Waves - The Principle of Superposition of Waves

Two waves of same frequency have amplitudes $3\text{ cm}$ and $4\text{ cm}$. Find the resultant amplitude when the phase difference between them is $\frac{\pi}{2}$.

Given $A_1 = 3$, $A_2 = 4$, and $\phi = \frac{\pi}{2}$. Using the formula: $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$ $A = \sqrt{3^2 + 4^2 + 2(3)(4) \cos(\frac{\pi}{2})}$ $A = \sqrt{9 + 16 + 24(0)}$ $A = \sqrt{25} = 5\text{ cm}$

Since the waves are at a phase difference of $90^\circ$ ($\frac{\pi}{2}$), the cosine term becomes zero, and the resultant amplitude follows the Pythagorean theorem.

Two coherent sources of light have an intensity ratio of $81:1$. Calculate the ratio of maximum to minimum intensity in their interference pattern.

Given $\frac{I_1}{I_2} = \frac{81}{1}$. Since $I \propto A^2$, the amplitude ratio is: $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{81}{1}} = 9$ Let $A_1 = 9k$ and $A_2 = 1k$. $\frac{I_{max}}{I_{min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$ $\frac{I_{max}}{I_{min}} = \left( \frac{9k + 1k}{9k - 1k} \right)^2 = \left( \frac{10}{8} \right)^2 = \left( \frac{5}{4} \right)^2 = \frac{25}{16}$

The maximum intensity occurs during constructive interference ($A_1 + A_2$) and minimum during destructive interference ($A_1 - A_2$). The ratio is the square of the sums/differences of the amplitudes.