Question 55: Wave Optics

Question

An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at $30^{\circ}$ and $90^{\circ}$, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at $60^{\circ}$ with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is $\_\_\_\_$ .

Accepted Answer

Let the intensity of the unpolarized incident light be $I_0$.
We use
MalusÃ¢ÂÂ law
:
If plane polarized light passes through a polarizer making angle $	heta$ with the direction of polarization, then transmitted intensity is
$$ I = I_{	ext{in}} \cos^2 	heta $$
Also, when
unpolarized light
passes through a polarizer, its intensity becomes
$$ \frac{I_0}{2} $$
1) Without the third polarizer
The two polarizers have transmission axes at $30^\circ$ and $90^\circ$.
So, the angle between their axes is
$$ 90^\circ - 30^\circ = 60^\circ $$
After passing through the first polarizer, intensity becomes
$$ I_1 = \frac{I_0}{2} $$
Now this light passes through the second polarizer, so
$$ I_{	ext{without}} = I_1 \cos^2 60^\circ $$
Since
$$ \cos 60^\circ = \frac{1}{2} $$
therefore,
$$ I_{	ext{without}} = \frac{I_0}{2} \cdot \left(\frac{1}{2}ight)^2 = \frac{I_0}{2} \cdot \frac{1}{4} = \frac{I_0}{8} $$
2) With the third polarizer inserted between them
Now the polarizers are at $30^\circ$, $60^\circ$, and $90^\circ$.
After first polarizer:
$$ I_1 = \frac{I_0}{2} $$
From first to third polarizer:
Angle difference is
$$ 60^\circ - 30^\circ = 30^\circ $$
So intensity after the third polarizer is
$$ I_2 = I_1 \cos^2 30^\circ $$
Since
$$ \cos 30^\circ = \frac{\sqrt{3}}{2} $$
we get
$$ I_2 = \frac{I_0}{2} \cdot \left(\frac{\sqrt{3}}{2}ight)^2 = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3I_0}{8} $$
From third to second polarizer:
Angle difference is again
$$ 90^\circ - 60^\circ = 30^\circ $$
So final intensity is
$$ I_{	ext{with}} = I_2 \cos^2 30^\circ $$
$$ I_{	ext{with}} = \frac{3I_0}{8} \cdot \frac{3}{4} = \frac{9I_0}{32} $$
3) Required ratio
$$ \frac{I_{	ext{with}}}{I_{	ext{without}}} = \frac{\frac{9I_0}{32}}{\frac{I_0}{8}} = \frac{9}{32} \cdot 8 = \frac{9}{4} $$
So, the required ratio is
$$ \boxed{\frac{9}{4}} $$
Hence, the correct option is
Option C
.

Answer

3/4

Answer

4/3

Answer

9/4

Answer

4/9

Wave Optics

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