IAL phy U2 caqs/PP mistakes [2020-2023]

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28 Terms

1
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B is correct answer

constructive interference:
path diff = nλ (where n = whole number)

phase difference = 2n(π)


destructive interference:
Path difference = (n +1/2)λ

phase diff = (2n+1)π


NOTE: for phase difference, we need even INTEGER multiples of π,
so phase diff of 0.5π, 3.5π etc give neither constructive nor destructive interference, thats why C is wrong.
(they give partial interference)

<p><strong>B is correct answer</strong><br><br>constructive interference: <br><strong><u>path diff = nλ </u></strong>(where n = whole number)</p><p><strong><u>phase difference = 2n(π) </u></strong><br><br><br>destructive interference: <br><strong><u>Path difference = (n +1/2)λ </u></strong><br><br>phase diff = (2n+1)π<br><br><br>NOTE: for phase difference, we need <strong><u>even INTEGER multiples of </u></strong>π,<br>so<strong> phase diff of 0.5π, 3.5π etc give neither constructive nor destructive interference</strong>, thats why C is wrong.<br>(they give partial interference)<br></p>
2
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define polarised light



polarised light is when light is restricted so it has oscillations in a single direction which is perpendicular to direction of wave travel

OR in terms of plane:
“ “so it oscillates in a single plane which includes direction of wave travel

NOTE:
say the blue part/plane was flitered, it still wudnt be right to say perpendicular when talking about plane, as the oscilation includes the direction of wave travel

if you talk about the blue and red lines(wave directions), then yes the blue/red wave is perpendicular too direction of wave travel

<p><br></p><p>polarised light is when light is restricted so it has<strong><u> oscillations in a single direction<mark data-color="yellow" style="background-color: yellow; color: inherit;"> which is perpendicular to direction of wave travel</mark></u></strong><br><br>OR in terms of plane:<br>“ “so it oscillates in a <strong><u>single plane which <mark data-color="yellow" style="background-color: yellow; color: inherit;">includes direction of wave travel</mark></u></strong><br><br>NOTE:<br><span>say the blue part/plane was flitered, it still wudnt be right to say perpendicular when talking about plane, as the oscilation includes the direction of wave travel</span><br><br>if you talk about the blue and red lines(wave directions), then yes the blue/red wave is perpendicular too direction of wave travel</p>
3
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C is correct answer

in diffraction grating experiment:
-diffraction grating is set upparallel to screen
- diffraction grating is setup perpendicular to light source


- diffraction grating is the square thingy with 1000s of line holes or slits

- - we find theta using trigonometry NOT PROTRACTOR
- distance between the bright dots(maxima) is best measured using a metre rule

4
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in series, voltage is shared so if V2 increases v1 must decrease to yk balance out the ratios of share


<p>in series, voltage is shared so if V2 increases v1 must decrease to yk balance out the ratios of share</p><p></p>
5
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Explain how the ultrasound scan of the unborn baby is produced using pulse-echo technique (6marks)


  • a device/transdurcer is used that produces ultrasound and also detects the ultrasound reflected back

  • the produced ultrasound is the pulse which is reflected when it hits a boundary, and the reflected ultrasound is the echo

  • they get reflected due to change in density (like from body fluid to soft tissue)


  • speed of ultrasound is known and the time taken for pulse to return is measured

  • the distance from transducer to boundary is found with equation:

s = v x t, but here its 2d = v x t, therefore distance = (vxt) / 2
(the distance is found to create an accurate mapping of the image)

  • the echoes are converted to electrical signanls and computer uses the echo intensity and time delays to display as an ultrasound scan/image



btw reflected due to change in denisty is cuz of acoustic impedence,a quantity determines how much sound waves are resisted to flow past a medium. but this is not in syllabus so you can ignore it

6
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define work function

minimum energy required to emitt/remove an electron from the surface of a metal

7
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In an experiment involving photocell, the intensity of UV light emitted was increased.
Explain what happened to the reading on ammeter (3marks)


  • increasing intensity means more photons hit metal surface per second

  • so more electrons(photoelectrons) are emitted per second

  • therefore current increases (or reading on ammeter increases)

(as rate of flow of charge increased)

8
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9
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Explain, using Huygens' construction, how diffraction occurs as waves pass through a gap.

[2marks]


  • in Hyugen's construction, points along the wavefront (called point sources) acts as a source for secondary waves/wavelets

  • so when waves pass slit they bend and spread out (diffraction)


10
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here you should have figured out that there are sever maximums or maximas, central one being largest and others smaller

these maxima are due to (constructive) interference between the secondary wavelets
dark patches due to destructive interference

<p>here you should have figured out that there are sever maximums or maximas, central one being largest and others smaller<br><br>these maxima are due to (constructive) interference between the secondary wavelets<br>dark patches due to destructive interference</p>
11
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define emf


define potential difference

emf = energy supplied per unit charge by a source

p.d = energy transferred per unit charge between two points in a cicuit

<p>emf = energy<strong><u> supplied </u></strong>per unit charge by a source<br><br>p.d = energy <strong><u><mark data-color="yellow" style="background-color: yellow; color: inherit;">transferred</mark></u></strong> per unit charge between two points in a cicuit</p>
12
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  • Plot a graph V(y-axis) against I (x-axis)

  • Determine the gradient from graph

  • Gradient is −r, so r is - gradient


<ul><li><p>Plot a graph V(y-axis) against I (x-axis)<br></p></li><li><p> Determine the gradient from graph<br></p></li><li><p> Gradient is −r, so r is - gradient</p></li></ul><p></p>
13
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The calculated value for resistivity of air in lightning channel is 0.2 and accepted value is 1016

Suggest why the value calculated from the data is so much less than the accepted value.

Lightning channel unlikely to have a uniform diameter / CSA
(cuz lightning is non uniform)

14
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emf is the work done moving a unit charge around the whole circuit
basically emf is the total of all the voltages present

<p>emf is the work done moving a unit charge around the <strong><u>whole circuit</u></strong><br><strong><u>basically emf is the total of all the voltages present</u></strong></p>
15
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For semiconductors (thermistors):
(1) as temperature increases

(2) number density of charge carriers (electrons) increases

(3) the number of conduction electrons per unit volume increases which results in a decrease in resistance (and resistivity)

btw D is factually correct but B has more priority/effect i.e, as when temperature increases ions do vibrate more but here as it’s semi conductor no. of charge carrier electrons have more effect

<p>For semiconductors (thermistors):<br>(1) as<strong><u> temperature increases</u></strong><br><br>(2)<strong><u> number density of charge carriers (electrons) increases</u></strong><br><br>(3) the number of conduction electrons per unit volume increases which results in a <strong><u>decrease in resistance</u></strong> (and resistivity)<br><br>btw D is factually correct but B has more priority/effect i.e, as when temperature increases ions do vibrate more but here as it’s semi conductor no. of charge carrier electrons have more effect</p>
16
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A student suggests that the drift velocity will double if the length of wire used in the circuit is halved.
Comment on this suggestion

  • Halving length will cause resistance to half

  • it causes current to double

  • since I = nqvA, the drift velocity also doubles so student is correct


<ul><li><p>Halving length will cause <strong><u>resistance to half</u></strong></p></li><li><p><strong><u>it causes current to double</u></strong></p></li><li><p>since I = nqvA, the drift velocity also doubles so student is correct</p></li></ul><p></p>
17
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  • transducer sends ultrasound which is reflected

  • find the time taken (for signal to return)

  • Calculate (expected) time taken for pulse to return without air gap & then time taken with air gap

  • If time for pulse to return (with air gap) < time calculated (without air gap)
    air gap is present


18
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Ultrasound is a sound wave with a frequency greater than 20kHz.
The frequency of ultrasound used by the transducer in this method is 5MHz.

Explain why a much higher frequency than 20kHz is needed in this method
[3 marks]

  • Higher frequency gives smaller wavelength

  • smaller wavelength allows high level of resolution (detail)

  • which means small objects can be detected (i.e air gaps)

(with lower frequency like 20KHz, wavelengths would be larger so it may not give sufficient resolution to detect small objects like air gaps or even the air gaps might be smaller than the wavelength)

19
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Explain how observations from the photoelectric effect demonstrate the particle nature of light.

[6marks]

  • In particle theory/ nature, one photon interacts with one elctron

  • a minimum threshold frequency is required for electrons to be emitted
    [but light as a wave predicted that any frequency of light should eventually eject electrons from a metal if given enough time]

  • in particle nature, electrons are released instantenously

  • Photon energy is related to frequency, E = hf

  • Increasing the frequency increases the Ek max of emitted electrons
    [Ek = hf - work function]

  • In particle nature, increasing intensity, increases the number of photons incident per second thus increasing number of electrons emmited not Ek max of electron

    [ wave nature predicted Higher intensity = more energy transferred/ more Ek max]


<ul><li><p><strong><u><mark data-color="yellow" style="background-color: yellow; color: inherit;">In particle theory/ nature, one photon interacts with one elctron</mark></u></strong><br></p></li><li><p><strong><u>a minimum threshold frequency is required </u></strong>for electrons to be emitted<br>[but light as a wave predicted that<strong> </strong>any frequency of light should eventually eject electrons from a metal if given enough time<strong>]</strong><br></p></li><li><p>in particle nature, <strong><u><mark data-color="yellow" style="background-color: yellow; color: inherit;">electrons are released instantenously</mark></u></strong><br></p></li><li><p>Photon energy is related to frequency, E = hf<br></p></li><li><p><strong><u><mark data-color="yellow" style="background-color: yellow; color: inherit;">Increasing the frequency increases the Ek max of emitted electrons</mark></u></strong><br>[Ek = hf - work function]<br></p></li><li><p>In particle nature, <strong><u><mark data-color="yellow" style="background-color: yellow; color: inherit;">increasing intensity, increases the number of photons incident per second thus increasing number of electrons emmited </mark></u></strong>not Ek max of electron<br><br>[ wave nature predicted <strong><u>Higher intensity = more energy transferred/ more Ek max</u></strong>]</p></li></ul><p></p>
20
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Explain why a series of bright dots is seen on the screen after diffraction grating (3marks)



  • there are bright spots where waves are in phase

(and path difference of nλ)

  • these waves undergo superposition

  • and constructive interference occurs


21
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The monochromatic laser is replaced by a source producing a parallel beam of bright white light. Suggest what would now be observed on the screen

[2 marks]

  • for white light, there will be a bright white spot at the centre(zero order maxima or O)

  • and on either sides of center a continous spectra of visible spectra (rainbow)


<ul><li><p>for white light, there will be a <strong><u><mark data-color="yellow" style="background-color: yellow; color: inherit;">bright white spot at the centre</mark></u></strong>(zero order maxima or O)</p></li><li><p>and on <strong><u>either sides of center a continous spectra</u></strong> of visible spectra (rainbow)</p></li></ul><p></p>
22
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  • The light is incident along the normal (The angle of incidence is 0°)



  • draw the blue highlighted arrow, i did black which was wrong for reflected

    the blue arrow can continue out of air or undergo TIR


<ul><li><p>The light is incident along the normal (The angle of incidence is 0°)<br><br><br></p></li><li><p>draw the blue highlighted arrow, i did black which was wrong for reflected<br><br>the blue arrow can continue out of air or undergo TIR</p></li></ul><p></p>
23
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<p>basic trig maths to find angle of incidence, silly mistake</p>

basic trig maths to find angle of incidence, silly mistake

24
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  • since when filter is rotated intenisty varies, that means oscillation of light is getting restricted to one plane

  • hence light is polarized and ONLY transverse waves can be polarized

(thus light is transverse)

25
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got everything right except forgot the equation:

Q = ne

where n is number of electrons (or charge carrier)
Q is charge and e is charge of one electron/ elementery charge (1.60 × 10-19 )


so answer is 4.2 × 1018 electrons in one second

26
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27
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<p></p>
28
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  • atoms like hydrogen have discrete energy level

  • only certain energy level transitions are possible so photon with discrete energy and therefore wavelength are emitted