Question

Plane polarised light is passed through an analyser and the intensity of emerging light is reduced by $75\%$. Optical vibrations make an angle $\theta$ with the axis of analyser. Then $\theta$ is

• A. $60^{\circ }$
• B. $45^{\circ }$
• C. $30^{\circ }$
• D. $58^{\circ }$

Answer 1

Answer: (A)

Suppose intensity of light incident on the polariser be $I_0$. When it passes through polariser, its intensity is reduced to $\frac{I_0}{2}$
$I=\frac{I_0}{2}{\cos }^2\alpha$ (Malus's law)
where a is the angle between the plane of transmission analyser and polariser
According to the question,
$I=75\%$ of $\frac{I_0}{2}=\frac{3}{4}\times \frac{I_0}{2}$
Hence, ${\cos }^2\alpha =\frac{3}{4}$ or $\cos \alpha =\frac{\sqrt{3}}{2}=\cos 30^{\circ }$
$\therefore \alpha =30^{\circ }$