3.PhotonLaser

Question 1

C3 Explain in terms of the processes Absorption and Stimulated Emission why the laser would not work unless there was a population inversion.

Answer 1

Population inversion needed for stimulated emission to be more probable, or frequent, than [or predominant over] absorption (1)

This ensures light amplification or photon number increase or without population inversion no amplification or equivalent or by implication (1)

Question 2

Discuss whether or not the graph confirms the equation.

Answer 2

• Points lie on straight line [as required]
• [But] too few data points to form a valid conclusion
• Accepted value of h outside range of uncertainty
• Need to check if graph goes through [true] origin

（gradient，scatter，intercept）

Question 3

The difference in refractive index between the cladding and core is decreased. Explain carefully how this will affect the maximum frequency of data transmission along the optical fibre.

Answer 3

A lower n means that θ ↑ (or equivalent) (1)

Therefore there is less lag time↓ by different routes (1)

Therefore there will be a greater frequency↑ (1)

Question 4

A10PH2 (II) Deduce from the change in direction whether the waves travel faster or slower in shallow water. Give a reason. Calculations are not wanted. [2]

Answer 4

The waves travel more slowly in the shallow water (1) as the propagation direction bends towards the normal (or equiv) (1)

Question 5

A10PH2 (iii) Thick glass fibres Explain why these fibres are unsuitable for the transmission of rapid streams of data encoded in the light. [2]

Answer 5

allow light to travel in zigzag paths, with a range of angles, and also in straight paths parallel to the fibre axis.

Light takes longer by zigzag paths [accept ‘multimode dispersion’] [Accept – different paths give different times] (1)

A piece of data will be ‘smeared out’ over time on arrival or may overlap other pieces of data (1)

[Accept ‘pulse broadening’ only if first mark gained by reference to zigzag paths, i.e. not ‘multimode dispersion’ + ‘pulse broadening’ only (2)]

Question 6

A10PH2 (i) Explain, in terms of interference, phase and path difference, how the bright fringes arise. [4]

Answer 6

At [centres of] bright fringes:

• Path lengths from slits differ by 0, λ, 2λ… [if sources in phase]
• Waves arrive in phase or sketch graphs of in-phase waves
• Waves interfere constructively or displacements add to make larger displacement.
• Assume slits act as coherent sources or waves diffract at slits

Question 7

A10PH2 (c) Suppose that you have to make your own measurements to find the wavelength of light from a laser. Discuss whether you would choose the Young’s fringes method, or the diffraction grating method, if you wanted an accurate value for the wavelength. [2]

Answer 7

More uncertainty with Young’s method (1)…. because….. either fringe separation is small and difficult to measure [whereas grating beams are well spaced] or fringes are not sharp compared to the beams (1) [accept: d can be measured more accurately for grating [because there are more slits]

Question 8

A10PH2 (ii) Describe briefly how you could check this experimentally. [2]

Answer 8

Use stroboscope (1) and adjust flash frequency for slow motion / expect to see A moving up as C moves down etc. (1) [Or: Use a video camera and replay in slow motion / expect to see A moving up as C moves down etc.]

Question 9

A10PH3 difference between progressive and stationary waves. [2]

Answer 9

Progressive waves transfer energy through the medium; stationary waves do not

Either:Amplitude constant [or falls off] for progressive wave (1) as we go through the medium; goes up and down [regularly] form stationary wave (1)

Or: Phase changes steadily with distance for progressive waves (1); reverses at nodes [otherwise constant] form stationary waves (1) [“Stationary waves have nodes, progressive waves don’t” → 1]

Question 10

Explain how, in the set-up above, the stationary wave can be thought of as arising from progressive waves. [2]

Answer 10

Reflections give rise to waves propagating in both directions (1); interference between these [progressive] waves gives stationary wave (1)

Question 11

A student mistakenly thinks that the ‘minus’ sign should be a ‘plus’ sign. Explain, in terms of electrons and photons, why the equation must be correct as written above. [3]E _{k max} = hf – ø.

Answer 11

E_{k max} is the maximum KE of emitted electron (1)

φ is the minimum energy for an electron to escape (1).

What is left over of the photon’s energy after the escape is its kinetic energy. (1)

Question 12

Explain what is meant by the wavelength of the waves. [2]

Answer 12

Distance [along the direction of wave propagation] between two [consecutive] point (1) oscillating in phase (1) [“Distance between two peaks / troughs → 1]

Question 13

(i) Explain what part diffraction plays in the formation of this pattern. [2]

Answer 13

Wavefronts [or waves] from each slit spread out (1) [accept: waves diffract at each slit] …….and overlap (1) [or superpose or interfere].

Question 14

(I) Explain what is meant by in-phase sources. [1]

(II) State one feature of the diagram which confirms that S1 and S2 are in-phase. [1]

Answer 14

I. Sources which emit waves, which are at the same point in their cycle at the same time [accept: “emit peaks at the same time”]

II. A maximum on central axis or microwave source central w.r.t. S1 and S2.

Question 15

(I) What can be said about the phase of the waves from S1 and S2 when they arrive at point P? Justify your answer. [2]

(II) Calculate the path difference, S1P–S2P, explaining your reasoning. [3]

Answer 15

I. Constructive interference at P (1) [accept: waves reinforce] So waves are in phase (1) [Accept: phase difference = 2πn etc]

II. S1P − S2P = nλ [for n = 0, ±1, ±2….]

(1) [n = 0 for central maximum, n = 1 for next one out from centre], n = 2 at P. (1)

So S1P − S2P = 0.024 m (1) [Geometric method based upon Pythagoras 333 if correct]

Question 16

The microwave source of part (a) emits polarised waves. Describe how you would demonstrate this. [2]

Answer 16

Interpose a grille of parallel metal rods and rotate. (1)

The signal strength varies. (1) [Accept rotation of the sensor / ærial]

Question 17

Potatoes can be heated quickly in a microwave oven. Which properties of the microwave radiation account for this? [2]

Answer 17

• the radiation penetrates the potato
• absorbed within the potato, _heating interio_r
• waves transfer energy [or equiv]
• water content heated / water molecules made to vibrate more

Question 18

(ii) Explain why light initially travelling at an angle to the axis greater than ! will not reach the end of the fibre. [3]

Answer 18

Some enters the cladding (1)

…. and is lost (1)

Some is reflected but lost on subsequent reflections (1).

Question 19

(d) Modern communications systems require very high data transmission rates, for which mono-mode optical fibres are needed. Explain why optical fibres with thick cores (multimode fibres) are not suitable. [3]

Answer 19

Paths at different angles to the axis are of different lengths (1).

Data travelling on different paths arrive different times or by clear implic.

so data is muddled / smeared out / data pulses overlap (1)

Question 20

(a) (i) When the blinds are opened a little, so that sunlight falls on the caesium surface, the ammeter registers a continuous current. Explain, in terms of photons and electrons, why this happens. [3]

Answer 20

Photons hit the caesium surface. (1)

Electrons knocked out (1)

• Electrons cross vacuum to collecting electrode
• returned to the caesium via cell and metercycle
• constituting an electric current
• aided by [p.d. of] cell 3

Question 21

(ii) PHOTOELECTRICWhat difference, if any, would be observed if the blinds were adjusted so that a greater intensity of light fell on the caesium surface? Give your reasoning. [2]

Answer 21

Larger current (1) because more photons arrive [per second] (1)

Question 22

PHOTOELECTRIC (i) State two ways in which the apparatus would need to be modified in order to measure the maximumKE kinetic energy of the emitted electrons. [2]

Answer 22

• Power supply+- polarity needs reversing
• Voltage needs to be variable
• voltmeter needed

Question 23

maximumKE kinetic energy of the emitted electrons does not depend on the intensity of the light (for a given frequency). Explain in terms of photons, why this non-dependence is to be expected. [1]

Answer 23

Intensity doesn’t affect individual photon energies [or equiv.]

Question 24

(iii) This radiation is produced by stimulated emission. Explain what is meant by stimulated emission. [Your answer should include statements about photon energy and phase.] [3]

Answer 24

[Incident] photon causes emission of a photon (1)

+ 2 × (1) of: • Incident photon energy needs to be EA − EB [or equiv.]

• Emitted photon has same energy (or λ or f) as incident photon.
• Emitted photon in phase with incident photon

Question 25

(iv) Explain briefly, in terms of photons, why stimulated emission gives rise to ‘light amplification’. [1]

Answer 25

Two photons where there was one before [and the process repeats]

Question 26

(ii) Explain why a population inversion is needed for the laser to work. [1]

Answer 26

If more electrons in B than A, absorption of photons is more likely than stimulated emission.

Question 27

(iii) In this 4-level laser system, level B is above the ground state. How does this make the population inversion easier to establish than in a 3-level system? [2]

Answer 27

B almost empty [because electrons ‘fall’ from B to ground state] (1)

So _not many e_lectrons needed in A to cause population inversion. (1)

Question 28

(iii) What changes would occur to the diagram above if the frequency×4 of the wave were increased by a factor of 4? No calculations are needed. [2]

Answer 28

λ ↓

less spreading a~λ(comparable)

side beams

Question 29

Interference Explain why there is a maximum at point P. [2]

Answer 29

Constructive interference at P / waves arrive in phase at P (1)

Same path length from sources / AP = BP / no path difference (1)

Question 30

(iii) Explain the advantage of monomode fibres over multimode fibres for communicating a rapid sequence of data encoded as light pulses. [3]

Answer 30

Monomode: parallel to axis (accept straight) Multimode: zig-zag paths as well (1) or some paths involve reflections [1]

(iii) Only one route for data (1) [no zig-zag routes]

Each pulse [data element etc] arrives [at other end of fibre] at same time (1)

No overlapping of pulses (1) [even over long distances]

Question 31

) Here is a summary of a theory (now considered incorrect) to account for the photoelectric effect: “The electrons in a surface gradually gain energy from light waves falling on the surface. After a time they will have gained enough energy to escape. The greater the intensity of the light waves the greater the maximum kinetic energy of the emitted electrons.” State some ways in which Einstein’s explanation (in terms of photons) of the photoelectric effect differs from the theory above. [4]

Answer 31

light [energy] in discrete packets

one electron ejected by one photon OR photons don’t cooperate

energy not accumulated [by electron] over time or emission from instant light shines

intensity has no effect on Ekmax or accept intensity affects number emitted per second

wave theory doesn’t predict Einstein’s equation or doesn’t predict threshold frequency

Question 32

Explain in terms of electrons and photons (i) absorption [1]

(ii) stimulated emission [4]
(iii) spontaneous emission

Answer 32

(i) Photon disappears and the electron gains its energy or electron promoted from G to U [1]
(ii) 1. [Passing] photon energy 2.26 x 10-19 [J] or O = 880 [nm] or equivalent 3. Causes electron to drop [from U to G] 4. Releasing additional photon 5. Identical to or in phase or polarised in the same direction or travelling in the same direction with the incident photon Award (1) mark for each of statements 1, 3 and 4 Award the 4th mark for either statement 2 or 5. [4]
(iii) Electron drops [from U to G] by itself (or randomly or without stimulation…), with emission of photon

Question 33

(i) Explain what is meant by pumping in a laser. [1] (ii) Explain why pumping is essential to the operation of the laser. [2]

Answer 33

(i) Raising electrons to higher level or causing population inversion [1]
(ii) So _more e_lectrons in higher level than lower (1). So stimulated emission more probable than absorption (1).

Question 34

(c) A stationary wave is equivalent to a superposition of progressive waves of equal amplitude travelling in opposite directions. Why is this condition not exactly met in a laser emitting a beam of light? [2]

Answer 34

Less amplitude [or fewer photons…] reflected back from [partially reflecting] mirror than arrive at it. (1)

+ (1) of the following: Mirror not a proper node

Amplitudes of progressive waves travelling in opposite directions not equal. (Except near fully reflecting mirror).

Question 35

(iii) Suggest how this plastic rod might be used as part of a device to give a warning when the water level in a tank falls below a certain height. [1]

Answer 35

[Sensor at] Q receives more light when water level drops and exposes lower end of rod to the air. No ecf if paths badly wrong

Question 36

(ii) Explain why the pulse is spread out over time when it arrives at B. A sketched diagram may help your explanation. [2]

Answer 36

Zig-zag routes [take] longer than straight. (1)

(1) For one of the following: Good diagram (angles equal by eye)

A continuous range of zig-zag routes, all of different lengths

Question 37

State, in terms of energy, the meaning of each term in Einstein’s photoelectric equation

EK_max = hf –Ø

Answer 37

KE_max:Maximum k.e. of emitted / photo electrons

hf:Energy of a photon[s]

Ø:[Minimum] energy needed to remove electron [from surface]. Don’t accept from an atom

Question 38

(iii) Explain in detail how light amplification takes place. [4]

Answer 38

Incident (or by implication) photons (1) causes an electron to drop (1).

Emitting photon: so two photons where one previously (or by implication) (1).

(1) For one of the following: x Atom / electron drops [from U] to O. x Incident photon energy must be 2.10 x 10-19 J or equivalent x Process happens repeatedly as photons traverse cavity to and fro x Stimulated photon in phase with incident photon

Question 39

(c) In a 4-level laser the light output results from a transition to a lower level which is above the ground state. Explain the advantage over a 3-level system. [2]

Answer 39

Electrons in lower level drop [spontaneously] to ground state (1) (accept de-excite)

Making population inversion easier to maintain or lowering number of electrons in lower level or making photon absorption less likely. (1)