## wave-optics Question 43

Question: Q. 7. Two polaroids, $P_{1}$ and $P_{2}$ are ‘set-up’ so that their ‘pass-axes’ are ‘crossed’ with respect to each other. A third polaroid, $P_{3}$, is now introduced between these two so that its ‘pass-axis’ makes an angle $\theta$ with ‘pass-axis’ of $P_{1}$.

A beam of unpolarised light, of intensity $I$, is incident on $P_{1}$. If the intensity of light, that gets transmitted through this combination of three polaroids, is $I^{\prime}$, find the ratio $\left(\frac{I^{\prime}}{I}\right)$ when $\theta$ equals :

(i) $30^{\circ}$, (ii) $45^{\circ}$.

A[Delhi Comptt., 2016]

Intensity of unpolarised light is given as $I$, It becomes $\frac{I}{2}$ on passing through polaroid $P_{1}$

$\therefore$ Using law of Malus, for the intensity passing through $P_{3}$

$$ I_{1}=\left(\frac{I}{2}\right) \cos ^{2} \theta $$

This intensity $I_{1}$ is incident on $P_{2}$

Hence using law of Malus for Polaroid $P_{2}$

$I^{\prime}=I_{1} \cos ^{2}\left(90^{\circ}-\theta\right)$

$=\frac{I}{2} \cos ^{2} \theta\left[\cos ^{2}\left(90^{\circ}-\theta\right)\right]$

$=\frac{I}{8}(2 \cos \theta \sin \theta)^{2}$

$1 / 2$

$=\frac{I}{8}(\sin 2 \theta)^{2}$

$1 / 2$

(i) When

$\theta=30^{\circ}$

Then

$I^{\prime}=\frac{I}{8}\left(\sin 60^{\circ}\right)^{2}$

$\Rightarrow \quad \frac{I^{\prime}}{I}=\frac{3}{32}$

(ii) When $\quad \theta=45^{\circ}$

$\Rightarrow \quad \frac{I^{\prime}}{I}=\frac{1}{8}$

[CBSE Marking Scheme 2016]

## Commonly Made Error

- Many students used a final long and tedius method to obtain final Intensity.

Q. 8. (a) When an unpolarized light of intensity $I_{0}$ is passed through a polaroid, what is the intensity of the linearly polarized light? Does it depend on the orientation of the polaroid? Explain your answer.

(b) A plane polarized beam of light is passed through a polaroid. Show graphically the variation of the intensity of the transmitted light with angle of rotation of the polaroid in complete one rotation.

[CBSE Comptt. 2018]

## Show Answer

Solution:

Ans. (a) Intensity of linearly polarized light $1 / 2$ Dependence on orientation Explanation

(b) Graphical representation

(a) The intensity of the linearly polarized light would $\frac{I_{0}}{2}$.

$1 / 2$

No; it does not depend on the orientation. $1 / 2$

Explanation : The polaroid will let the component of the unpolarized light, parallel to its pass axis, to pass through it irrespective of its orientation.

(b) We have

$$ I=I_{0} \cos ^{2} \theta $$

$\therefore$ The graph is as shown below

[CBSE Marking Scheme 2018]

## Long Answer Type Questions

(5 marks each)