (e).

11. The figure shows total internal reflection, which happens only when n2 < n1. Having n2 = n1is like having no boundary at all. The answer is (b).

12. The ray travelling across the oil is incident at the oil-air interface at an angle

θ = tan−1(

2.00 m− 0.90 m

1.88 m

)= 30.33◦

If this is the critical angle at that interface, then θc = sin−1(nair/noil) −→ noil = 1.98. Theanswer is (e).


13. For a double-slit d sin θ′m = (m + 1/2)λ. Increasing d decreases θ, and thus the separationbetween minima decreases. The answer is (d).

14. Ideally, all the bright fringes of a double-slit interference pattern are equally intense. Theanswer is (a).

15. 450 lines/mm means a grating spacing d = 2.22 × 10−6 m. The most deviated colour is theone that corresponds to the longest wavelength. We can find the order at which red (700 nm)deviates at 90◦: m = d sin 90◦/λ = 3.17. Rounding down, m = 3 is the highest completeorder. The answer is (b).

16. For a circular aperture θ1 ≈ 1.22λ/D. Doubling the size of the aperture will halve the valueof θ1, but the means the resolving power doubles (the smaller θ1, the greater the resolvingpower). The answer is (c).

17. The number of photons N = Etotal/Ephoton = Etotalλ/hc = 2.8 × 1018 photons. The time isnot needed. The answer is (e).

18. Increasing the intensity increases the number of photons hitting the cathode, but has no effecton the energy carried by each photon. So more electrons are released from the cathode, butwith the same maximum energy. The answer is (b).

19. According to Wien’s law λpeak = 2.90 × 106 nm K/T . Doubling the temperature halves thepeak wavelength. The answer is (b).

20. Using the hydrogen energy levels

hc

λ=−13.6 eV

n2− −13.6 eV

22−→ n = 5

The answer is (c).


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