A beam of linearly polarised light of intensity I is incident on an...

According to Malus's law, the intensity of transmitted light is given by:

$I' = I \cos^2 \theta$

Given that the transmitted intensity is $I' = \dfrac{I}{4}$, we can substitute this into the equation:

$\dfrac{I}{4} = I \cos^2 \theta$

$\cos^2 \theta = \dfrac{1}{4}$

$\cos \theta = \dfrac{1}{2}$

The angle whose cosine is $\dfrac{1}{2}$ is $60^{\circ}$.

Choosing $30^{\circ}$ is a common mistake resulting from using sine instead of cosine, and $45^{\circ}$ comes from incorrectly evaluating the inverse trigonometric function.

Answer: $60^{\circ}$