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)