Answer: \(\tan\theta_B = n\) (Brewster’s law); slit separation \(d = 5.1\times10^{-4}\) m \(= 0.51\) mm.
(i) Brewster’s angle: When unpolarised light is incident on a transparent medium of refractive index \(n\) at a particular angle \(\theta_B\), the reflected light is completely plane polarised. Brewster found that at this angle the reflected and refracted rays are perpendicular to each other.
[Figure: unpolarised light incident at angle \(\theta_B\) on the surface of a glass slab; the reflected ray (plane polarised, vibrations perpendicular to the plane of incidence) and the refracted ray (angle of refraction \(\theta_r\)) are at 90° to each other.]
From the figure, \(\theta_B + 90^\circ + \theta_r = 180^\circ\), so
\[\theta_r = 90^\circ - \theta_B\]
By Snell’s law (light going from air into the medium):
\[n = \frac{\sin\theta_B}{\sin\theta_r} = \frac{\sin\theta_B}{\sin(90^\circ - \theta_B)} = \frac{\sin\theta_B}{\cos\theta_B}\]\[\boxed{\tan\theta_B = n}\]
This is Brewster’s law; \(\theta_B\) is the Brewster angle (polarising angle). For glass, \(n = 1.5\) gives \(\theta_B \approx 56.3^\circ\).
(ii) Numerical: \(\lambda = 5100\) Å \(= 5.1\times10^{-7}\) m, \(D = 2\) m. The overall separation of 10 fringes is 2 cm, so the fringe width is
\[X = \frac{2\ \mathrm{cm}}{10} = 0.2\ \mathrm{cm} = 2\times10^{-3}\ \mathrm{m}\]
Fringe width \(X = \dfrac{\lambda D}{d}\), so
\[d = \frac{\lambda D}{X} = \frac{5.1\times10^{-7}\times2}{2\times10^{-3}} = 5.1\times10^{-4}\ \mathrm{m} = 0.51\ \mathrm{mm}\]
