Polarisation of Light and Malus' Law Questions in English

(A) When unpolarised light of intensity $I_0$ passes through the first polariser,it becomes linearly polarised with intensity $I_1 = I_0 / 2$.
Since the first and second polaroids were initially crossed (angle $90^{\circ}$),the third polaroid is placed at an angle $\theta$ with the first and $(90^{\circ} - \theta)$ with the second.
According to Malus' Law,the intensity after the second (middle) polariser is $I_2 = I_1 \cos^2 \theta = (I_0 / 2) \cos^2 \theta$.
The intensity after the third (last) polariser is $I_3 = I_2 \cos^2(90^{\circ} - \theta) = (I_0 / 2) \cos^2 \theta \sin^2 \theta$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 2\theta = 4 \sin^2 \theta \cos^2 \theta$.
Therefore,$I_3 = \frac{I_0}{2} \cdot \frac{\sin^2 2\theta}{4} = \frac{I_0}{8} \sin^2 2\theta$.

(C) Let $I_0 = 32 \ W m^{-2}$ be the intensity of unpolarised light.
After passing through the first polaroid,the intensity becomes $I_1 = I_0 / 2 = 32 / 2 = 16 \ W m^{-2}$.
Let $\theta$ be the angle between the axes of the first and second polaroid. The intensity after the second polaroid is $I_2 = I_1 \cos^2 \theta = 16 \cos^2 \theta$.
The angle between the second and third polaroid is $(90^{\circ} - \theta)$ because the first and third are crossed.
The intensity after the third polaroid is $I_3 = I_2 \cos^2(90^{\circ} - \theta) = I_2 \sin^2 \theta$.
Substituting $I_2$,we get $I_3 = 16 \cos^2 \theta \sin^2 \theta = 16 (\sin \theta \cos \theta)^2 = 16 (\sin(2\theta) / 2)^2 = 4 \sin^2(2\theta)$.
Given $I_3 = 3 \ W m^{-2}$,we have $4 \sin^2(2\theta) = 3$,so $\sin^2(2\theta) = 3/4$.
Thus,$\sin(2\theta) = \sqrt{3}/2$,which implies $2\theta = 60^{\circ}$ or $120^{\circ}$.
Therefore,$\theta = 30^{\circ}$ or $60^{\circ}$.

(B) In double refraction (birefringence),a crystal splits an unpolarized light beam into two rays: the ordinary ray ($O$-ray) and the extraordinary ray ($E$-ray).
Statement $A$ is correct: The refractive index of the $E$-ray depends on the direction of propagation relative to the optic axis,which effectively means it varies with the angle of incidence.
Statement $B$ is correct: Both the $O$-ray and $E$-ray are plane-polarized. Polarization is the phenomenon where light waves acquire 'one-sidedness' (vibrations restricted to a single plane).
Therefore,both statements are correct.

(B) Initially,the two polarisers $P_1$ and $P_2$ are crossed,meaning the angle between their pass-axes is $90^{\circ}$,which blocks all light.
When an unpolarised beam of intensity $I_0$ passes through the first polariser $P_1$,the intensity of the emerging light is $I_1 = \frac{I_0}{2}$.
$A$ third polariser $P_3$ is introduced between $P_1$ and $P_2$ at an angle $\theta_1 = 60^{\circ}$ with respect to $P_1$.
Using Malus's Law,the intensity $I_3$ emerging from $P_3$ is $I_3 = I_1 \cos^2(60^{\circ}) = \frac{I_0}{2} \times (\frac{1}{2})^2 = \frac{I_0}{8}$.
The angle between the pass-axes of $P_3$ and $P_2$ is $\theta_2 = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
The final intensity $I_f$ emerging from $P_2$ is $I_f = I_3 \cos^2(30^{\circ}) = \frac{I_0}{8} \times (\frac{\sqrt{3}}{2})^2 = \frac{I_0}{8} \times \frac{3}{4} = \frac{3}{32} I_0$.

(C) The light is partially polarized. The intensities along the principal axes are $I_y = I_0$ and $I_x = \frac{2I_0}{3}$.
When a polaroid is placed with its pass axis at an angle $\theta = 45^{\circ}$ to the $y$-axis,the transmitted intensity $I$ is given by the sum of the components of the intensities along the pass axis:
$I = I_y \cos^2(45^{\circ}) + I_x \cos^2(45^{\circ})$
Substituting the given values:
$I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 + \frac{2I_0}{3} \left(\frac{1}{\sqrt{2}}\right)^2$
$I = I_0 \cdot \frac{1}{2} + \frac{2I_0}{3} \cdot \frac{1}{2}$
$I = \frac{I_0}{2} + \frac{I_0}{3} = \frac{3I_0 + 2I_0}{6} = \frac{5I_0}{6}$

(D) Longitudinal waves cannot be polarized. Polarization is a property that applies only to transverse waves,where the oscillations occur perpendicular to the direction of wave propagation. In longitudinal waves,the oscillations occur parallel to the direction of wave propagation. Since there is no asymmetry in the direction of oscillation relative to the direction of propagation,longitudinal waves cannot be restricted to a single plane of vibration.

(C) Let $I_0$ be the initial intensity of unpolarized light.
$1$. Without the third polarizer: The first polarizer reduces intensity to $I_1 = I_0/2$. The second polarizer at $90^\circ$ relative to the first (which is at $30^\circ$) has an angle $\theta = 90^\circ - 30^\circ = 60^\circ$. The output intensity is $I_{final} = I_1 \cos^2(60^\circ) = (I_0/2) \times (1/4) = I_0/8$.
$2$. With the third polarizer at $60^\circ$: $I_1 = I_0/2$. Intensity after the third polarizer (angle $60^\circ - 30^\circ = 30^\circ$): $I_2 = I_1 \cos^2(30^\circ) = (I_0/2) \times (3/4) = 3I_0/8$. Intensity after the second polarizer (angle $90^\circ - 60^\circ = 30^\circ$): $I_{final}' = I_2 \cos^2(30^\circ) = (3I_0/8) \times (3/4) = 9I_0/32$.
Ratio = $(9I_0/32) / (I_0/8) = 9/4$.

(A) The intensity of unpolarized light after passing through a polarizer is $I_p = \frac{I_0}{2}$.
Let $\theta$ be the angle of rotation produced by the optically active solution.
The light incident on the analyzer is polarized at an angle $\theta$ relative to the transmission axis of the analyzer (since the polarizer and analyzer are initially at $0^{\circ}$ relative to each other).
According to Malus's Law,the intensity of light emerging from the analyzer is $I = I_p \cos^2(\theta) = \frac{I_0}{2} \cos^2(\theta)$.
Given $I = \frac{3}{8} I_0$,we have $\frac{I_0}{2} \cos^2(\theta) = \frac{3}{8} I_0$.
Simplifying this,$\cos^2(\theta) = \frac{3}{4}$,which gives $\cos(\theta) = \frac{\sqrt{3}}{2}$.
Therefore,$\theta = 30^{\circ}$.