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, \( \theta_2 = 30^\circ \).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).