Question 305: Waves Optics

Question

Three identical polaroids $P_1, P_2$ and $P_3$ are placed one after another. The pass axis of $P_2$ and $P_3$ are inclined at an angle $60^{\circ}$ and $90^{\circ}$ with respect to axis of $P_1$. The source has an intensity $I_0$. The intensity of transmitted light through $P_3$ is $\left(\cos 60^{\circ}=0.5, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)$

Accepted Answer

The source intensity is $ I_0 $.
The pass axis of $ P_2 $ is at an angle of $ 60^\circ $ with respect to the axis of $ P_1 $.
The pass axis of $ P_3 $ is at an angle of $ 90^\circ $ with respect to the axis of $ P_1 $.
Transmission through $ P_1 $
:
The light intensity after passing through $ P_1 $ is reduced to half, according to the properties of a polaroid.
Therefore, $ I = \frac{I_0}{2} $.
Transmission through $ P_2 $
:
Use Malus' Law: $ I = I_0 \cos^2 	heta $.
At $ P_2 $, $ 	heta = 60^\circ $.
Hence, $ I_2 = \frac{I_0}{2} \cos^2 60^\circ = \frac{I_0}{2} 	imes (0.5)^2 = \frac{I_0}{8} $.
Transmission through $ P_3 $
:
Between $ P_2 $ and $ P_3 $, the angle is effectively $ 30^\circ $ due to the relative orientations.
Use MalusÃ¢ÂÂ law again: $ I_3 = \frac{I_0}{8} \cos^2 30^\circ $.
With $ \cos 30^\circ = \frac{\sqrt{3}}{2} $, the calculation becomes:
$    I_3 = \frac{I_0}{8} 	imes \left(\frac{\sqrt{3}}{2}ight)^2 = \frac{I_0}{8} 	imes \frac{3}{4} = \frac{3I_0}{32}    $
This calculation shows how the intensity $ I_3 $ of the transmitted light through $ P_3 $ is reduced to $ \frac{3I_0}{32} $.

Answer

$\frac{\mathrm{I}_0}{8}$

Answer

$\frac{3 \mathrm{I}_0}{16}$

Answer

$\frac{3 \mathrm{I}_0}{32}$

Answer

$\frac{\mathrm{I}_0}{32}$

Waves Optics

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