The power radiated by a linear antenna is proportional to the square of the current and the square of the antenna's length. Additionally, the radiated power depends inversely on the square of the wavelength of the electromagnetic wave. Thus, the proportionality relation is: \[ P \propto \left( \frac{l}{\lambda} \right)^2. \]
Final Answer: The power radiated from a linear antenna is proportional to: \[ \boxed{\left( \frac{l}{\lambda} \right)^2}. \]
Two plane polarized light waves combine at a certain point, whose "E" components are: \[ E_1 = E_0 \sin \omega t, \quad E_2 = E_0 \sin \left( \omega t + \frac{\pi}{3} \right) \] Find the amplitude of the resultant wave.

Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: