According to Brewster’s law, the reflected ray is completely polarized when the angle of incidence \( \theta_p \) satisfies:
\[ \tan \theta_p = n, \]where \( n \) is the refractive index of the medium relative to air.
Given \( \theta_p = 60^\circ \), we have:
\[ n = \tan 60^\circ = \sqrt{3}. \]Using Snell’s law \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), with \( n_1 = 1 \) (for air) and \( \theta_1 = 60^\circ \):
\[ \sin \theta_2 = \frac{\sin 60^\circ}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}. \]Therefore, \( \theta_2 = 30^\circ \).
List-I | List-II | ||
P | If \(n = 2\) and \(\alpha = 180°\), then all the possible values of \(\theta_0\) will be | I | \(30\degree\) or \(0\degree\) |
Q | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\theta_0\) will be | II | \(60\degree\) or \(0\degree\) |
R | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\phi_0\) will be | III | \(45\degree\) or \( 0\degree\) |
S | If \(n = \sqrt2\) and \(\theta_0 = 45°\), then all the possible values of \(\alpha\) will be | IV | \(150\degree\) |
\[0\degree\] |
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: