Question

When an unpolarized light of Intensity $$I_0$$ is incident on polarizing sheet, the Intensity of light which does not get transmitted is
(A) ${\displaystyle \frac{I_0}{2}}$
(B) ${\displaystyle \frac{I_0}{4}}$
(C) Zero
(D) $I_0$

Answer 1

Hint:We have to understand how polarized light waves can be formed. Unpolarized light waves become polarized when they pass through a polarizer.
Formula used :
We have to use the following formula to solve the problem :
$I=I_0{\cos }^2\theta$
Where $I$is the intensity of polarized light.
$I_0$ is the original intensity and $\theta$ is the angle between the direction of polarization and the axis of the filter.

Complete step by step answer:
Light is an electromagnetic wave.In ordinary light the electric field vectors are distributed in all directions in a light is called unpolarized light.Polarized light waves are those light waves which vibrates along a particular or fixed direction.And, a polarizer or polarizer sheet converts an unpolarized light to polarized light.The process of transforming unpolarized light waves into polarized light waves is known as polarization.
As we have discussed, in case of unpolarized light falling on the polarizer sheet, an amount of the light becomes polarized.
In this scenario, the theta value is $$45$$degrees.
So from the equation we can write $I=I_0{\cos }^2\theta$
$I=I_0{\cos }^2{45}^0$
$$I=I_0(\sqrt{{\displaystyle \frac{1}{2}}}{)}^2$$
$$I={\displaystyle \frac{I_0}{2}}$$
Hence the intensity of polarized light $$={\displaystyle \frac{I_0}{2}}$$
There will be an amount of light that does not get transmitted through the polarizer.
That Intensity Of Unpolarized light can be calculated by taking difference$$=\left(Intensityofunpolarizedlightbeforetransmission-Intensityofpolarizedlightaftertransmission\right)$$ Hence the Intensity of unpolarized light in which does not get transmitted is ${\displaystyle \frac{I_0}{2}}$
So the correct answer is option (A) ${\displaystyle \frac{I_0}{2}}$
Additional information :
1) White light is an unpolarized light.
2) Sun light is an unpolarized light.
3) A flame is an unpolarized light source
4) Reflected light from glass is a polarized light.

Note:Student should take care while putting the value. In the equation ${\cos }^2\theta$ is there. By mistake don’t consider $\cos \theta$.After putting the value of $\theta$ we will get the value of intensity of polarized light. So, to get the answer ( the intensity of unpolarized light) we have to subtract the intensity of polarized light from intensity of unpolarized light. Students can make silly mistakes here.