Question

A mixture of polarised and unpolarised light is incident on a polariser. As the polariser is rotated through ${360}^{\circ }$, it is found that the minimum and the maximum of the transmitted intensity is in the ratio $1:q$. The ratio of intensities of polarised to unpolarised light in the incident beam is

• A. $\frac{1}{2}(q-1):1$
• B. $1:\frac{1}{2}(q-1)$
• C. $\frac{1}{2}q:1$
• D. $1:\frac{1}{2}q$

Answer 1

The correct option is A $\frac{1}{2}(q-1):1$

Malus law is mathematically stated as, I${}_1$ = I${}_0$Cos${}^2$q

where q is the angle between the polarization direction of the light and the transmission axis of the polarizer.

I${}_0$ is incident intensity on polarized light

I am the transmitted intensity

I = I${}_0$Cos${}^2$360 i.e the value of this equation is 1.

I = I${}_0$

Hence the value of unpolarized light is 1

For polarised light;

According to malus law,

When unpolarized light is polarized with two polarizers, the intensity becomes I=Iocos2(θ)I=Iocos2(θ)(Malus's law). But when unpolarized light is polarized with only one polarizer, the intensity is reduced to half the intensity of the unpolarized light.

I = $\frac{1}{2\pi }$${\int }_0^{2\pi }$I${}_0$Cos($\theta$)${}^2$ = $\frac{1}{2}$. Hence the value of polarised is $\frac{1}{2}$.

Since the intensity is reduced by half the value of q i.e polarising angle will be subtracted by 1. Hence its value will become $\frac{1}{2}$(q - 1)

Thus the ratio of intensities of polarised to unpolarised light in the incident beam is $\frac{1}{2}(q-1):1$