A polarised beam of light with intensity \(I\) is incident normally on a polarising filter.
The transmitted light has intensity \(I\).
The filter is rotated about the normal axis through an angle \(\theta\).
The transmitted light has intensity \(0.75I\).
What is the angle \(\theta\)?
Correct Answer: \( \boxed{\mathrm{A}} \)
Malus’ law states that
\(I=I_0\cos^2\theta\).
Given \(0.75I=I\cos^2\theta\),
\(\cos^2\theta=0.75=\dfrac{3}{4}\).
Hence, \(\cos\theta=\dfrac{\sqrt{3}}{2}\), giving
\(\theta=30^\circ\).
Therefore, the correct answer is (A).
Which group contains only waves that can be polarised?
Correct Answer: \( \boxed{\mathrm{B}} \)
Only transverse waves can be polarised.
Visible light, microwaves, and radio waves are all electromagnetic waves and are transverse, so they can be polarised.
Sound waves are longitudinal waves and cannot be polarised.
Therefore, the correct answer is (B).
Vertically polarised light of intensity \(I_0\) is incident normally on a polarising filter.
The transmission axis of the filter is at an angle \(\theta\) to the plane of polarisation of the light.
The intensity of the light after passing through the filter is \(\dfrac{I_0}{3}\).
What is \(\theta\)?
Correct Answer: \( \boxed{\mathrm{B}} \)
Malus’ law states
\(I=I_0\cos^2\theta\).
Given
\(\dfrac{I_0}{3}=I_0\cos^2\theta\).
Hence,
\(\cos^2\theta=\dfrac{1}{3}\).
\(\cos\theta=\dfrac{1}{\sqrt3}\).
\(\theta=\cos^{-1}\!\left(\dfrac{1}{\sqrt3}\right)\approx54.7^\circ\).
\(\theta\approx55^\circ\).
Therefore, the correct answer is (B).
