Question

In an experiment of polarization by polarizers, two polarizing sheets placed are together with their axis oriented such that no light is transmitted. A third sheet is inserted between them with its transmission axis at an angle of $${45}^{\circ }$$ with respect to each of the other axes. The fraction of incident unpolarized light intensity transmitted by the three sheet combination is
A. $${\displaystyle \frac{1}{2}}$$
B. $${\displaystyle \frac{1}{4}}$$
C. $${\displaystyle \frac{1}{8}}$$
D. $${\displaystyle \frac{1}{16}}$$

Answer 1

Hint: The intensity of light passing through the first polarizer is equal to half of the incident intensity. Use the formula for transmitted intensity of light passing through the polarizer to determine the intensity of light passing through the second and third polarizer. The ratio of intensity of light transmitted through the third polarizer and incident intensity of light is the answer.

Formula Used: $$I=I_o{\cos }^2\left(\theta \right)$$

Here, $$I_o$$ is the intensity of incident light on the polarizer and $$\theta$$ is the angle of the transmission axis with the former polarizer.

Complete step by step answer:
We know that, when two polarizers are placed in front of each other such that their transmission axes are at angle $${90}^{\circ }$$ with each other, no light can pass through the second polarizer.

Now, we have given that the third polarizer is inserted between them. We know that the intensity of light passing through the first polarizer is half of the intensity incident on it irrespective of the angle of the transmission axis.
Let the intensity of light incident of the first polarizer is $$I_0$$. So, the intensity of light coming out from the first polarizer is $${\displaystyle \frac{I_0}{2}}$$.

We know the formula for intensity of light transmitted from the polarizer is,
$$I=I_o{\cos }^2\left(\theta \right)$$

Here, $$I_o$$ is the intensity of incident light on the polarizer and $$\theta$$ is the angle of the transmission axis with the former polarizer.

Therefore, the intensity of light transmitted from the second polarizer is,
$$I_2={\displaystyle \frac{I_o}{2}}{\cos }^2\left({45}^{\circ }\right)$$

Here, $${\displaystyle \frac{I_o}{2}}$$ is the intensity of light coming out from the first polarizer which is incident light for the second polarizer.
$$I_2={\displaystyle \frac{I_o}{2}}\left({\displaystyle \frac{1}{2}}\right)$$
$$\Rightarrow I_2={\displaystyle \frac{I_o}{4}}$$

This will be the incident intensity of light for the third polarizer.

The intensity of light transmitted from the third polarizer is,
$$I_3={\displaystyle \frac{I_o}{4}}{\cos }^2\left({45}^{\circ }\right)$$
$$\Rightarrow I_3=\left({\displaystyle \frac{I_o}{4}}\right)\left({\displaystyle \frac{1}{2}}\right)$$
$$\Rightarrow I_3={\displaystyle \frac{I_o}{8}}$$

Therefore, the fraction of light transmitted through combination of three polarizers is,
$$f={\displaystyle \frac{I_3}{I_0}}$$
$$\Rightarrow f={\displaystyle \frac{{\displaystyle \frac{I_0}{8}}}{I_0}}$$
$$\Rightarrow f={\displaystyle \frac{1}{8}}$$

So, the correct answer is option (C).

Note:You should not use the formula, $$I=I_o{\cos }^2\left(\theta \right)$$ to calculate the intensity of light passing through the first polarizer. The intensity of light passing through the first polarizer is always half of the intensity incident on it.