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(D) Given, refractive index $\mu = 1.414 = \sqrt{2}$.For complete polarisation, the angle of incidence is the Brewster's angle $i_p$, where $	an i_p

Vedclass · Accepted Answer

$(iv), (iii), (i), (ii)$

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

$A$. 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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