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(A) The optical rotation $	heta$ produced by a sugar solution is given by the formula: $	heta = S \cdot l \cdot C$, where $S$ is the specific rotati

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$	heta$

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(A) The optical rotation $	heta$ produced by a sugar solution is given by the formula: $	heta = 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, $	heta = 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, $	heta = 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 $	heta_1$ is given by $	heta_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 $	heta_1 = 	heta$.

$A$. Angle of reflection	$(i)$ $2 \sin^{-1} \left(\sqrt{\frac{2}{3}}\right)$
$B$. Angle of refraction	$(ii)$ $\sin^{-1} \left(\sqrt{\frac{2}{3}}\right) - \sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$
$C$. Angle between incident and completely polarised light	$(iii)$ $\sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$
$D$. Angle of deviation of the incident ray	$(iv)$ $\cos^{-1} \left(\frac{1}{\sqrt{3}}\right)$

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