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PHYSICS, M7 2025 HSC 12 MC

Which graph correctly represents Malus' Law?
 

 

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\(A\)

Show Worked Solution
  • Malus’ Law: \(I=I_{\text{max}}\cos^{2}\theta\)
  • Light intensity is therefore directly proportional to \(\cos^{2}\theta\) (noting \(I_{\text{max}}\) is a constant).

\(\Rightarrow A\)

♦♦ Mean mark 34%.

PHYSICS, M7 2019 VCE 14*

Students have set up a double-slit experiment using microwaves. The beam of microwaves passes through a metal barrier with two slits, shown as \(\text{S}_1\) and \(\text{S}_2\) in the diagram. The students measure the intensity of the resulting beam at points along the line shown. They determine the positions of maximum intensity to be at the points labelled \(\text{P}_0,\) \(\text{P}_1\), \(\text{P}_2\) and \(\text{P}_3\). 
 

The distance from \(\text{S}_1\) to \(\text{P}_3\) is 72.3 cm and the distance from \(\text{S}_2\) to \(\text{P}_3\) is 80.6 cm.

  1. What is the frequency of the microwaves transmitted through the slits? Show your working.   (2 marks)
  2. The signal strength is at a minimum approximately midway between points \(\text{P}_0\) and \(\text{P}_1\).
  3. Explain the reason why the signal strength would be a minimum at this location.   (2 marks)

  4. The microwaves from the source are polarised.
  5. Explain what is meant by the term 'polarised'. You may use a diagram in your answer.   (2 marks)

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a.    \(1.08 \times 10^{10}\ \text{Hz}\)

b.    Signal strength at midpoints:

  • Halfway between \(\text{P}_0\) and \(\text{P}_1\ \ \Rightarrow\)  path difference = \(\frac{\lambda}{2}\).
  • Therefore, destructive interference will occur and the signal strength will be a minimum.

c.    Polarised:

  • Light is a transverse wave that can oscillate in any direction.
  • Light becomes polarised when its plane of oscillation is only in a single direction, as seen in the diagram below.
      

Show Worked Solution

a.    The path difference to \(P_3\) is equal to 3 wavelengths of the microwaves.

\(3\lambda\) \(=0.806-0.723\)  
\(\lambda\) \(=\dfrac{0.083}{3}=0.0277\ \text{m}\)  

 
\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{0.0277}=1.08 \times 10^{10}\ \text{Hz}\)
 

♦♦ Mean mark (a) 35%.
COMMENT: Common error was not converting cm to m.

b.    Signal strength at midpoints:

  • Halfway between \(\text{P}_0\) and \(\text{P}_1\ \ \Rightarrow\)  path difference = \(\frac{\lambda}{2}\).
  • Therefore, destructive interference will occur and the signal strength will be a minimum.
♦♦ Mean mark (b) 42%.

c.    Polarised:

  • Light is a transverse wave that can oscillate in any direction.
  • Light becomes polarised when its plane of oscillation is only in a single direction, as seen in the diagram below.

 

♦ Mean mark (c) 50%.
Comment: Students are encouraged to use diagrams when possible.

PHYSICS, M7 2021 VCE 11

The diagram shows a system of two ideal polarising filters, \(\text{F}_1\) and \(\text{F}_2\), in the path of an initially unpolarised light beam. The polarising axis of the first filter, \(\text{F}_1\), is parallel to the \(y\)-axis and the polarising axis of the second filter, \(\text{F}_2\), is parallel to the \(x\)-axis.  

Will any light be observed at point \(\text{P}\)? Give your reasoning.  (2 marks)

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  • Light will initially pass through \(F_1\) and so will be polarised in the vertical plane with the intensity of the light will reducing to 50%.
  • \(F_2\) will polarise light in the horizontal direction, but as the light beam is already polarised in the vertical direction, none of the light will be able to pass through \(F_2\).
  • Therefore, there will be no light visible at \(P\).
Show Worked Solution
  • Light will initially pass through \(F_1\) and so will be polarised in the vertical plane with the intensity of the light will reducing to 50%.
  • \(F_2\) will polarise light in the horizontal direction, but as the light beam is already polarised in the vertical direction, none of the light will be able to pass through \(F_2\).
  • Therefore, there will be no light visible at \(P\).

PHYSICS, M7 2022 VCE 16 MC

Which one of the following phenomena best demonstrates that light waves are transverse?

  1. polarisation
  2. interference
  3. dispersion
  4. diffraction
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\(A\)

Show Worked Solution
  • Polarisation requires the transfer of energy through a wave to be perpendicular to the direction of oscillation of the wave.
  • As this is a property of a transverse wave, light waves must be traverse.

\(\Rightarrow A\)

PHYSICS, M7 2023 VCE 14 MC

Polarisation of visible light provides evidence that electromagnetic radiation can be explained using a

  1. standing wave model for light.
  2. transverse wave model for light.
  3. mechanical wave model for light.
  4. longitudinal wave model for light.
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\(B\)

Show Worked Solution
  • For polarisation to occur, the plane of vibration of the wave must be perpendicular to direction of motion. 
  • This is a property of the transverse wave model of light.

\(\Rightarrow B\)

PHYSICS, M7 2023 HSC 29

When light from an incandescent lamp is passed through a plane polarising filter, the intensity of the light is reduced.

Explain this phenomenon.  (4 marks)

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  • Light from an incandescent lamp is unpolarised meaning it is composed of many different planes of light waves as defined by the plane of oscillation of the electric fields of the EM waves.
  • The different planes of light can be resolved into components that are perpendicular and parallel to the plane of the polarising filter.
  • Plane polarising filters are commonly made up of long iodine molecular chains that only allow light waves with oscillating electric fields parallel to the plane of polarisation to pass through them.
  • The component of light with oscillating electric fields perpendicular to the plane of polarisation will be absorbed by the filter and so completely blocked from passing through the filter.
  • Therefore, the intensity of the light measured after passing through the polarising filter is reduced to around 50% of the original intensity of the light.

Show Worked Solution

  • Light from an incandescent lamp is unpolarised meaning it is composed of many different planes of light waves as defined by the plane of oscillation of the electric fields of the EM waves.
  • The different planes of light can be resolved into components that are perpendicular and parallel to the plane of the polarising filter.
  • Plane polarising filters are commonly made up of long iodine molecular chains that only allow light waves with oscillating electric fields parallel to the plane of polarisation to pass through them.
  • The component of light with oscillating electric fields perpendicular to the plane of polarisation will be absorbed by the filter and so completely blocked from passing through the filter.
  • Therefore, the intensity of the light measured after passing through the polarising filter is reduced to around 50% of the original intensity of the light.
♦ Mean mark 54%.

PHYSICS, M7 EQ-Bank 24

Parallel light rays of intensity `I_0` pass through two polarising filters `P_1` and `P_2` to a detector. The filters are initially aligned so that they produce the maximum amount of light, then filter `P_2` is slowly rotated through 180° as shown.
 

  1. On the axes provided sketch a graph showing how the intensity of light at the detector, `I`, changes as `P_2` rotates from zero to 180°.
     

  1. `P_2` is now rotated to a position such that no light reaches the detector. Without moving `P_1` or `P_2`, a third polarising filter is inserted between `P_1` and `P_2` and rotated at an angle of 30° from `P_1`.
  2. Explain, with the aid of calculations, why the light intensity at the detector is no longer zero.
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a.   

b.   Light intensity at detector:

  • `P_(2)` must be rotated 90° from `P_(1)` (no light reaches detector)
  • Calculate the light passing through the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(m)=I_(1)cos^(2)30°=0.75 xxI_(1)`
  • `P_(2)` is a 60° rotation from the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(2)=0.75xxI_(1)cos^(2)60°=0.1875xxI_(1)`
  • Light intensity at detector is therefore no longer zero.
Show Worked Solution

a.   

b.   Light intensity at detector:

  • `P_(2)` must be rotated 90° from `P_(1)` (no light reaches detector)
  • Calculate the light passing through the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(m)=I_(1)cos^(2)30°=0.75 xxI_(1)`
  • `P_(2)` is a 60° rotation from the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(2)=0.75xxI_(1)cos^(2)60°=0.1875xxI_(1)`
  • Light intensity at detector is therefore no longer zero.

PHYSICS, M7 EQ-Bank 21

  1. A student was given a smartphone with a light sensor and an angle sensor, and a computer screen which emitted polarised light. A polariser was fixed over the top of the light sensor in the smartphone.
     

  1. The student wants to use this equipment to investigate Malus's Law of polarised light. Describe a procedure that is suitable for carrying out this investigation.   (3 marks)
  1. An experiment was conducted to demonstrate Malus's Law for plane polarisation of light. The results are shown in the graph.
     

  1. Based on the graph shown, how effective was the experiment in meeting its aim?   (3 marks)
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a.   Procedure steps:

  • Secure the smartphone so it can rotate at a measured and fixed distance from the computer screen.
  • Set the computer to emit a constant intensity of light.
  • Find the position of the smartphone at which light of maximum intensity is obtained. Set this position to be 0° and record the intensity.
  • Rotate the smartphone through 360° and record the intensity at regular intervals of 10°.
  • Plot obtained results on a graph. 

b.    Experiment effectiveness:

  • Malus’ Law gives the equation  `I=I_(max)cos^2theta`
  • The graph shape that is produced by the experimental data is consistent with the equation. 
  • The experiment uses a sufficient range of data to show a relationship and uses a sufficient number of data points.
  • However, the equipment used must have been incorrectly calibrated, or the measurements were not taken correctly. Measurements appear to be offset by 30° as the maximum intensity of light should occur at 0° rather than 30°. 
Show Worked Solution

a.   Procedure steps:

  • Secure the smartphone so it can rotate at a measured and fixed distance from the computer screen.
  • Set the computer to emit a constant intensity of light.
  • Find the position of the smartphone at which light of maximum intensity is obtained. Set this position to be 0° and record the intensity.
  • Rotate the smartphone through 360° and record the intensity at regular intervals of 10°.
  • Plot obtained results on a graph. 

b.    Experiment effectiveness:

  • Malus’ Law gives the equation  `I=I_(max)cos^2theta`
  • The graph shape that is produced by the experimental data is consistent with the equation. 
  • The experiment uses a sufficient range of data to show a relationship and uses a sufficient number of data points.
  • However, the equipment used must have been incorrectly calibrated, or the measurements were not taken correctly. Measurements appear to be offset by 30° as the maximum intensity of light should occur at 0° rather than 30°. 

PHYSICS, M7 EQ-Bank 5 MC

Anna and Bo carried out independent experiments to investigate Malus's Law. They graphed the results of their experiments. The graphs are shown below.
 

Based on the two graphs, which of the following is correct?

  1. Anna has taken more measurements but Bo has used a better data range.
  2. Bo's graph is more precise as the angles in Anna's graph are too small.
  3. Anna's graph is more valid as Bo's graph shows a straight line relationship.
  4. Anna's measurements are more reliable than Bo's as a line of best fit cannot be drawn for Bo's graph.
Show Answers Only

`A`

Show Worked Solution

By Elimination:

  • Anna’s graph uses too narrow an angle range whereas Bo’s graph does not include enough measurements to show a valid relationship (possibly A).
  • It is not possible to determine precision or validity without knowing the equipment used or experimental design of both experiments (eliminate B and C).
  • Reliability refers to the consistency of results and has no impact on the ability to draw a line of best fit (eliminate D).

`=>A`

PHYSICS M7 2022 HSC 17 MC

Unpolarised light of intensity `I_0` is incident upon a vertically polarised filter. The filtered light then passes through a pair of glasses. The glass have polarising filters, with one side polarised vertically and the other horizontally.

The filter undergoes one complete 360° rotation around point `P`, as shown.
 


 

Which of the following correctly compares `I_y` to the intensity at other positions?

  1. `I_y` never equals `I_x`
  2. `I_y` never equals `I_z`
  3. `I_y` sometimes equals `I_z`
  4. `I_y` sometimes equals `I_0`
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`C`

Show Worked Solution
  • Once the filter rotates 45° to the vertical:
  •    `I_(y)=I_(x)cos^(2)45^(@)=(1)/(2)I_(x)`
  •    `I_(z)=I_(x)cos^(2)45^(@)=(1)/(2)I_(x)`
  • `I_(y)` sometimes equals `I_(z)`.

`=>C`


Mean mark 55%.

PHYSICS, M7 2019 HSC 10 MC

A beam of light passes through two polarisers. The second polariser has a transmission axis at an angle of 30° to that of the first polariser. The intensity of the light beam before and after the second polariser is \( I_0\) and \(I_B\) respectively.
 

Which row of the table correctly identifies the value of \( \dfrac{I_B}{I_0} \), and the model of light demonstrated by this investigation?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-3.5ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Value of \(\dfrac{I_{ B }}{I_0}\)}\rule[-1ex]{0pt}{0pt} \quad & \textit{Model of light demonstrated} \\
\hline
\rule{0pt}{2.5ex}0.750\rule[-1ex]{0pt}{0pt}&\text{Wave model}\\
\hline
\rule{0pt}{2.5ex}0.750\rule[-1ex]{0pt}{0pt}& \text{Particle model}\\
\hline
\rule{0pt}{2.5ex}0.866\rule[-1ex]{0pt}{0pt}& \text{Wave model} \\
\hline
\rule{0pt}{2.5ex}0.866\rule[-1ex]{0pt}{0pt}& \text{Particle model} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

Applying Malus’ law:

\(I_{B} \) \( =I_{0} \cos^{2} \theta \)
\( \dfrac{I_B}{I_0} \) \(=\cos^{2} 30^{\circ}=0.750 \)

 
Polarisation is a wave property of light.

\( \Rightarrow A\)

PHYSICS, M7 2021 HSC 15 MC

Unpolarised light is incident upon two consecutive polarisers as shown. The second polariser has a fixed transmission axis which cannot be rotated. `I_1` is the intensity of light after the first polariser, and `I_2` is the intensity of light after the second polariser.
 

How would `I_1` and `I_2` be affected if the transmission axis of the first polariser was rotated?

  1. Both would change.
  2. Only `I_1` would change.
  3. Only `I_2` would change.
  4. Neither would change.
Show Answers Only

`C`

Show Worked Solution

`I_(1)`  is constant at 50% of the original intensity.

`I_(2)`  varies depending on the angle between the two polarisers (Malus’ Law).

`=>C`


♦ Mean mark 42%.