Physics - Wave Optics Question with Solution | TestHub

The intensity of unpolarised light is $I_0$.

When this light passes through the first polaroid, the intensity of the transmitted light becomes $I_1=\frac{I_0}{2}$.

The second polaroid is placed such that no light is emitted from it, which means its transmission axis is perpendicular to that of the first polaroid. The angle between their axes is $90^{\circ }$.

A third polaroid is placed between them with its axis at an angle $\theta$ with respect to the first polaroid.

According to Malus's law, the intensity of light after passing through the third polaroid is:

$I_3=I_1{\cos }^2\theta =\frac{I_0}{2}{\cos }^2\theta$

The angle between the transmission axis of the third polaroid and the second polaroid is $90^{\circ }-\theta$.

The intensity of light emerging from the last (second) polaroid is:

$I_2=I_3{\cos }^2(90^{\circ }-\theta )$

$I_2=\left(\frac{I_0}{2}{\cos }^2\theta \right){\sin }^2\theta$

Using the trigonometric identity $\sin 2\theta =2\sin \theta \cos \theta$, we get ${\sin }^2\theta {\cos }^2\theta =\frac{(\sin 2\theta {)}^2}{4}$.

Substituting this into the equation for $I_2$:

$I_2=\frac{I_0}{2}\times \frac{(\sin 2\theta {)}^2}{4}=\frac{I_0}{8}(\sin 2\theta {)}^2$

Answer: $\frac{I_0}{8}(\sin 2\theta {)}^2$