Question

On a polarizing sheet a mixture of plane polarised and unpolarised light falls normally. On rotating the polarising sheet about the direction of the incident beam, the transmitted maximum and minimum intensities vary by a factor 4. The ratio of the intensities of polarised and unpolarised light is ?

Answer 1

Answer on Question#41320, Physics, Optics

On a polarizing sheet a mixture of plane polarized and unpolarized light falls normally. On rotating the polarizing sheet about the direction of the incident beam, the transmitted maximum and minimum intensities vary by a factor 4. The ratio of the intensities of polarized and unpolarized light is?

Solution:

If a beam contains a mix of plane polarized light and unpolarized light the fraction of the polarized light can be determined by continuously measuring the transmittance through a polarizer as this is rotated through an angle of $90{}^{\circ}$ about the beam axis. At maximum all the polarized and half the unpolarized light is transmitted, while at the minimum simply half the unpolarized light is transmitted.

Thus,

$$I_{max} = I_p + \frac{I_u}{2}$$ $$I_{min} = \frac{I_u}{2}$$

where $I_p$ is the intensity of plane polarized light, and $I_u$ is the intensity of unpolarized light.

We have that

$$\frac{I_{max}}{I_{min}} = 4$$

Thus,

$$\frac{I_p + \frac{I_u}{2}}{\frac{I_u}{2}} = 4$$ $$I_p + \frac{I_u}{2} = 4 \frac{I_u}{2}$$ $$I_p = \frac{3I_u}{2}$$

So,

$$\frac{I_p}{I_u} = \frac{3}{2} = 1.5$$

Answer, $\frac{I_p}{I_u} = \frac{3}{2} = 1.5$.

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