Question

A polaroid making an angle ${{60}^{0}}$ with electric vector is then reduced by a factor of –

& \text{A) }\dfrac{1}{4} \\\ & \text{B) }\dfrac{3}{4} \\\ & \text{C) }\dfrac{1}{2} \\\ & \text{D) }\dfrac{1}{3} \\\ \end{aligned}$$

Answer 1

We need to analyse and understand the laws and relations between the angle made by a polaroid when kept in the path of an unpolarised light. The amount of intensity reduced is given by the Malu’s law which can be used to solve this problem.

Complete answer:
We are given a polaroid which is aligned at an angle of ${{60}^{0}}$ with the electric vector of the electromagnetic waves falling on it. We know that the polaroid or polarizer is a device which is used to polarise an incident light.
The polarization is the process of removing the unwanted light rays which travel in directions which are not required for a particular experiment. There are solutions like sucrose which are natural polarizers which can pass the particular components of lights.
The polarized lights, as we can assume, will have lower intensity as a part of the light is removed from the beam. The amount of light reduced is dependent on the angle at which the light rays are incident on the polariser.
In the present situation, the polariser is kept such that it forms ${{60}^{0}}$ with the electric vector of the light falling on it. According to the Malu’s law, the intensity is proportional to the square of the cosine of the angle between the polariser and the component of light. We can write it as –

& I\propto {{\cos }^{2}}\theta \\\ & \Rightarrow I={{I}_{0}}{{\cos }^{2}}\theta \\\ & \Rightarrow I={{I}_{0}}{{\cos }^{2}}({{60}^{0}}) \\\ & \therefore \dfrac{I}{{{I}_{0}}}=\dfrac{1}{4} \\\ \end{aligned}$$ The ratio between the final intensity of the light to the initial intensity is 1:4. **The correct answer is option A.** **Note:** The polarization was observed in the chemical compounds very lately. After the discovery, the light properties of the chemical substances became a basic property which can be used to distinguish the various types of the same chemical composition.