Polarisation of Light and Malus' Law Questions in English

(C) When an already polarised beam passes through a polariser,the intensity $I$ of light obtained is given by Malus' law: $I = I_0 \cos^2 \theta$,where $I_0$ is the intensity of incident polarised light and $\theta$ is the angle between the plane of polarisation and the transmission axis of the polariser.
Given that each polariser absorbs $10 \%$ of the light,the transmission factor is $k = 0.9$.
$1$. For the first polariser $(P_1)$: The incident light is vertical. The transmission axis is at $30^{\circ}$ to the vertical. So,$\theta_1 = 30^{\circ}$.
$I_1 = k \cdot I_0 \cdot \cos^2(30^{\circ}) = 0.9 \cdot 100 \cdot (\sqrt{3}/2)^2 = 0.9 \cdot 100 \cdot 0.75 = 67.5 \, W/m^2$.
$2$. For the second polariser $(P_2)$: The light incident on $P_2$ is polarised at $30^{\circ}$. The transmission axis of $P_2$ is at $60^{\circ}$ to the vertical. The angle between the incident light and the axis is $\theta_2 = 60^{\circ} - 30^{\circ} = 30^{\circ}$.
$I_2 = k \cdot I_1 \cdot \cos^2(30^{\circ}) = 0.9 \cdot 67.5 \cdot 0.75 = 45.5625 \, W/m^2$.
$3$. For the third polariser $(P_3)$: The light incident on $P_3$ is polarised at $60^{\circ}$. The transmission axis of $P_3$ is at $90^{\circ}$ to the vertical. The angle between the incident light and the axis is $\theta_3 = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
$I_3 = k \cdot I_2 \cdot \cos^2(30^{\circ}) = 0.9 \cdot 45.5625 \cdot 0.75 \approx 30.75 \, W/m^2$.
The final intensity is approximately $30 \, W/m^2$.

(A) Since the beam incident on the first polariser is unpolarised,its intensity is reduced to half upon passing through the first polariser. Malus' law is not applicable here.
Therefore,the intensity $I_{1}$ after the first polariser is:
$I_{1} = \frac{I_{0}}{2} = \frac{20}{2} = 10 \,W/m^{2}$
As the light emerging from the first polariser is linearly polarised,Malus' law is applicable for the second polariser.
The intensity $I_{2}$ obtained after the second polariser is given by:
$I_{2} = I_{1} \cdot \cos^{2} \theta$
where $\theta$ is the angle between the transmission axes of the first and second polarisers.
Given the angles are $30^{\circ}$ and $60^{\circ}$ from the vertical,the angle between them is $\theta = 60^{\circ} - 30^{\circ} = 30^{\circ}$.
Thus,$I_{2} = 10 \times \cos^{2} 30^{\circ}$
$I_{2} = 10 \times \left(\frac{\sqrt{3}}{2}\right)^{2} = 10 \times \frac{3}{4} = 7.5 \,W/m^{2}$

(C) When an unpolarised light of intensity $I_0$ passes through the first polariser,the intensity of the transmitted light becomes $I_1 = \frac{I_0}{2}$.
Now,this polarised light passes through the second polariser,which is at an angle $\theta = 30^{\circ}$ with respect to the first one. According to Malus' Law,the intensity of the emergent light $I'$ is given by:
$I' = I_1 \cos^2 \theta$
Substituting the values:
$I' = \frac{I_0}{2} \times \cos^2 30^{\circ}$
$I' = \frac{I_0}{2} \times \left( \frac{\sqrt{3}}{2} \right)^2$
$I' = \frac{I_0}{2} \times \frac{3}{4}$
$I' = \frac{3 I_0}{8}$

(C) Let the intensity of the incident light be $I_0$. According to Malus' law,the intensity of light after passing through a polariser is $I = I_{in} \cos^2 \theta$,where $\theta$ is the angle between the incident light's polarisation plane and the polariser's axis.
Since the first polariser's axis is parallel to the plane of polarisation of the incident light,the angle $\theta_1 = 0^{\circ}$. Thus,the intensity after the first polariser is $I_1 = I_0 \cos^2(0^{\circ}) = I_0$.
For the subsequent polarisers,each is rotated by $30^{\circ}$ relative to the previous one. Therefore,the angle between the light's polarisation plane and the next polariser is $30^{\circ}$.
Intensity after the second polariser: $I_2 = I_1 \cos^2(30^{\circ}) = I_0 (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} I_0$.
Intensity after the third polariser: $I_3 = I_2 \cos^2(30^{\circ}) = (\frac{3}{4} I_0) \times \frac{3}{4} = (\frac{3}{4})^2 I_0$.
Intensity after the fourth polariser: $I_4 = I_3 \cos^2(30^{\circ}) = (\frac{3}{4})^2 I_0 \times \frac{3}{4} = (\frac{3}{4})^3 I_0$.
Intensity after the fifth polariser: $I_5 = I_4 \cos^2(30^{\circ}) = (\frac{3}{4})^3 I_0 \times \frac{3}{4} = (\frac{3}{4})^4 I_0$.
Calculating the value: $I_5 = (0.75)^4 I_0 = 0.3164 I_0 \approx 31.6 \% I_0$.
Comparing with the given options,the intensity is closest to $30 \%$ of the incident light.

(D) The correct answer is $D$.
Polarisation is a phenomenon that occurs only in transverse waves,where the oscillations of the medium particles (or electric field vectors in electromagnetic waves) are restricted to a specific plane perpendicular to the direction of wave propagation.
Light is an electromagnetic wave,which is transverse in nature,and therefore it exhibits the phenomenon of polarisation.
Sound waves in an air column are longitudinal waves,meaning the particles of the medium oscillate parallel to the direction of wave propagation. Longitudinal waves cannot be polarised because they lack the necessary transverse components to restrict oscillation to a single plane.
Reflection,diffraction,and refraction are properties exhibited by both light waves and sound waves.

(D) The angle of incidence $i$ is the angle between the incident ray and the normal. From the diagram,the angle between the incident ray and the surface is $33^{\circ}$. Therefore,the angle of incidence $i = 90^{\circ} - 33^{\circ} = 57^{\circ}$.
Given the refractive index $\mu = 1.54$ and $\tan 57^{\circ} = 1.54$,we have $\mu = \tan i_p$,where $i_p$ is the Brewster's angle.
Since the light is incident at Brewster's angle,the reflected ray $OB$ is completely plane-polarized.
When this plane-polarized light passes through a rotating polaroid,the intensity $I$ of the transmitted light varies according to Malus's Law: $I = I_0 \cos^2 \theta$,where $\theta$ is the angle between the transmission axis of the polaroid and the plane of polarization of the light.
As the polaroid is rotated,the intensity will gradually decrease to zero (when $\theta = 90^{\circ}$) and then increase again as the polaroid continues to rotate.

(C) Option $C$ is correct. When sunlight enters the Earth's atmosphere,the atoms and molecules scatter the light. For light scattered at an angle of $\frac{\pi}{2}$ ($90$ degrees) relative to the incident beam,the scattered light is found to be plane-polarised. Option $A$ is incorrect because plane-polarised light remains plane-polarised after passing through a polaroid (its intensity may change). Option $B$ is incorrect because the refracted light at Brewster's angle is partially polarised,not fully linearly polarised. Option $D$ is incorrect because natural sunlight is unpolarised.

(A) Let the initial intensity of the unpolarized light from source $S$ be $I_0 = 256 \text{ W/m}^2$.
When unpolarized light passes through the first polaroid $P_1$,the intensity becomes $I_1 = \frac{I_0}{2} = \frac{256}{2} = 128 \text{ W/m}^2$.
According to Malus' Law,when polarized light passes through a second polaroid with its pass axis at an angle $\theta$ to the polarization direction of the incident light,the transmitted intensity is $I = I_{incident} \cos^2(\theta)$.
For $P_2$,the angle with $P_1$ is $\theta_1 = 60^{\circ}$. So,$I_2 = I_1 \cos^2(60^{\circ}) = 128 \times (\frac{1}{2})^2 = 128 \times \frac{1}{4} = 32 \text{ W/m}^2$.
For $P_3$,the angle with $P_2$ is $\theta_2 = 90^{\circ} - 60^{\circ} = 30^{\circ}$. So,$I_3 = I_2 \cos^2(30^{\circ}) = 32 \times (\frac{\sqrt{3}}{2})^2 = 32 \times \frac{3}{4} = 24 \text{ W/m}^2$.
Thus,the intensity at point $O$ is $24 \text{ W/m}^2$.

(C) When unpolarised light of intensity $I_0$ passes through the first polaroid $A$,the intensity of the transmitted light is $I_A = \frac{I_0}{2}$.
The pass-axis of polaroid $C$ makes an angle of $45^{\circ}$ with the pass-axis of $A$. According to Malus' Law,the intensity of light after passing through $C$ is $I_C = I_A \cos^2(45^{\circ}) = \frac{I_0}{2} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{4}$.
The pass-axis of polaroid $B$ is perpendicular to $A$,so it makes an angle of $45^{\circ}$ with the pass-axis of $C$. The intensity of light after passing through $B$ is $I_B = I_C \cos^2(45^{\circ}) = \frac{I_0}{4} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{8}$.

(B) When unpolarized light of intensity $I$ passes through the first polarizing sheet,the intensity becomes $I_1 = \frac{I}{2}$.
For subsequent sheets,we use Malus' Law: $I_{k} = I_{k-1} \cos^2(\theta)$,where $\theta = 45^{\circ}$.
Since $\cos^2(45^{\circ}) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$,the intensity after each subsequent sheet is halved.
After the first sheet,the intensity is $I_1 = \frac{I}{2}$.
After the second sheet,$I_2 = I_1 \cos^2(45^{\circ}) = \frac{I}{2} \cdot \frac{1}{2} = \frac{I}{2^2}$.
After the $n^{\text{th}}$ sheet,the intensity is $I_n = \frac{I}{2} \cdot (\frac{1}{2})^{n-1} = \frac{I}{2^n}$.
Given $I_n = \frac{I}{64}$,we have $\frac{I}{2^n} = \frac{I}{64}$.
$2^n = 64 = 2^6$.
Therefore,$n = 6$.

(C) According to Malus's law,the intensity of light transmitted through an analyser is given by $I_2 = I_1 \cos^2 \theta$.
Here,$I_1 = I$ is the intensity of the incident linearly polarised light.
The angle between the transmission axes of the polariser and the analyser is $\theta = 30^{\circ}$.
Substituting these values into the formula:
$I_2 = I \cos^2(30^{\circ})$
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have:
$I_2 = I \left( \frac{\sqrt{3}}{2} \right)^2 = I \left( \frac{3}{4} \right) = \frac{3I}{4}$.

(D) Let the intensity of unpolarised light be $I_0 = 32 \, W m^{-2}$.
After passing through the first polaroid,the intensity becomes $I_1 = \frac{I_0}{2} = 16 \, W m^{-2}$.
Let $\theta$ be the angle between the pass axis of the first and second polaroid.
The intensity after the second polaroid is $I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$.
The angle between the second and third polaroid is $(90^{\circ} - \theta)$ because the first and third polaroids are perpendicular.
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 $I_3 = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta = \frac{I_0}{8} (2 \sin \theta \cos \theta)^2 = \frac{I_0}{8} \sin^2(2 \theta)$.
Given $I_3 = 3 \, W m^{-2}$ and $I_0 = 32 \, W m^{-2}$,we have $3 = \frac{32}{8} \sin^2(2 \theta) = 4 \sin^2(2 \theta)$.
$\sin^2(2 \theta) = \frac{3}{4} \implies \sin(2 \theta) = \frac{\sqrt{3}}{2}$.
Thus,$2 \theta = 60^{\circ}$ or $120^{\circ}$,which gives $\theta = 30^{\circ}$ or $60^{\circ}$.

(D) Let $I_0$ be the intensity of unpolarized light incident on the first polaroid.
$I_1 = I_0 / 2$ is the intensity of light transmitted from the first polaroid.
Let $\theta$ be the angle between the first and second polaroid,and $\phi$ be the angle between the second and third polaroid.
Since the first and third polaroids are crossed,$\theta + \phi = 90^{\circ}$,so $\phi = 90^{\circ} - \theta$.
The intensity transmitted from the second polaroid is $I_2 = I_1 \cos^2 \theta$.
The intensity transmitted from the third polaroid is $I_3 = I_2 \cos^2 \phi = I_1 \cos^2 \theta \cos^2 (90^{\circ} - \theta) = I_1 \cos^2 \theta \sin^2 \theta$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get $I_3 = I_1 (\sin 2\theta / 2)^2 = (I_0 / 2) \cdot (\sin^2 2\theta / 4) = (I_0 / 8) \sin^2 2\theta$.
$I_3$ is maximum when $\sin^2 2\theta = 1$,which implies $2\theta = 90^{\circ}$,so $\theta = 45^{\circ}$.

(A) $1$. When unpolarised light of intensity $I_0$ passes through the first polaroid $A$,the intensity of the transmitted light becomes $I_1 = I_0 / 2$.
$2$. According to Malus' Law,when this polarised light passes through a second polaroid $B$ whose transmission axis is at an angle $\theta = 45^{\circ}$ with respect to the first polaroid,the final intensity $I_2$ is given by $I_2 = I_1 \cos^2 \theta$.
$3$. Substituting the values: $I_2 = (I_0 / 2) \cos^2(45^{\circ})$.
$4$. Since $\cos(45^{\circ}) = 1 / \sqrt{2}$,then $\cos^2(45^{\circ}) = 1/2$.
$5$. Therefore,$I_2 = (I_0 / 2) \times (1 / 2) = I_0 / 4$.

(D) The intensity of unpolarized light passing through a single polarizer is reduced to half: $I_{P} = I_B / 2$.
According to Malus' Law,when this light passes through a second polarizer with an angle $\theta = 45^{\circ}$ between the pass-axes,the intensity becomes $I_B' = I_P \cos^2(45^{\circ})$.
Substituting the values: $I_B' = (I_B / 2) \times (1 / \sqrt{2})^2 = (I_B / 2) \times (1 / 2) = I_B / 4$.
The new ratio $r'$ is given by $r' = I_A / I_B' = I_A / (I_B / 4) = 4 \times (I_A / I_B)$.
Given $I_A / I_B = 2$,we get $r' = 4 \times 2 = 8$.

(B) When unpolarized light of intensity $I_0$ passes through a polarizer,the transmitted intensity is $\frac{I_0}{2}$.
Thus,the intensity of light emerging from $P_1$ and $P_2$ is $I_1 = I_2 = \frac{I_0}{2}$.
When this light passes through $P_3$ (whose axis is at $45^{\circ}$ to both $P_1$ and $P_2$),the intensity of each wave becomes $I' = I_1 \cos^2(45^{\circ}) = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$.
At a point where the path difference is $\Delta x = \frac{\lambda}{3}$,the phase difference is $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
The resultant intensity $I_{res}$ is given by $I_{res} = I' + I' + 2\sqrt{I' I'} \cos(\Delta \phi) = 2I' + 2I' \cos(\frac{2\pi}{3})$.
Substituting $I' = \frac{I_0}{4}$ and $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$:
$I_{res} = 2(\frac{I_0}{4}) + 2(\frac{I_0}{4})(-\frac{1}{2}) = \frac{I_0}{2} - \frac{I_0}{4} = \frac{I_0}{4}$.

(A) Let the intensity of light incident on $P_1$ be $I_0$. Since $P_1$ and $P_2$ are crossed,the angle between them is $90^{\circ}$ or $\frac{\pi}{2}$.
Let the angle between $P_1$ and $P_3$ be $\theta$. Then the angle between $P_3$ and $P_2$ is $(\frac{\pi}{2} - \theta)$.
According to Malus' Law,the intensity after passing through $P_3$ is $I_1 = I_0 \cos^2 \theta$.
The intensity after passing through $P_2$ is $I_{\text{net}} = I_1 \cos^2(\frac{\pi}{2} - \theta) = I_0 \cos^2 \theta \sin^2 \theta$.
$I_{\text{net}} = I_0 (\sin \theta \cos \theta)^2 = I_0 (\frac{\sin 2\theta}{2})^2 = \frac{I_0}{4} \sin^2(2\theta)$.
For maximum intensity,$\sin^2(2\theta)$ must be $1$,so $2\theta = 90^{\circ}$ or $\theta = 45^{\circ}$ or $\frac{\pi}{4}$.
The angle between $P_3$ and $P_2$ is $\frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(C) Let the intensity of light after passing through the first polaroid be $I_0$.
When a second polaroid is placed at an angle $\theta = 22.5^{\circ}$ with the first,the intensity of light transmitted through it is given by Malus' Law: $I_1 = I_0 \cos^2 \theta$.
The third polaroid is placed at an angle $(90^{\circ} - \theta)$ with respect to the second polaroid.
The intensity of light transmitted through the third polaroid is $I_2 = I_1 \cos^2(90^{\circ} - \theta) = I_0 \cos^2 \theta \sin^2 \theta$.
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we have $\sin^2(2\theta) = 4 \sin^2 \theta \cos^2 \theta$,so $\sin^2 \theta \cos^2 \theta = \frac{\sin^2(2\theta)}{4}$.
Substituting this into the expression for $I_2$: $I_2 = I_0 \frac{\sin^2(2\theta)}{4}$.
Given $\theta = 22.5^{\circ}$,then $2\theta = 45^{\circ}$.
$I_2 = \frac{I_0}{4} \sin^2(45^{\circ}) = \frac{I_0}{4} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{4} \times \frac{1}{2} = \frac{I_0}{8}$.

(D) According to Malus' Law,the intensity of transmitted light $I$ through a polaroid is given by $I = I_0 \cos^2(\phi)$,where $I_0$ is the maximum intensity and $\phi$ is the angle between the polarizing directions of the two polaroids.
Initially,the polarizing directions are parallel,so $\phi = 0^\circ$ and $I = I_0$.
We want the new intensity $I'$ to be half of the initial intensity,so $I' = \frac{I_0}{2}$.
Substituting this into Malus' Law: $\frac{I_0}{2} = I_0 \cos^2(\phi)$.
This simplifies to $\cos^2(\phi) = \frac{1}{2}$,which means $\cos(\phi) = \frac{1}{\sqrt{2}}$.
Therefore,$\phi = 45^\circ$.

(D) When natural light of intensity $I_0$ passes through the first polaroid,the transmitted intensity is $I_1 = \frac{I_0}{2}$.
For subsequent polaroids,Malus' Law states $I_n = I_{n-1} \cos^2 \theta$,where $\theta = 60^{\circ}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,then $\cos^2 60^{\circ} = \frac{1}{4}$.
For $P_2$: $I_2 = I_1 \cos^2 60^{\circ} = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$.
For $P_3$: $I_3 = I_2 \cos^2 60^{\circ} = \frac{I_0}{8} \times \frac{1}{4} = \frac{I_0}{32}$.
For $P_4$: $I_4 = I_3 \cos^2 60^{\circ} = \frac{I_0}{32} \times \frac{1}{4} = \frac{I_0}{128}$.
For $P_5$: $I_5 = I_4 \cos^2 60^{\circ} = \frac{I_0}{128} \times \frac{1}{4} = \frac{I_0}{512}$.
The fraction of incident intensity is $\frac{I_5}{I_0} = \frac{1}{512}$.

(C) Let the intensity of the incident unpolarized light be $I_0$.
When unpolarized light passes through the first polarizer,the intensity of the transmitted polarized light becomes $I_1 = \frac{I_0}{2}$.
According to Malus' Law,when this polarized light passes through the second polarizer whose principal plane is at an angle $\theta = 60^{\circ}$ to the first,the transmitted intensity $I_2$ is given by $I_2 = I_1 \cos^2 \theta$.
Substituting the values: $I_2 = \left(\frac{I_0}{2}\right) \cos^2 60^{\circ} = \left(\frac{I_0}{2}\right) \left(\frac{1}{2}\right)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$.
The percentage of incident light that passes through the system is $\frac{I_2}{I_0} \times 100 = \frac{I_0/8}{I_0} \times 100 = \frac{100}{8} = 12.5 \%$.

(A) tourmaline crystal is a dichroic material. When unpolarized light passes through it,the crystal selectively absorbs the ordinary ray (the component of light vibrating perpendicular to the optic axis) and transmits the extraordinary ray (the component of light vibrating parallel to the optic axis). Therefore,the light emerging from the crystal is plane-polarized.

(A) Let the initial intensity of unpolarized light be $I_0 = 256 \ W/m^2$.
When unpolarized light passes through the first polaroid $P_1$,the intensity of the transmitted light becomes $I_1 = I_0 / 2 = 256 / 2 = 128 \ W/m^2$.
The pass axis of $P_2$ is at an angle $\theta_1 = 60^{\circ}$ with respect to $P_1$. According to Malus' Law,the intensity after $P_2$ is $I_2 = I_1 \cos^2(\theta_1) = 128 \times \cos^2(60^{\circ}) = 128 \times (0.5)^2 = 128 \times 0.25 = 32 \ W/m^2$.
The pass axis of $P_3$ is at an angle of $90^{\circ}$ with respect to $P_1$. The angle between the pass axes of $P_2$ and $P_3$ is $\theta_2 = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Applying Malus' Law again for $P_3$,the final intensity at point $O$ is $I_3 = I_2 \cos^2(\theta_2) = 32 \times \cos^2(30^{\circ}) = 32 \times (\sqrt{3}/2)^2 = 32 \times (3/4) = 24 \ W/m^2$.

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light is $I_1 = I_0 / 2$.
For subsequent polaroids,we use Malus' Law: $I_n = I_{n-1} \cos^2 \theta$,where $\theta = 30^{\circ}$.
Intensity after the second polaroid: $I_2 = I_1 \cos^2 30^{\circ} = (I_0 / 2) \times (\sqrt{3} / 2)^2 = (I_0 / 2) \times (3 / 4) = 3 I_0 / 8$.
Intensity after the third polaroid: $I_3 = I_2 \cos^2 30^{\circ} = (3 I_0 / 8) \times (3 / 4) = 9 I_0 / 32$.
Intensity after the fourth polaroid: $I_4 = I_3 \cos^2 30^{\circ} = (9 I_0 / 32) \times (3 / 4) = 27 I_0 / 128$.
Thus,the correct option is $B$.

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the emergent light is $I_1 = \frac{I_0}{2}$.
According to Malus' Law,the intensity of light emerging from a polaroid is $I = I_{incident} \cos^2 \theta$,where $\theta$ is the angle between the pass axis of the polaroid and the plane of polarisation of the incident light.
For the second polaroid,the angle between its pass axis and the pass axis of the first polaroid is $\theta_1 = 30^{\circ}$. Thus,the intensity of light emerging from the second polaroid is $I_2 = I_1 \cos^2 30^{\circ} = \left(\frac{I_0}{2}\right) \left(\frac{\sqrt{3}}{2}\right)^2 = \left(\frac{I_0}{2}\right) \left(\frac{3}{4}\right) = \frac{3 I_0}{8}$.
For the third polaroid,the angle between its pass axis and the pass axis of the second polaroid is $\theta_2 = 90^{\circ} - 30^{\circ} = 60^{\circ}$. Thus,the intensity of light emerging from the third polaroid is $I_3 = I_2 \cos^2 60^{\circ} = \left(\frac{3 I_0}{8}\right) \left(\frac{1}{2}\right)^2 = \left(\frac{3 I_0}{8}\right) \left(\frac{1}{4}\right) = \frac{3 I_0}{32}$.

(B) $1$. When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light becomes $I_1 = \frac{I_0}{2}$.
$2$. The first and second polaroids are crossed (angle between axes is $90^\circ$),so no light passes through the second polaroid initially.
$3$. $A$ third polaroid is placed between them at an angle $\theta$ with the first polaroid. The angle between the third and the second polaroid is $(90^\circ - \theta)$.
$4$. Intensity after the third polaroid: $I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$.
$5$. Intensity after the second (last) polaroid using Malus' Law: $I_3 = I_2 \cos^2(90^\circ - \theta) = I_2 \sin^2 \theta$.
$6$. Substituting $I_2$: $I_3 = (\frac{I_0}{2} \cos^2 \theta) \sin^2 \theta = \frac{I_0}{2} (\sin \theta \cos \theta)^2 = \frac{I_0}{2} (\frac{\sin 2 \theta}{2})^2 = \frac{I_0}{8} \sin^2 2 \theta$.

(D) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light becomes $I_1 = I_0 / 2$.
This light is now plane-polarised.
When this light passes through the second polaroid whose transmission axis makes an angle $\theta = 30^{\circ}$ with the first,the intensity of the transmitted light $I_2$ is given by Malus' Law: $I_2 = I_1 \cos^2 \theta$.
Substituting the values: $I_2 = (I_0 / 2) \cos^2 30^{\circ}$.
Since $\cos 30^{\circ} = \sqrt{3} / 2$,we have $\cos^2 30^{\circ} = 3 / 4$.
Therefore,$I_2 = (I_0 / 2) \times (3 / 4) = 3 I_0 / 8$.
Calculating the fraction: $3 / 8 = 0.375$,which is $37.5 \%$.

(B) Let the intensity of the incident light be $I_0$.
When light passes through the first polaroid,its intensity becomes $I_1 = I_0/2$ (assuming unpolarized light).
For the second polaroid,the angle between the transmission axis and the incident light is $\theta = 60^{\circ}$. According to Malus' Law,$I_2 = I_1 \cos^2(60^{\circ}) = (I_0/2) \times (1/2)^2 = I_0/8$.
For the third polaroid,the angle between its transmission axis and the second polaroid's axis is also $\theta = 60^{\circ}$.
Thus,$I_3 = I_2 \cos^2(60^{\circ}) = (I_0/8) \times (1/2)^2 = I_0/32$.
The fraction of the incident light intensity that passes through the system is $I_3/I_0 = 1/32$.

(C) According to Malus' law,the intensity of light transmitted through a polarizer is $I = I_{in} \cos^2 \theta$,where $\theta$ is the angle between the polarization direction of incident light and the pass axis of the polarizer.
$1$. When unpolarized light of intensity $I_0$ passes through the first polaroid $P_1$,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.
$2$. The light from $P_1$ is now polarized at an angle $0^{\circ}$ relative to $P_1$'s axis. $P_2$ is at $60^{\circ}$ to $P_1$. Thus,the intensity after $P_2$ is:
$I_2 = I_1 \cos^2(60^{\circ}) = \frac{I_0}{2} \times (0.5)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$.
$3$. The light from $P_2$ is polarized at $60^{\circ}$ relative to $P_1$. $P_3$ is at $90^{\circ}$ relative to $P_1$. The angle between the polarization of light from $P_2$ and the pass axis of $P_3$ is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
$4$. The intensity after $P_3$ is:
$I_3 = I_2 \cos^2(30^{\circ}) = \frac{I_0}{8} \times \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I_0}{8} \times \frac{3}{4} = \frac{3 I_0}{32}$.

(A) According to Malus' Law,the intensity of light transmitted through a polarizer is given by $I = I_{max} \cos^2 \theta$,where $\theta$ is the angle between the transmission axes of the two polarizers.
Initially,the principal planes are parallel,so $\theta = 0^{\circ}$ and the intensity is $I_0 = I_{max}$.
When crystal $B$ is rotated by $45^{\circ}$,the new angle between the transmission axes becomes $\theta = 45^{\circ}$.
Substituting the values into the formula:
$I = I_0 \cos^2(45^{\circ})$
$I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2$
$I = I_0 \left(\frac{1}{2}\right) = \frac{I_0}{2}$.

(C) When unpolarised light passes through the first polaroid, it becomes linearly polarised with intensity $I_0 = I_{in} / 2$.
When this polarised light hits the second polaroid, the intensity transmitted is given by Malus' Law: $I = I_0 \cos^2 \theta$.
Since the polaroids are crossed, the angle between their transmission axes is $\theta = 90^{\circ}$.
Substituting this value, we get $I = I_0 \cos^2(90^{\circ}) = I_0 \times 0 = 0$.
Therefore, the light is completely blocked by the second polaroid.

(A) When unpolarized light of intensity $I_{0}$ is incident on a polarizing sheet, the intensity of the transmitted light is given by $I_{t} = \frac{I_{0}}{2}$.
Since the total incident intensity is $I_{0}$ and the transmitted intensity is $\frac{I_{0}}{2}$, the intensity of the light that does not get transmitted is the difference between the incident and transmitted intensities.
Intensity of untransmitted light = $I_{0} - I_{t} = I_{0} - \frac{I_{0}}{2} = \frac{I_{0}}{2}$.

(C) Let the initial intensity of the light be $I_{0}$.
According to Malus's Law,the final intensity $I$ of the light after passing through the tourmaline plate is given by:
$I = I_{0} \cos^{2} \theta$
Given that the angle $\theta = 60^{\circ}$,we have:
$I = I_{0} \cos^{2} 60^{\circ} = I_{0} \left(\frac{1}{2}\right)^{2} = \frac{I_{0}}{4}$
The percentage difference between the initial and final intensities is calculated as:
$\text{Percentage Difference} = \left(\frac{I_{0} - I}{I_{0}}\right) \times 100$
Substituting the value of $I$:
$\text{Percentage Difference} = \left(\frac{I_{0} - \frac{I_{0}}{4}}{I_{0}}\right) \times 100 = \left(\frac{\frac{3I_{0}}{4}}{I_{0}}\right) \times 100 = \frac{3}{4} \times 100 = 75 \%$

(D) When unpolarized light of intensity $I_1$ is incident on a polaroid,the intensity of the emergent polarized light $I_2$ is given by Malus's Law for unpolarized light.
According to the law,the intensity of light transmitted through a polarizer is half the intensity of the incident unpolarized light.
Therefore,$I_2 = \frac{I_1}{2}$.
Rearranging this equation,we get $I_1 = 2I_2$.

(D) According to Malus's Law,the intensity of the emergent light is given by:
$I = I_{0} \cos^{2} \theta$
where $I$ is the intensity of the emergent polarized light,$I_{0}$ is the intensity of the incident polarized light,and $\theta$ is the angle between the pass axes of the two polarizers.
Given that $\theta = 60^{\circ}$,we substitute this value into the formula:
$I = I_{0} \cos^{2}(60^{\circ})$
Since $\cos(60^{\circ}) = 1/2$,we have:
$I = I_{0} \times (1/2)^{2}$
$I = I_{0} \times (1/4)$
$I = I_{0} / 4$
Thus,the intensity of the emergent polarized light from the second polarizer is $I_{0} / 4$.

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.
The angle between the pass axes of the first and second polaroid is $\theta_{12} = 30^{\circ}$. According to Malus' Law,the intensity of light transmitted through the second polaroid is $I_2 = I_1 \cos^2(30^{\circ}) = \frac{I_0}{2} \times (\frac{\sqrt{3}}{2})^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3 I_0}{8}$.
The angle between the pass axes of the second and third polaroid is $\theta_{23} = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
According to Malus' Law,the intensity of light transmitted through the third polaroid is $I_3 = I_2 \cos^2(60^{\circ}) = \frac{3 I_0}{8} \times (\frac{1}{2})^2 = \frac{3 I_0}{8} \times \frac{1}{4} = \frac{3 I_0}{32}$.

(A) When unpolarised light of intensity $I$ passes through the first polaroid,its intensity becomes $\frac{I}{2}$.
According to Malus' Law,the intensity $I^{\prime}$ of light emerging from the second polaroid is given by:
$I^{\prime} = I_{0} \cos^2 \theta$
where $I_{0} = \frac{I}{2}$ is the intensity of light incident on the second polaroid and $\theta$ is the angle between the pass axes.
Given $I^{\prime} = \frac{I}{4}$,we have:
$\frac{I}{4} = \frac{I}{2} \cos^2 \theta$
$\cos^2 \theta = \frac{1}{2}$
$\cos \theta = \frac{1}{\sqrt{2}}$
$\theta = 45^{\circ}$

(C) When unpolarised light of intensity $I_{0}$ passes through the first polaroid $P_{1}$,the intensity of the transmitted light becomes $I_{1} = \frac{I_{0}}{2}$.
Given $I_{0} = 128 \ Wm^{-2}$,so $I_{1} = \frac{128}{2} = 64 \ Wm^{-2}$.
According to Malus's law,when light of intensity $I_{1}$ passes through a polaroid whose pass axis makes an angle $\theta$ with the incident light's polarization direction,the transmitted intensity is $I = I_{1} \cos^{2} \theta$.
For $P_{2}$,the angle between the pass axis of $P_{1}$ and $P_{2}$ is $\theta_{1} = 45^{\circ}$.
Intensity after $P_{2}$ is $I_{2} = I_{1} \cos^{2} 45^{\circ} = 64 \times (\frac{1}{\sqrt{2}})^{2} = 64 \times \frac{1}{2} = 32 \ Wm^{-2}$.
For $P_{3}$,the angle between the pass axis of $P_{2}$ and $P_{3}$ is $\theta_{2} = 45^{\circ}$.
Intensity after $P_{3}$ is $I_{3} = I_{2} \cos^{2} 45^{\circ} = 32 \times (\frac{1}{\sqrt{2}})^{2} = 32 \times \frac{1}{2} = 16 \ Wm^{-2}$.

(A) According to Malus's Law,the intensity of light transmitted through a polariser is given by:
$I = I_{0} \cos^{2} \theta$
Where $I_{0}$ is the intensity of light incident on the second polariser (which is the intensity of light emerging from the first polariser),and $\theta$ is the angle between the pass axes of the two polarisers.
Given that the intensity of light from the second polariser $(I)$ is half of the intensity from the first polariser $(I_{0})$,we have:
$I = \frac{I_{0}}{2}$
Substituting this into Malus's Law:
$\frac{I_{0}}{2} = I_{0} \cos^{2} \theta$
$\frac{1}{2} = \cos^{2} \theta$
$\cos \theta = \frac{1}{\sqrt{2}}$
$\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}$
Therefore,the angle between the pass axes of the polarisers is $45^{\circ}$.

(A) The optical rotation $\theta$ produced by a sugar solution is given by the formula: $\theta = S \cdot l \cdot C$, where $S$ is the specific rotation, $l$ is the length of the tube, and $C$ is the concentration of the solution.
In the first case, $\theta = S \cdot l \cdot C$, where $l = 0.2 \,m$ and $C = \frac{m}{V}$, where $m$ is the mass of sugar and $V$ is the volume of the tube $(V = \pi R^2 l)$.
Thus, $\theta = S \cdot l \cdot \frac{m}{\pi R^2 l} = \frac{S \cdot m}{\pi R^2}$.
In the second case, the same amount of sugar $m$ is placed in a tube of length $l_1 = 0.3 \,m$ with the same radius $R$. The volume of the new tube is $V_1 = \pi R^2 l_1 = \pi R^2 (0.3)$.
The concentration of the solution in the second case is $C_1 = \frac{m}{V_1} = \frac{m}{\pi R^2 (0.3)}$.
The new rotation $\theta_1$ is given by $\theta_1 = S \cdot l_1 \cdot C_1 = S \cdot (0.3) \cdot \frac{m}{\pi R^2 (0.3)} = \frac{S \cdot m}{\pi R^2}$.
Comparing the two expressions, we find $\theta_1 = \theta$.

(C) The phenomenon of polarization is a characteristic property of transverse waves. Longitudinal waves,such as sound waves,cannot be polarized because their oscillations occur along the direction of propagation. Since light can be polarized,it confirms that light waves are transverse in nature.

(A) When an unpolarised light beam of intensity $I_{0}$ passes through a polaroid,it becomes plane-polarised.
According to the properties of polaroids,the intensity of the emergent plane-polarised light is exactly half of the intensity of the incident unpolarised light.
Therefore,the intensity of the emergent light is $I = \frac{I_{0}}{2}$.

(B) dichroic crystal is a material that exhibits the property of dichroism,which is the selective absorption of light based on its polarization direction.
Some crystals,such as tourmaline,have the property of strongly absorbing light with vibrations perpendicular to a specific direction (called the transmission axis) while transmitting light with vibrations parallel to it.
This selective absorption of light is known as dichroism.
Therefore,tourmaline is a well-known example of a dichroic crystal.

(D) For solution $A$: $L_{1} = 20 \ cm$,$\theta_{1} = +38^{\circ}$. Let concentration be $C_{1}$. The specific rotation is $\alpha_{1} = \frac{\theta_{1}}{L_{1} C_{1}} = \frac{38^{\circ}}{20 C_{1}}$.
For solution $B$: $L_{2} = 30 \ cm$,$\theta_{2} = -24^{\circ}$ (left-handed). Let concentration be $C_{2}$. The specific rotation is $\alpha_{2} = \frac{\theta_{2}}{L_{2} C_{2}} = \frac{-24^{\circ}}{30 C_{2}}$.
In the mixture,the volume ratio is $1:2$. Thus,the new concentrations are $C_{1}' = \frac{C_{1}}{3}$ and $C_{2}' = \frac{2C_{2}}{3}$.
The total optical rotation $\theta$ for a path length $l = 30 \ cm$ is given by $\theta = (\alpha_{1} C_{1}' + \alpha_{2} C_{2}') l$.
Substituting the values: $\theta = \left( \frac{38^{\circ}}{20 C_{1}} \cdot \frac{C_{1}}{3} + \frac{-24^{\circ}}{30 C_{2}} \cdot \frac{2 C_{2}}{3} \right) \times 30$.
$\theta = \left( \frac{38^{\circ}}{60} - \frac{48^{\circ}}{90} \right) \times 30 = \left( \frac{19^{\circ}}{30} - \frac{16^{\circ}}{30} \right) \times 30 = 19^{\circ} - 16^{\circ} = +3^{\circ}$.
Since the result is positive,it is a right-handed rotation of $3^{\circ}$.

(C) Let the intensity of the unpolarised incident light be $I_0$.
After passing through the first polariser, the intensity becomes $I_1 = I_0 / 2$.
According to Malus' Law, the intensity of light emerging from a polariser is $I = I_{in} \cos^2 \theta$, where $\theta$ is the angle between the incident light's polarisation axis and the polariser's axis.
For the second sheet, the angle between the first and second sheet is $\theta = 30^{\circ}$. Thus, $I_2 = I_1 \cos^2(30^{\circ}) = (I_0 / 2) \times (\sqrt{3} / 2)^2 = (I_0 / 2) \times (3 / 4) = 3I_0 / 8$.
For the third sheet, the angle between the second and third sheet is also $\theta = 30^{\circ}$. Thus, $I_3 = I_2 \cos^2(30^{\circ}) = (3I_0 / 8) \times (3 / 4) = 9I_0 / 32$.
The ratio of intensities emerging from the second and third sheets is $I_2 / I_3 = (3I_0 / 8) / (9I_0 / 32) = (3 / 8) \times (32 / 9) = 4 / 3$.
Therefore, the ratio is $4: 3$.

(D) Let $I_0$ be the intensity of the unpolarised light incident on the first polaroid.
When unpolarised light passes through the first polaroid,the intensity of the transmitted polarised light is $I_1 = \frac{I_0}{2}$.
According to Malus' Law,when this polarised light passes through the second polaroid,the intensity of the light emerging from it is $I_2 = I_1 \cos^2 \theta$,where $\theta$ is the angle between the axes of the two polaroids.
Given that $I_2 = 37.5 \% \text{ of } I_0 = 0.375 I_0$.
Substituting $I_1 = \frac{I_0}{2}$ into the equation: $0.375 I_0 = \frac{I_0}{2} \cos^2 \theta$.
$0.375 = 0.5 \cos^2 \theta$.
$\cos^2 \theta = \frac{0.375}{0.5} = 0.75 = \frac{3}{4}$.
Taking the square root on both sides: $\cos \theta = \frac{\sqrt{3}}{2}$.
Therefore,$\theta = 30^{\circ}$.

(D) The intensity of unpolarized light incident on the polariser is $I$.
After passing through the polariser,the intensity of the polarized light becomes $I_1 = \frac{I}{2}$.
According to Malus' Law,the intensity of light emerging from the analyser is $I_2 = I_1 \cos^2 \theta$.
Given $\theta = 45^{\circ}$,we have $I_2 = \frac{I}{2} \cos^2 45^{\circ}$.
Substituting the value of $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we get $I_2 = \frac{I}{2} \times (\frac{1}{\sqrt{2}})^2 = \frac{I}{2} \times \frac{1}{2} = \frac{I}{4}$.

(D) The relation $I = I_0 \cos^2 \theta$ is known as Malus's law.
This law states that when completely plane-polarized light is incident on an analyser,the intensity of the light transmitted by the analyser is directly proportional to the square of the cosine of the angle between the transmission axis of the analyser and the plane of polarization of the incident light.
Here,$I_0$ represents the maximum intensity of the incident light,$I$ represents the intensity of the emergent light,and $\theta$ is the angle between the polarization direction and the analyser axis.

(D) Given, refractive index $\mu = 1.414 = \sqrt{2}$.
For complete polarisation, the angle of incidence is the Brewster's angle $i_p$, where $\tan i_p = \mu = \sqrt{2}$.
Thus, $\sin i_p = \sqrt{\frac{2}{3}}$ and $\cos i_p = \frac{1}{\sqrt{3}}$.
From Snell's law, $\sin i_p = \mu \sin r$, so $\sin r = \frac{\sin i_p}{\mu} = \frac{\sqrt{2/3}}{\sqrt{2}} = \frac{1}{\sqrt{3}}$.
Therefore, $r = \sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$. This matches $(iii)$.
The angle of reflection $\theta$ is equal to the angle of incidence $i_p$. Since $\cos i_p = \frac{1}{\sqrt{3}}$, we have $i_p = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right)$. This matches $(iv)$.
The angle of deviation $\delta$ of the refracted ray is $i_p - r = \sin^{-1} \left(\sqrt{\frac{2}{3}}\right) - \sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$. This matches $(ii)$.
The angle $\phi$ between the incident ray and the reflected (polarised) ray is $i_p + \theta = 2i_p = 2 \sin^{-1} \left(\sqrt{\frac{2}{3}}\right)$. This matches $(i)$.
The correct matching is $A-(iv), B-(iii), C-(i), D-(ii)$.