Solution:

Ans. (i) In an unpolarized light, the oscillations, of the electric field, are in random directions. For a polarized light, the oscillations are aligned along one particular direction

Alternatively,

Polarized light can be distinguished, from unpolarized light, when it is allowed to pass through a polaroid. Unpolarized light can show change in its intensity, on passing through a Polaroid; intensity remains same in case of unpolarized light.

When unpolarized light wave is incident on a polaroid, then the electric vector along the direction of its aligned molecules, get absorbed; the electric vector, oscillating along a direction perpendicular to the aligned molecules, pass through. This light is called linearly polarized light.

(ii) According to Malus’s Law

$$I=I_0{\cos }^2\theta$$

$$\therefore I_1=\left(\frac{I_0}{2}\right){\cos }^2\theta$$

where, $I_0$ is the intensity of unpolarized light. $\mathbf{1}$

$$\begin{array}{}\text{(1)}& \theta & ={60}^{\circ }{I}_2& =\frac{I_0}{2}{\cos }^2{60}^{\circ }& =\frac{I_0}{2}\times {\left(\frac{1}{2}\right)}^2& =\frac{I_0}{8}\end{array}$$

[CBSE Marking Scheme 2017]

AI Q. 3. (i) Why does unpolarized light from a source show a variation in intensity when viewed through a polaroid which is rotated? Show with the help of diagram how unpolarized light from sun gets linearly polarised by scattering ?

(ii) Three identical polaroid sheet $P_{1^{\prime }}{P}_2$ and $P_3$ are oriented so that the pass axes of $P_2$ and $P_3$ are inclined at the angle of ${60}^{\circ }$ and ${90}^{\circ }$ respectively with the pass axis of $P_1$. A monochromatic source $S$ of unpolarized light of intensity $I_0$ is kept in front of polaroid sheet $P_1$ as shown in figure. Determine the intensities of light as observed by the observer at $O$, when the polaroid $P_3$ is rotated with respect to $P_2$ at angle $\theta ={30}^{\circ }$ and ${60}^{\circ }$.

[OD 2016]

Ans. (i) As per Malus’ law,

Transmitted intensity $I=I_0{\cos }^2\theta$

$\therefore$ The transmitted intensity will show a variation as per ${\cos }^2\theta$.

[Note : If the student writes that “unpolarized light will not show any variation in intensity, when viewed through a polaroid, which is rotated” award this $1/2$ mark ]

The electric field, of the incident wave, makes the electrons of the air molecules, acquire both components of motion (perpendicular as wellas parallel).

Charges accelerating parallel to perpendicular component, do not radiate energy towards the observer. Hence the radiation, scattered towards observer gets linearly polarised.

(ii) Try yourself similar to Q. 8 SATQ

[CBSE Marking Scheme 2016]

[AI Q. 4. (i) Distinguish between linearly polarised and unpolarized light.

(ii) Show that the light waves are transverse in nature.

(iii) Why does light from a clear blue portion of the sky show a rise and fall of intensity when viewed through a polaroid which is rotated? Explain by drawing the necessary diagram.

U] [Delhi Comptt. I, II, III 2014]

Ans. (i) A light wave, in which the electric vector oscillates in all possible directions is known as unpolarized light.

If the oscillations of the electric vectors are restricted to only one direction, in a plane perpendicular to the direction of propagation, the corresponding light is known as linearly polarised light.

(ii)

Unpolarized light passing through the polaroid $P_1$ gets linearly polarized.

[As the electric field vector components parallel to the pass-axis of $P_1$ are transmitted, whereas the others are blocked.]

When this polarised light is incident on the Polaroid $P_2$, kept crossed with respect to $P_1$, then these components also get blocked and no light is transmitted beyond $P_2$.

(iii) It is due to the scattering of light by the molecules of earth’s atmosphere.

Incident Sunlight

(Unpolarised)

1

Ûnder the influence of the electric field of the incident (unpolarized) wave, the electrons in the molecules acquire components of motion in both these directions. The charges, accelerating parallel to the double arrows, do not radiate energy towards the observer since their acceleration has no transverse component.

The radiation scattered by the molecule is therefore represented by dots, i.e., it is polarised perpendicular to plane of figure.

[CBSE Marking Scheme 2014]