Question

A polaroid is placed at 45° to an incoming light of intensity ${{I}_{0}}$ now the intensity of light passing through the polaroid after polarization would be:
A. ${{I}_{0}}$
B. $\dfrac{{{I}_{0}}}{2}$
C. $\dfrac{{{I}_{0}}}{4}$
D. Zero

Answer 1

Hint: Polaroids are the materials which restrict the vibrations of waves from many planes into a single plane. The light passing through a polaroid is governed by a simple formula. We are going to use this formula to solve the problem.

Formula used:
$I={{I}_{0}}{{\cos }^{2}}\theta $

Complete answer:
We know that the intensity of light passing through a polaroid after polarization is given by:
$I={{I}_{0}}{{\cos }^{2}}\theta $ ----(i)
Where, $I$ is the intensity of the light incident on the polaroid before polarization.
${{I}_{0}}$ is the intensity of light coming out of the polaroid after polarization.
And, $\theta $ is the angle of incidence of the light on the polaroid.
Here, according to question, $\theta =45{}^\circ $
Hence, $I={{I}_{0}}{{\cos }^{2}}45$
$\Rightarrow I={{I}_{0}}{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}$
$\Rightarrow I=\dfrac{{{I}_{0}}}{2}$ ---(ii)

So, the correct answer is “Option B”.

Additional Information:
In unpolarized light the vibrations are not confined to any particular plane. These vibrations are unwanted in certain scenarios. So we need to restrict these vibrations to a particular desired plane as per our need. This restriction of the vibrations of the light wave from many planes into a single desired plane is done with the help of polaroid. Polaroid only allows those waves to pass through it which vibrate in a particular plane while restricting others from passing through it.

There are many applications of these polaroids. For example, these polaroids are used in certain eye wears so that the glare of the sun could be reduced. Hardness of a molecule could also be determined with the help of polaroids.

Note:
The angle of incidence on a polaroid is measured from the normal to the polaroid to the incident wave. Do not take the angle of incidence as the angle from the plane of polaroid to the incident wave. Also, note that the formula of output intensity has ${{\cos }^{2}}$ and not $\cos $.