Polarisation

Plane polarised light of intensity 12 W.m^-2 is incident at a Polaroid. Calculate the intensity of the transmitted light when the angle between the plane of polarisation of the incident light and the transmission axis of the Polaroid is
a. 45°.
b. 60°.

a. Use Malus’s law:
`I = I_0xx cos^2θ= 12 xx cos^2 45° = 12 xx 0.5 = 6 " W.m"^-2`
b. Use Malus’s law:
`I = I_0xx cos^2θ= 12 xx cos^2 60° = 12 xx 0.25 = 3 " W.m"^-2`

Plane polarised light is incident at a Polaroid. Calculate the angle θ, which gives transmitted light of intensity 30% that of the incident intensity of light.

Use Malus’s Law:
`I = I_0xx cos^2θ`
I is 30% of I₀

This mean `I / I_0 = 0.30`

Therefore:

`cos^2θ = I / I_0 = 0.30`

`cosθ = sqrt(0.30)`
So `θ = 57°`

a. State what is meant by plane polarised light.
b. Reflected light from the surface of water is partially plane polarised. Describe briefly how you could demonstrate this.
c. Vertically plane polarised light is incident on three polarising filters. The transmission axis of the first Polaroid is vertical. The transmission axis of the second filter is 45° to the vertical and the transmission axis of the last filter is horizontal. Show that the intensity of light emerging from the final filter is not zero.

a. A plane polarised wave is a transverse wave with oscillations (of the electric field) in just one plane.
b. View the reflected light using a polarising (Polaroid). Rotate the filter about the horizontal axis: the intensity of the light passing through the filter will change, reaching a minimum value when the transmission axis of the filter is at right angles to the plane of polarisation of the reflected light.
c.  The intensity of transmitted light from the first polarising filter = I₀ (the same as the incident intensity).
The intensity of light from the second filter will be: 
`I = I_0xx cos^2θ = I_0 xxcos^2 45° = 0.5xx I_0`
The intensity of light from the last filter will be: 
`I = I_0 xxcos^2θ = [0.5xx I_0] xxcos^2 45° = 0.5xx I_0 xx 0.5 = 0.25xx I_0`

Two polarising filters are placed next to each other so that their planes are parallel.
The first polarising filter has its transmission axis at an angle of 50° to the vertical.
The second polarising filter has its transmission axis at an angle of 20° to the vertical.
The angle between the transmission axes of the two polarising filters is 30°.
A beam of vertically polarised light of intensity 8.0 W·m⁻² is incident normally on the first polarising filter.
What is the intensity of the light that is transmitted from the second polarising filter?
A. Zero.
B. 2.5 W.m⁻².
C. 2.9 W.m⁻².
D. 6 W.m⁻².

Answer: B
First polarising filter: Angle = 50° to vertical.
Incident light is vertically polarised → the angle between the light and the filter = 50°
Intensity after 1st filter:
`I_1 = I_0xx cos^2(50°)=8xx cos^2(50°)≈ 3.3 " W.m"^-2`
Second polariser: At angle 20° to vertical.
⇒ Angle between the first and second filters: `θ = |50° - 20°| = 30°`
Final intensity (after second filter): 
`I_2 = 3.3 xx cos^2(30°) = 3.3 xx 0.75 = 2.5 " W m"^-2`

Polarisation is a phenomenon associated with a certain type of wave. Which condition must be fulfilled if a wave is to be polarised?
A. It must be a light wave.
B. It must be a longitudinal wave.
C. It must be a radio wave.
D. It must be a transverse wave.

Answer: D
A. Incorrect: Polarisation is not limited to light waves. It applies to all transverse waves, including radio and water waves. Light waves can be polarised, but being a light wave is not a necessary condition for polarisation.
B. Incorrect: Longitudinal waves (like sound) oscillate parallel to the direction of wave travel, so they cannot be polarised. Polarisation is not possible for longitudinal waves.
C. Incorrect: Radio waves can be polarised, but they are just one example of transverse waves. The requirement for polarisation is that the wave must be transverse, not that it be radio specifically.
D. Correct: 
Polarisation can only occur if the wave has oscillations perpendicular to its direction of travel, which is a property of transverse waves.
Transverse waves (light, radio waves,...) can be polarised because their vibrations occur in multiple planes perpendicular to propagation - and polarisation limits these to one plane.

Which phenomenon is associated with transverse waves but not longitudinal waves?
A. Polarisation.
B. Reflection.
C. Refraction.
D. Superposition.

Answer: A
A. Correct: 
Polarisation occurs when the vibrations of a wave are restricted to one plane perpendicular to the direction of travel.
This is only possible in transverse waves, where vibrations are already perpendicular to motion.
Longitudinal waves, such as sound, vibrate parallel to the direction of travel and cannot be polarised.
B. Incorrect: Both transverse and longitudinal waves can be reflected.
Example: Light waves (transverse) and sound waves (longitudinal) reflect off surfaces.
C. Incorrect: Refraction, the change in wave direction due to a change in speed, happens in both wave types.
Example: Light bends in glass (transverse); sound changes speed in different air densities (longitudinal).
D. Incorrect: Superposition applies to all types of waves. When two or more waves overlap, they combine to form a new resultant wave.

Which statement describes a situation when polarisation could not occur?
A. Light waves are reflected.
B. Light waves are scattered.
C. Microwaves pass through a metal grid.
D. Sound waves pass through a metal grid.

Answer: D
A. Incorrect: Light is a transverse wave. When light reflects off certain surfaces (like water or glass), polarisation can occur.
B. Incorrect: Scattering of light (e.g., in the atmosphere) can result in partial polarisation.
C. Incorrect: Microwaves are also transverse electromagnetic waves. A metal grid can act as a polariser for microwaves.
D. Correct: 
Polarisation can only occur in transverse waves, where the vibrations are perpendicular to the direction of wave travel.
Sound waves are longitudinal, meaning their vibrations are parallel to the direction of wave travel. Therefore, sound waves cannot be polarised.

A student looked at the liquid crystal display on their calculator with a polarising film. They observed that as they rotated the film, the display changed. Which property of the radiation from the calculator display is described correctly?
A. The emitted radiation is a transverse wave.
B. The emitted radiation is a wave with 3 cm wavelength.
C. The emitted radiation is unpolarised.
D. The emitted radiation is a longitudinal wave.

Answer: A
A. Correct: 
Polarisation is the process by which oscillations of transverse waves are made to occur in one plane only.
Transverse waves can be polarised as they can oscillate in many planes.
The radiation emitted from the calculator’s display is visible light.
Visible light is part of the electromagnetic spectrum, which is an example of a transverse wave.
B. Incorrect: a wave with a 3 cm wavelength would be a microwave.
C. Incorrect: as the display is changed as the film rotated this is evidence that the light emitted from the calculator is polarised.
D. Incorrect: a longitudinal wave can’t be polarised as it oscillates parallel to the direction of travel.

Unpolarised light of intensity 8 W.m^-2 is incident upon a polariser. The light is then transmitted to a second polariser at an angle of 25° to that of the first polariser. What is the intensity of the light beam emerging from the second?

When unpolarised light is incident on the first polariser, the intensity is reduced by half:

`I_1 = (1/2) I_0`

`I_1 = (1/2) xx 8 = 4 " W m"^-2`

There is a 25° angle between the first and second polariser so Malus’s Law should be used: 
`I = I_0 cos^2(theta)`
As the light incident on the second polariser comes from the first, I will be replaced with I₁
`I = I_1xx cos^2(25°) = 4 xx cos^2(25°) ≈ 3.3 " W m"^-2`

A beam of vertically polarised monochromatic light is incident on a polarising filter, as shown in figure:The transmission axis of the filter is initially vertical and the transmitted light beam has the same intensity as the incident light beam.
The filter may be rotated about the direction of the light beam to change the angle of the transmission axis against the vertical.
State one angle of the transmission axis to the vertical that results in no transmitted light beam.

Answer: 90^o
Explanation:
The light is vertically polarized.
The filter's axis is initially vertical, so θ = 0°, and $=1$ → full transmission.
When the filter is rotated to 90°, its axis becomes horizontal.
`I = I_0xx cos^2(90°) = 0`