Question

The intensity of direct sunlight on the surface normal to the rays is ${I_0}$. What is the intensity of direct sunlight on a surface whose normal makes an angle ${60^ \circ }$ with the rays of the sun.
A. ${I_0}$
B. ${I_0}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
C. $\dfrac{{{I_0}}}{2}$
D. $2{I_0}$

Answer 1

Transverse light has a property to undergo a phenomenon known as polarization wherein its vibrations are confined to one plane. The concept and principle of polarization and the Law of Malus is applied in order to determine the intensity of the light rays. The components of the inclined ray of light are split into its respective components to find the answer.

Complete step by step answer:
The above problem revolves around the concept of polarization. In order to find the intensity of the light rays at an angle to the normal we first need to know the concept behind polarization. Polarization is a phenomenon where unpolarised light is converted to plane polarized light by restricting the vibrations or the oscillations of the light to just one direction in the transverse plane which is known as the plane polarized light.

Unpolarised light is said to be the transverse wave in which vibrations are present in all possible directions, in a plane which is perpendicular to the direction of propagation. The rays from the sunlight which is given to be directly incident on a surface is said to be unpolarised because it will have its vibrations in all the directions.

This light ray will have a certain amount of intensity. When it gets incident on the surface there will be a variation of its intensity based on the polarization power of the surface and this intensity is what we are asked to find. Since there will be a change in intensity the surface on which it is incident on may be considered to be a polarizer which converts this unpolarised light into a plane polarized light.

The surface on which light is being incident on is given to be perpendicular to the light rays. Since a light of intensity ${I_0}$ is incident at an angle to the normal, there will be two components which are taken into consideration. The light ray will be split into its horizontal component given by cos and the vertical component given by sine.

Since the sine component is the vertical component which means that it is perpendicular to the plane of transmission of light to the polarizer or the surface it is being incident on this component vanishes. Hence, only the cos component of the light ray is considered.Let, $I$ be the intensity of this light ray which is being incident at an angle. Hence, this intensity is given by:
$I = {I_0}\cos \theta$ ---------($1$)
Where, $\theta$ is the angle it makes with the rays of the sun. This angle is already given to be ${60^ \circ }$.
Thus, by substituting this value in equation ($1$) we get:
$I = {I_0}\cos {60^ \circ }$
We know that $\cos {60^ \circ } = \dfrac{1}{2}$.
Hence we get:
$\therefore I = \dfrac{{{I_0}}}{2}$
Therefore, the intensity of the direct sun rays whose normal makes an angle ${60^ \circ }$ is given as $\dfrac{{{I_0}}}{2}$.

Hence the correct answer is option C.

Note: The sine component of light is not taken into consideration which may be a common error that can be made in this problem. Usually, the angle made by the light rays between the polarizer and the analyzer is considered as the $\theta$ angle. However, there is no analyzer that is considered in the above problem and hence un-polarized light is said to directly fall on a surface as given which is also a point to be noted.