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(D) According to Brewster's law,for complete polarization of reflected light,the angle of incidence $\phi$ must satisfy the condition $\mu = 	an \phi

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the intensity gradually reduces to zero and then again increases

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(D) According to Brewster's law,for complete polarization of reflected light,the angle of incidence $\phi$ must satisfy the condition $\mu = 	an \phi$.Given $\mu = 1.54$,we have $\phi = 	an^{-1}(1.54) = 57^{\circ}$.From the figure,the angle between the incident ray and the surface is $33^{\circ}$. Therefore,the angle of incidence (the angle with the normal) is $i = 90^{\circ} - 33^{\circ} = 57^{\circ}$.Since the angle of incidence is equal to Brewster's angle,the reflected light is completely plane-polarized.When plane-polarized light is viewed through a rotating Nicol prism,the intensity of the transmitted light varies according to Malus's Law,$I = I_0 \cos^2 	heta$. As the prism rotates,the intensity gradually reduces to zero (when the transmission axis is perpendicular to the plane of polarization) and then increases again.

$A$ beam of light $AO$ is incident on a glass slab $(\mu = 1.54)$ in a direction as shown in the figure. The reflected ray $OB$ is passed through a Nicol prism. On viewing through the rotating Nicol prism,we find that $(\tan 57^{\circ} = 1.54)$.

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