Polarisation of Light and Malus' Law Questions in English

(D) The phenomena of interference,diffraction,and reflection are exhibited by both transverse and longitudinal waves.
Polarisation is a property that is unique to transverse waves because it involves the orientation of oscillations perpendicular to the direction of wave propagation.
Longitudinal waves,where oscillations occur parallel to the direction of propagation,cannot be polarised.
Therefore,polarisation is the correct property to distinguish between the two types of waves.

(C) Crystals are classified based on their optical properties into isotropic,uniaxial,and biaxial crystals.
$1$. Isotropic crystals (e.g.,glass,rock salt) have the same refractive index in all directions.
$2$. Uniaxial crystals (e.g.,Calcite,Quartz,Tourmaline) have one optic axis.
$3$. Biaxial crystals (e.g.,Selenite,Mica,Aragonite) have two optic axes.
Among the given options,Selenite is a biaxial crystal,while Calcite,Quartz,and Tourmaline are uniaxial crystals. Therefore,the correct option is $C$.

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

(A) Polarisation is a phenomenon that occurs only in transverse waves.
In transverse waves,the oscillations of the medium particles (or the electric and magnetic field vectors in the case of light) occur perpendicular to the direction of wave propagation.
Since light waves are electromagnetic waves and their electric field vectors oscillate perpendicular to the direction of travel,they exhibit the property of polarisation.
Longitudinal waves,such as sound waves,cannot be polarised because their oscillations occur along the direction of propagation.

(C) Light waves are transverse waves,whereas sound waves are longitudinal waves.
Polarisation is a phenomenon that occurs only in transverse waves.
Since sound waves are longitudinal,they cannot be polarised.
Therefore,polarisation is the characteristic that distinguishes light waves from sound waves.

(D) The intensity of light transmitted through a polaroid is given by Malus' Law: $I = I_0 \cos^2 \theta$,where $\theta$ is the angle between the plane of polarization of the incident light and the transmission axis of the polaroid.
As the polaroid is rotated by $2\pi$ $(360^\circ)$,the angle $\theta$ varies from $0$ to $2\pi$.
- The intensity $I$ becomes maximum $(I = I_0)$ when $\theta = 0$ and $\theta = \pi$ (twice).
- The intensity $I$ becomes zero when $\theta = \pi/2$ and $\theta = 3\pi/2$ (twice).
Therefore,during one complete rotation,the intensity of light becomes maximum twice and zero twice.

(A) Nicol's prism is an optical device made from a calcite crystal that produces plane-polarized light.
When unpolarized light enters the Nicol's prism,it undergoes double refraction into ordinary ($O$-ray) and extraordinary ($E$-ray) rays.
The $O$-ray is removed by total internal reflection at the Canada balsam layer,while the $E$-ray emerges as plane-polarized light,not elliptically polarized light.
Therefore,the statement in option $A$ is incorrect.

(A) According to Brewster's Law,when unpolarised light is incident on a transparent surface at the polarising angle,the reflected light is completely plane-polarised.
The electric field vector of this reflected light vibrates in a direction perpendicular to the plane of incidence.
In the given figure,the glass plate is placed vertically on a horizontal table. The plane of incidence is the plane containing the incident ray and the normal to the surface. Since the light is incident at an angle with the normal on a vertical surface,the plane of incidence is horizontal.
Therefore,the electric vector of the reflected light,being perpendicular to the horizontal plane of incidence,will vibrate in a vertical plane.

(A) Polaroid glass cuts glare and haze by reducing the light intensity to half on account of polarisation. This makes the eyes more comfortable and allows the viewer to see better,which is why they are used in sunglasses.

(C) The phenomenon of $Polarisation$ of light establishes that light waves are transverse in nature. Before this, it was widely believed that light waves were longitudinal, similar to sound waves. Since only transverse waves can be polarized, the observation of polarization confirms the transverse nature of light.

(D) According to the law of Malus for amplitudes,the amplitude of the light transmitted through an analyser is given by $A' = A \cos \theta$,where $A$ is the initial amplitude and $\theta$ is the angle between the transmission axes of the polariser and the analyser.
Given,$A = A$ and $\theta = 60^o$.
Substituting the values,we get $A' = A \cos(60^o)$.
Since $\cos(60^o) = 1/2$,the amplitude of the light transmitted by the analyser is $A' = A \times (1/2) = A/2$.

(D) In a doubly refracting (birefringent) crystal,the incident unpolarised light splits into two plane-polarised rays.
$1$. The ordinary ray ($O$-ray) obeys the standard laws of refraction and its vibrations are perpendicular to the principal section (plane of incidence).
$2$. The extraordinary ray ($E$-ray) does not obey the standard laws of refraction and its vibrations occur within the principal section (plane of incidence).
Therefore,the $E$-ray is polarised in the plane of incidence,and the $O$-ray is polarised perpendicular to the plane of incidence.
Thus,the correct option is $(D)$.

(B) The rotation produced by a polarimeter tube 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.
For the first tube (dextrorotatory): $\theta_1 = S_1 \cdot l \cdot c_1 = 0.01 \times 0.29 \times 60 = 0.174 \ rad$.
For the second tube (laevorotatory): $\theta_2 = S_2 \cdot l \cdot c_2 = 0.02 \times 0.29 \times 30 = 0.174 \ rad$.
Since the first solution is dextrorotatory (positive rotation) and the second is laevorotatory (negative rotation),the net rotation is $\theta_{net} = \theta_1 - \theta_2 = 0.174 - 0.174 = 0 \ rad$.
Thus,the net rotation produced is $0^\circ$.

(C) In double refraction,an incident light ray splits into two rays: the ordinary ray ($O$-ray) and the extraordinary ray ($E$-ray).
For a negative uniaxial crystal like calcite,the velocity of the extraordinary ray $(V_E)$ is greater than the velocity of the ordinary ray $(V_o)$,which implies $\mu_E < \mu_o$ or $\mu_o > \mu_E$.
Thus,for calcite,$V_o < V_E$ and $\mu_o > \mu_E$.
For a positive uniaxial crystal like quartz,the velocity of the extraordinary ray $(V_E)$ is less than the velocity of the ordinary ray $(V_o)$,which implies $\mu_E > \mu_o$.
Therefore,option $(c)$ correctly describes the relationship for calcite.

(D) According to Malus' Law,when a plane-polarised light of intensity $I_0$ passes through an analyser,the intensity of the emerging light $I$ is given by $I = I_0 \cos^2 \theta$,where $\theta$ is the angle between the transmission axis of the analyser and the plane of polarisation of the incident light.
Initially,if the transmission axis is aligned with the plane of polarisation,$\theta = 0^{\circ}$,then $I = I_0 \cos^2(0^{\circ}) = I_0$ (Maximum intensity).
As the analyser is rotated through $90^{\circ}$,the angle $\theta$ changes from $0^{\circ}$ to $90^{\circ}$.
At $\theta = 90^{\circ}$,the intensity $I = I_0 \cos^2(90^{\circ}) = 0$ (Minimum intensity).
Therefore,the intensity varies continuously from a maximum value $(I_0)$ to zero as the analyser is rotated through $90^{\circ}$.
Hence,the correct answer is option $(D)$.

(A) Statement $A$ is correct: Nucleic acids are optically active molecules. Circularly polarized light interacts differently with the helical structure of these molecules,allowing researchers to study their conformation using techniques like Circular Dichroism.
Statement $B$ is correct: An optical axis is defined as a specific direction within a crystal along which the ordinary and extraordinary rays travel with the same velocity. It is a direction,not a fixed line in space,meaning any line parallel to this direction is also an optical axis.

(C) Let the intensity of the incident unpolarized light be $I_0$.
When unpolarized light passes through the first Nicol prism (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 Nicol prism (analyzer) oriented at an angle $\theta = 60^\circ$ with respect to the first,the transmitted intensity $I_2$ is given by $I_2 = I_1 \cos^2 \theta$.
Substituting the values: $I_2 = \frac{I_0}{2} \cos^2(60^\circ) = \frac{I_0}{2} \times (\frac{1}{2})^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$.
The percentage of incident light transmitted is $\frac{I_2}{I_0} \times 100 = \frac{1}{8} \times 100 = 12.5\%$.

(B) Let $I_0$ be the intensity of the unpolarized light.
When it passes through the first polarizer,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.
When this light passes through the second polarizer,the intensity of the final transmitted light $I_2$ is given by Malus' Law: $I_2 = I_1 \cos^2 \theta$,where $\theta$ is the angle between the characteristic directions of the sheets.
Given that the final intensity $I_2$ is one-third of the maximum intensity of the first transmitted beam $(I_1)$,we have $I_2 = \frac{1}{3} I_1$.
Substituting this into the equation: $\frac{1}{3} I_1 = I_1 \cos^2 \theta$.
This simplifies to $\cos^2 \theta = \frac{1}{3}$,which means $\cos \theta = \frac{1}{\sqrt{3}} \approx 0.577$.
Calculating the angle: $\theta = \cos^{-1}(0.577) \approx 54.7^o$,which rounds to $55^o$.

(B) The angle between the transmission axes of $P_1$ and $P_2$ is $\theta_1 = 30^\circ$ (given).
The transmission axis of the last polarizer $P_3$ is crossed with the first polarizer $P_1$,meaning the angle between $P_1$ and $P_3$ is $90^\circ$.
Therefore,the angle between the transmission axes of $P_2$ and $P_3$ is $\theta_2 = 90^\circ - 30^\circ = 60^\circ$.
The intensity of light transmitted by the first polarizer $P_1$ is $I_1 = \frac{I_0}{2} = \frac{32}{2} = 16 \ W m^{-2}$.
According to Malus' law,the intensity of light transmitted by $P_2$ is $I_2 = I_1 \cos^2(30^\circ) = 16 \times (\frac{\sqrt{3}}{2})^2 = 16 \times \frac{3}{4} = 12 \ W m^{-2}$.
Similarly,the intensity of light transmitted by $P_3$ is $I_3 = I_2 \cos^2(60^\circ) = 12 \times (\frac{1}{2})^2 = 12 \times \frac{1}{4} = 3 \ W m^{-2}$.

(B) Given the equation for optical rotation: $\theta = a + \frac{b}{\lambda^2}$.
At $\lambda_1 = 5000 \ \mathring{A}$,$\theta_1 = 30^\circ$ per mm: $30 = a + \frac{b}{(5000)^2} \quad (1)$.
At $\lambda_2 = 4000 \ \mathring{A}$,$\theta_2 = 50^\circ$ per mm: $50 = a + \frac{b}{(4000)^2} \quad (2)$.
Subtracting equation $(1)$ from equation $(2)$:
$50 - 30 = b \left( \frac{1}{4000^2} - \frac{1}{5000^2} \right) = b \left( \frac{5000^2 - 4000^2}{4000^2 \times 5000^2} \right)$.
$20 = b \left( \frac{(5000-4000)(5000+4000)}{16 \times 10^6 \times 25 \times 10^6} \right) = b \left( \frac{1000 \times 9000}{400 \times 10^{12}} \right) = b \left( \frac{9 \times 10^6}{400 \times 10^{12}} \right) = b \left( \frac{9}{400 \times 10^6} \right)$.
$b = \frac{20 \times 400 \times 10^6}{9} = \frac{8000 \times 10^6}{9}$.
Substitute $b$ into equation $(1)$:
$a = 30 - \frac{8000 \times 10^6}{9 \times 25 \times 10^6} = 30 - \frac{8000}{225} = 30 - \frac{320}{9} = \frac{270 - 320}{9} = - \frac{50^\circ}{9} \text{ per mm}$.

(B) In an electromagnetic wave,the electric field vector $\vec{E}$ at any point in space is given by $\vec{E} = E_0 \sin(kx - \omega t) \hat{n}$.
Here,$E_0$ is the amplitude,$k$ is the wave number,$\omega$ is the angular frequency,and $t$ is time.
The magnitude of the electric field vector is $|\vec{E}| = |E_0 \sin(kx - \omega t)|$.
Since the sine function oscillates between $-1$ and $1$,the magnitude of the electric field vector varies periodically with time at any fixed position $x$.

(A) An optically active compound is a substance that has the ability to rotate the plane of vibration of plane-polarised light.
When plane-polarised light passes through such a substance,the plane of polarisation is rotated by a certain angle about the direction of propagation of the light.
Therefore,the correct option is $A$.

(D) The helical structure of nucleic acids,such as $DNA$,is studied using $X$-ray crystallography.
In this technique,the diffraction of $X$-rays is the primary phenomenon used to determine the structure.
However,in the context of specific optical studies involving the helical nature of biological molecules,the property of polarization (specifically circular dichroism) is used to analyze the handedness and helical structure of these molecules.
Therefore,polarization is the correct property used for this specific study.

(A) The formula for specific rotation is given by $\alpha = \frac{\theta}{l \cdot c}$,where $\theta$ is the optical rotation,$l$ is the length of the tube,and $c$ is the concentration of the pure sugar.
Rearranging the formula to find the concentration of pure sugar: $c = \frac{\theta}{\alpha \cdot l}$.
Substituting the given values: $c = \frac{0.4}{0.01 \times 0.25} = \frac{0.4}{0.0025} = 160 \ kg/m^3$.
The percentage purity is calculated as the ratio of the concentration of pure sugar to the concentration of the impure sample multiplied by $100$.
Percentage purity $= \left( \frac{160}{200} \right) \times 100 = 80\%$.

(D) When natural (unpolarized) light of intensity $I_0$ passes through the first polaroid,the transmitted intensity becomes $I_1 = \frac{I_0}{2}$.
For the subsequent $5$ polaroids,we apply Malus' Law: $I_n = I_{n-1} \cos^2(\theta)$,where $\theta = 30^\circ$.
The final intensity $I_6$ after passing through all $6$ polaroids is given by:
$I_6 = I_1 \times (\cos^2 30^\circ)^5 = \frac{I_0}{2} \times (\cos^2 30^\circ)^5$.
Substituting $\cos 30^\circ = \frac{\sqrt{3}}{2}$:
$I_6 = \frac{I_0}{2} \times \left(\frac{3}{4}\right)^5 = \frac{I_0}{2} \times \frac{243}{1024} = \frac{243}{2048} I_0 \approx 0.1186 I_0$.
Converting to percentage: $0.1186 \times 100 \approx 11.86\% \approx 12\%$.

(A) According to Malus's law,the intensity of transmitted light is $I = I_0 \cos^2 \theta$,where $I_0$ is the incident intensity and $\theta$ is the angle between the polarization direction and the polarizer axis.
Since the polarizer is rotating,$\theta$ varies from $0$ to $2\pi$ in one revolution.
The average intensity $I_{av}$ is given by $I_{av} = \frac{1}{2\pi} \int_0^{2\pi} I_0 \cos^2 \theta \, d\theta = \frac{I_0}{2}$.
The incident power $P = 10^{-3} \, W$. The incident intensity $I_0 = \frac{P}{A} = \frac{10^{-3}}{3 \times 10^{-4}} = \frac{10}{3} \, W/m^2$.
The average power transmitted is $P_{av} = I_{av} \times A = \frac{I_0}{2} \times A = \frac{P}{2} = \frac{10^{-3}}{2} = 0.5 \times 10^{-3} \, W$.
The time period for one revolution is $T = \frac{2\pi}{\omega} = \frac{2 \times 3.14}{31.4} = 0.2 \, s$.
The energy passing through the polarizer per revolution is $E = P_{av} \times T = (0.5 \times 10^{-3}) \times 0.2 = 10^{-4} \, J$.

(C) When unpolarized light of intensity $I_0$ passes through the first Nicol prism (polarizer),the intensity of the transmitted light becomes $I_1 = \frac{I_0}{2}$.
According to Malus' Law,when this polarized light passes through the second Nicol prism (analyzer) inclined at an angle $\theta = 60^o$ to the first,the transmitted intensity $I$ is given by:
$I = I_1 \cos^2 \theta = \left(\frac{I_0}{2}\right) \cos^2(60^o)$.
Since $\cos(60^o) = \frac{1}{2}$,we have $\cos^2(60^o) = \frac{1}{4}$.
Therefore,$I = \left(\frac{I_0}{2}\right) \times \left(\frac{1}{4}\right) = \frac{I_0}{8}$.
The percentage of incident light transmitted is $\left(\frac{I}{I_0}\right) \times 100 = \left(\frac{I_0/8}{I_0}\right) \times 100 = \frac{100}{8} = 12.5\%$.

(B) polariser is an optical device that converts unpolarised light into polarised light by allowing only those light waves whose electric field vectors oscillate in a specific plane to pass through. Therefore,the primary function of a polariser is to produce polarised light.

(A) Polarization is a phenomenon that occurs only in transverse waves. Since light waves exhibit polarization,it confirms that light is a transverse wave in nature. Longitudinal waves,such as sound waves,cannot be polarized because their oscillations occur in the direction of propagation.

(B) According to Malus's Law,the intensity of light passing through a polaroid is given by $I = I_0 \cos^2 \theta$.
Here,the angle $\theta = 45^\circ$.
Substituting the value: $I = I_0 \cos^2(45^\circ)$.
Since $\cos(45^\circ) = 1/\sqrt{2}$,we have $\cos^2(45^\circ) = 1/2$.
Therefore,$I = I_0 \times (1/2) = I_0 / 2$.

(B) When unpolarized light is incident on the first nicol prism (polarizer),the intensity of the transmitted light is $I_1 = \frac{1}{2} I_0$,where $I_0$ is the intensity of the incident light.
Initially,the two nicols are crossed,meaning the angle between their transmission axes is $\theta_{initial} = 90^{\circ}$.
When one of them is rotated by $60^{\circ}$,the new angle between the transmission axes becomes $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
According to Malus' Law,the intensity of light emerging from the second nicol prism (analyzer) is $I = I_1 \cos^2 \theta$.
Substituting the values: $I = (\frac{1}{2} I_0) \cos^2 30^{\circ}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $\cos^2 30^{\circ} = \frac{3}{4}$.
Thus,$I = \frac{1}{2} I_0 \times \frac{3}{4} = \frac{3}{8} I_0$.
The percentage of incident light passing through the system is $\frac{I}{I_0} \times 100 = \frac{3}{8} \times 100 = 37.5\%$.

(C) Step $1$: When unpolarized light of intensity $I_0$ passes through the first polaroid $A$,the intensity of the transmitted light becomes half of the initial intensity.
$I_A = \frac{I_0}{2}$
Step $2$: According to Malus's Law,when this polarized light passes through the second polaroid $B$ whose transmission axis makes an angle $\theta = 45^{\circ}$ with the transmission axis of $A$,the intensity of the emerging light $I_B$ is given by:
$I_B = I_A \cos^2 \theta$
Step $3$: Substituting the values:
$I_B = \frac{I_0}{2} \cos^2(45^{\circ})$
Since $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,then $\cos^2(45^{\circ}) = \frac{1}{2}$.
$I_B = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$

(C) When unpolarized light of intensity $I_0$ is incident on a polarizing sheet,the transmitted intensity is given by Malus's Law concept for unpolarized light,which is $I = \frac{1}{2} I_0$.
Since the total incident intensity is $I_0$ and the transmitted intensity is $\frac{1}{2} I_0$,the intensity of light that does not pass through (absorbed intensity) is $I_{absorbed} = I_0 - \frac{1}{2} I_0 = \frac{1}{2} I_0$.

(A) According to Malus's Law,the intensity of transmitted light is given by $I = I_0 \cos^2 \theta$,where $I_0$ is the maximum intensity and $\theta$ is the angle between the transmission axes of the polarizer and the analyzer.
Initially,the intensity is maximum,meaning the axes are parallel $(\theta = 0^{\circ})$.
When the analyzer is rotated by $60^{\circ}$,the new angle becomes $\theta = 60^{\circ}$.
Substituting the values into the formula:
$I = I_0 \cos^2(60^{\circ})$
$I = I_0 \left( \frac{1}{2} \right)^2$
$I = \frac{I_0}{4}$

(C) Light waves are transverse in nature and can be polarized,whereas sound waves are longitudinal in nature and cannot be polarized. Therefore,polarization is the property that distinguishes light waves from sound waves.

(D) According to Malus's Law,the intensity of light emerging from an analyzer is $I = I_0 \cos^2 \theta$,where $I_0$ is the intensity of light incident on the analyzer and $\theta$ is the angle between the polarizer and the analyzer.
Since intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$,the amplitude of the light emerging from the analyzer is given by $A' = A_0 \cos \theta$,where $A_0$ is the amplitude of light incident on the analyzer.
Given that the light incident on the polarizer has amplitude $A$,the light emerging from the polarizer has amplitude $A_0 = A$.
Substituting the given values: $A' = A \cos(60^o) = A \times (1/2) = A/2$.

(A) When unpolarized light of intensity $I_0$ is incident on a polarizing sheet,the intensity of the transmitted polarized light is given by $I_t = I_0 / 2$.
The intensity of light that does not get transmitted (absorbed or reflected) is the difference between the incident intensity and the transmitted intensity.
Intensity of untransmitted light $= I_0 - I_t = I_0 - I_0 / 2 = I_0 / 2$.

(C) When unpolarized light of intensity $I_0$ passes through a polarizer,the intensity of the transmitted light is given by Malus's Law principle for unpolarized light,which is $I_{transmitted} = \frac{I_0}{2}$.
The intensity of the light that does not pass through the polarizer is the difference between the incident intensity and the transmitted intensity.
$I_{blocked} = I_0 - I_{transmitted} = I_0 - \frac{I_0}{2} = \frac{I_0}{2}$.

(C) The transverse nature of light means that the oscillations of the electric and magnetic fields occur perpendicular to the direction of wave propagation.
Interference,refraction,and dispersion can be explained by both longitudinal and transverse wave models.
Polarization is a phenomenon that occurs only in transverse waves,where the orientation of the oscillations can be restricted to a specific plane.
Since light exhibits polarization,it confirms that light is a transverse wave.

(C) When unpolarized light of intensity $I_0$ is incident on the first Nicol prism (polarizer),the intensity of the transmitted light is $I_1 = I_0 / 2$.
This light then passes through the second Nicol prism (analyzer) whose principal plane is at an angle $\theta = 60^o$ with respect to the first.
According to Malus's Law,the intensity of light transmitted through the analyzer is $I = I_1 \cos^2 \theta$.
Substituting the values: $I = (I_0 / 2) \cos^2(60^o) = (I_0 / 2) \times (1/2)^2 = I_0 / 8$.
To find the percentage of transmitted light: $(I / I_0) \times 100 = (1/8) \times 100 = 12.5\%$.

(A) Polarized sunglasses are used because they reduce glare caused by reflected light from surfaces like water,roads,or snow. By blocking the horizontally polarized component of light,they significantly reduce eye strain and improve visibility,preventing the eyes from being dazzled by intense light.

(D) According to Malus's Law,the intensity of transmitted light is given by $I = I_0 \cos^2 \theta$,where $\theta$ is the angle between the transmission axis of the Polaroid and the plane of polarization of the incident light.
As the Polaroid completes one full rotation ($0$ to $2\pi$),the angle $\theta$ changes continuously.
The intensity $I$ becomes maximum when $\theta = 0$ and $\theta = \pi$,and it becomes zero (minimum) when $\theta = \pi/2$ and $\theta = 3\pi/2$.
Therefore,the intensity of light becomes maximum twice and zero twice during one full rotation.

(D) When an unpolarized light beam of intensity $I_0$ is incident on polarizer $A$,the intensity of the emerging polarized light is $I_1 = I_0/2$.
Now,this polarized light of intensity $I_1 = I_0/2$ is incident on polarizer $B$ at an angle $\theta = 45^\circ$ with respect to its transmission axis.
According to Malus's Law,the intensity of light emerging from polarizer $B$ is given by $I = I_1 \cos^2 \theta$.
Substituting the values: $I = (I_0/2) \cos^2(45^\circ) = (I_0/2) \times (1/\sqrt{2})^2 = (I_0/2) \times (1/2) = I_0/4$.

(C) When a calcite crystal is rotated,the intensity of the transmitted light changes because the light from the sky is partially polarized due to scattering.
According to Rayleigh's law of scattering,the intensity of scattered light $I \propto 1/\lambda^4$.
Since the wavelength $\lambda$ of blue light is minimal,its scattering intensity $I$ is maximum.
Thus,light from the sky is polarized due to scattering by atmospheric particles.
Both statements are true,and Statement-$2$ provides the correct explanation for the polarization observed in Statement-$1$.

(B) The formula for optical rotation is $\theta = [\alpha]_\lambda^T \cdot l \cdot C$,where $\theta$ is the rotation,$[\alpha]_\lambda^T$ is the specific rotation,$l$ is the length of the tube in decimeters $(dm)$,and $C$ is the concentration in $g/cm^3$.
Given: $l = 300 \, mm = 30 \, cm = 3 \, dm$,$\theta = 10^\circ$,$[\alpha]_\lambda^T = 60^\circ$,and volume $V = 60 \, cm^3$.
First,calculate the concentration $C$: $C = \frac{\theta}{[\alpha]_\lambda^T \cdot l} = \frac{10}{60 \cdot 3} = \frac{10}{180} = \frac{1}{18} \, g/cm^3$.
The total amount of sugar $m$ is given by $m = C \cdot V$.
$m = \frac{1}{18} \cdot 60 = \frac{60}{18} = 3.33 \, g$.

(B) The intensity of unpolarised light incident on $P_1$ is $I_0$. After passing through $P_1$,the intensity becomes $I_1 = \frac{I_0}{2}$.
The intensity of light transmitted through $P_3$,which is at an angle $\theta_1 = 45^{\circ}$ with $P_1$,is given by Malus' Law: $I_2 = I_1 \cos^2(45^{\circ}) = \frac{I_0}{2} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{4}$.
The angle between the axes of $P_3$ and $P_2$ is $\theta_2 = 90^{\circ} - 45^{\circ} = 45^{\circ}$.
The intensity of light transmitted through $P_2$ is $I_3 = I_2 \cos^2(45^{\circ}) = \frac{I_0}{4} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{8}$.

(C) The intensity of light passing through a pair of polarizers is given by Malus' Law: $I = I_0 \cos^2 \theta$,where $\theta$ is the angle between the transmission axes of the two polarizers.
For unpolarized light incident on the first polarizer,the intensity becomes $I_1 = I_0 / 2$. The intensity after the second polarizer is $I = I_1 \cos^2 \theta = (I_0 / 2) \cos^2 \theta$.
To rank the pairs,we calculate the angle $\theta$ between the two sheets for each case:
$(i)$ The first sheet is at $30^\circ$ to the vertical,the second is at $60^\circ$ to the horizontal. The angle between them is $\theta = 90^\circ - 30^\circ - 60^\circ = 0^\circ$. Thus,$\cos^2(0^\circ) = 1$.
(ii) The first sheet is at $30^\circ$ to the vertical,the second is at $60^\circ$ to the horizontal. The angle between them is $\theta = 60^\circ$. Thus,$\cos^2(60^\circ) = 0.25$.
(iii) The first sheet is at $30^\circ$ to the horizontal,the second is at $60^\circ$ to the horizontal. The angle between them is $\theta = 30^\circ$. Thus,$\cos^2(30^\circ) = 0.75$.
(iv) The first sheet is at $30^\circ$ to the horizontal,the second is at $60^\circ$ to the horizontal. The angle between them is $\theta = 90^\circ$. Thus,$\cos^2(90^\circ) = 0$.
Comparing the values: $1 > 0.75 > 0.25 > 0$,which corresponds to $(i) > (iii) > (ii) > (iv)$.

(C) When unpolarized light of intensity $I_o$ is incident on a polarizing sheet,the intensity of the transmitted light is given by $I_t = \frac{I_o}{2}$.
The intensity of the light that does not get transmitted is the difference between the incident intensity and the transmitted intensity.
$\text{Intensity of untransmitted light} = I_o - I_t$
$\text{Intensity of untransmitted light} = I_o - \frac{I_o}{2} = \frac{I_o}{2}$.

(D) When unpolarised light of intensity $I_0$ passes through the first polaroid $A$,the intensity of the transmitted light becomes $I_1 = \frac{I_0}{2}$.
According to Malus' Law,when this polarised light passes through the second polaroid $B$ whose transmission axis makes an angle $\theta = 45^{\circ}$ with the axis of $A$,the intensity of the emergent light $I_R$ is given by:
$I_R = I_1 \cos^2(\theta)$
Substituting the values:
$I_R = \left(\frac{I_0}{2}\right) \cos^2(45^{\circ})$
Since $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have $\cos^2(45^{\circ}) = \frac{1}{2}$.
Therefore,$I_R = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$.

(C) According to Malus's law,the intensity of the emerging beam is given by $I = I_0 \cos^2 \theta$.
Let the initial intensity of beam $A$ be $I_A$ and beam $B$ be $I_B$. When the polaroid is at the position where beam $A$ has maximum intensity,the angle between the transmission axis of the polaroid and the plane of polarization of beam $A$ is $0^{\circ}$,and for beam $B$ it is $90^{\circ}$.
After rotating the polaroid by $30^{\circ}$,the new angle for beam $A$ is $\theta_A = 30^{\circ}$ and for beam $B$ is $\theta_B = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
The intensities of the beams after passing through the polaroid are:
$I_A' = I_A \cos^2(30^{\circ}) = I_A \left(\frac{\sqrt{3}}{2}\right)^2 = I_A \cdot \frac{3}{4}$
$I_B' = I_B \cos^2(60^{\circ}) = I_B \left(\frac{1}{2}\right)^2 = I_B \cdot \frac{1}{4}$
Given that the beams appear equally bright,$I_A' = I_B'$:
$I_A \cdot \frac{3}{4} = I_B \cdot \frac{1}{4}$
$\frac{I_A}{I_B} = \frac{1}{3}$