Question
State Brewster's law and show that when light is incident at polarizing angle the reflected and refracted rays are mutually perpendicular to each other.
✓
Answer
Brewster's law : The tangent of polarising angle is equal to the refractive index of the reflecting medium with respect to the surrounding $\left( n _2\right)$. If ip is the polarising angle. $\tan i_p={ }_1 n_2=\frac{n_2}{n_1} \ldots$ (1)
where $n _1$ is the absolute refractive index of the surrounding and $n _2$ is that of the reflecting medium. Figure shows a ray AO of unpolarised light incident on the interface separating two mediums. The degree of polarisation of the reflected ray OC varies with the angle of incidence and is a maximum for the angle of incidence equal to the polarising angle $i_p$ of the pair of mediums. For all angle of incidence, the refracted ray $O B$ is only partially polarised.
In figure, the angle of incidence is $i_p$, the angle of reflection is $r^{\prime}$ and the angle of refraction is r . By Snell's law, $\frac{n_2}{n_1}=\frac{\sin i_p}{\sin r} \ldots$ (2)
From, Eqs. (1) and (2)
$\tan i_p=\frac{\sin i_p}{\cos i_p=} \frac{\sin i_p}{\sin r}$
$\therefore \cos i_p=\sin r$
$\text { Now, } r^{\prime}=i_p$
$\therefore \cos r^{\prime}=i_p$
$\therefore \cos r^{\prime}=\operatorname{sir} r$
$\therefore \sin \left(90^{\circ}-r^{\prime}\right)=\sin r$
$\therefore r=90^{\circ}-r^{\prime} \text { or, } r^{\prime}+r=90^{\circ}$
$\therefore \angle \operatorname{coB}=90^{\circ}$
This indicates that, for complete polarisation of the reflected ray at the polarising angle, the reflected and the refracted rays are mutually perpendicular.
where $n _1$ is the absolute refractive index of the surrounding and $n _2$ is that of the reflecting medium. Figure shows a ray AO of unpolarised light incident on the interface separating two mediums. The degree of polarisation of the reflected ray OC varies with the angle of incidence and is a maximum for the angle of incidence equal to the polarising angle $i_p$ of the pair of mediums. For all angle of incidence, the refracted ray $O B$ is only partially polarised.
In figure, the angle of incidence is $i_p$, the angle of reflection is $r^{\prime}$ and the angle of refraction is r . By Snell's law, $\frac{n_2}{n_1}=\frac{\sin i_p}{\sin r} \ldots$ (2)
From, Eqs. (1) and (2)
$\tan i_p=\frac{\sin i_p}{\cos i_p=} \frac{\sin i_p}{\sin r}$
$\therefore \cos i_p=\sin r$
$\text { Now, } r^{\prime}=i_p$
$\therefore \cos r^{\prime}=i_p$
$\therefore \cos r^{\prime}=\operatorname{sir} r$
$\therefore \sin \left(90^{\circ}-r^{\prime}\right)=\sin r$
$\therefore r=90^{\circ}-r^{\prime} \text { or, } r^{\prime}+r=90^{\circ}$
$\therefore \angle \operatorname{coB}=90^{\circ}$
This indicates that, for complete polarisation of the reflected ray at the polarising angle, the reflected and the refracted rays are mutually perpendicular.
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