Question
Explain (/ Discuss) the variation in intensity of the light transmitted through two polaroids. And from that derive Malus' law.

Answer

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The light emitted from an oridinary source (like a sodium lamp) is unpolarised.
When such a light is passed through a polaroid sheet $P_1$, it is observed that its intensity is reduced by half. Rotating $P _{\text {, }}$ has no effect on the transmitted beam and transmitted intensity remains constant.
Now, let an identical piece of polaroid $P _2$ be placed before $P _1$.
As shown in Fig., initially $P_1$ and $P_2$ are arranged in such a way that their pass-axis are parallel to each other.
In this case, the intensity of the transmitted light is the same.
Now, if $P _1$ is rotated, there is variation seen in the light coming out of $P _2$.
As shown in a position in the fig., the intensity transmitted through $P_2$ followed by $P_1$ is nearly zero.
When $P _1$ is rotated by $90^{\circ}$, in one position for light coming from $P _2$, total intensity is absorbed by $P_1$. So, the intensity of light emerging from $P_1$ is zero.
Suppose, the pass axis of $P _2$ makes an angle $\theta$ with the pass axis of $P_1$, then when the polarised beam passes through the polaroid $P _2$, the component $E \cos \theta$ (along the pass-axis of $P_2$ ) will pass through $P_2$. Thus, as we rotate the polaroid $P _1$ (or $P _2$ ), the intensity will vary as :
$I = I _0 \cos ^2 \theta$
where $I_0$ is the intensity of the polarized light after passing through $P _1$.
This is known as Malus' law.
The above discussion shows that the intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from $50 \%$ to zero of the incident intensity by adjusting the angle between the pass axis of two polaroids.


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Coulomb's law states that the electrostatic force of attraction or repulsion acting between two stationary point charges is given by
Image

where F denotes the force between two charges $q _1$ and $q _2$ separated by a distance r in free space, $\varepsilon_0$ is a constant known as the permittivity of free space. Free space is a vacuum and may be taken to be air practically. If free space is replaced by a medium, then $\varepsilon_0$ is replaced by $\left(\varepsilon_0 k\right)$ or $\left(\varepsilon_0 \varepsilon_r\right)$ where k is known as dielectric constant or relative permittivity.

(i) In coulomb's law, $F =k \frac{q_1 q_2}{r^2}$, then on which of the following factors does the proportionality constant k depends?
(a) Nature of the medium between the two charges
(b) Distance between the two charges
(c) Electrostatic force acting between the two charges
(d) Magnitude of the two charges

(ii) Dimensional formula for the permittivity constant $\varepsilon_0$ of free space is
(a) $\left[ M ^{-1} L^3 T^2 A^2\right]$
(b) $\left[ ML ^{-3} T^4 A^2\right]$
(c) $\left[ M ^{-1} L^{-3} T^4 A^2\right]$
(d) $\left[ ML ^{-3} T^4 A^{-2}\right]$

(iii) The force of repulsion between two charges of 1 C each, kept 1m apart in vaccum is
(a) $\frac{1}{9 \times 10^9} N$
(b) $\frac{1}{9 \times 10^{12}} N$
(c) $9 \times 10^7 N$
(d) $9 \times 10^9 N$

(iv) Two identical charges repel each other with a force equal to 10 mgwt when they are 0.6 m apart in air. ( $g =$ $10 m s ^{-2}$ ). The value of each charge is
(a) 2 mC
(b) $2 \times 10^{-7} mC$
(c) $2 \mu C$
(d) 2 nC

OR

Coulomb's law for the force between electric charges most closely resembles with
(a) law of conservation of energy
(b) Newton's 2nd law of motion
(c) law of conservation of charge .
(d) Newton's law of gravitation
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