Physics - Sound Wave Question with Solution | TestHub

For maximum intensity, $$I_{\max }=I_1+I_2+2\sqrt{I_1{I}_2}\cos \delta$$ $$\begin{array}{rl}\delta & =0\\ I_{\max }& =L+\frac{L}{4}+2\sqrt{L\times \frac{L}{4}}\cdot \cos 0\\ & =\frac{5L}{4}+\frac{2L}{2}\\ & =\frac{5L+4L}{4}\\ & =\frac{9L}{4}\\ I_{\min }& =I_1+I_2-2\sqrt{I_1{I}_2}\cos \theta \\ \theta & =0^{\circ }\\ & =L+\frac{L}{4}-2\sqrt{L\times \frac{L}{4}}\times 1\\ & =\frac{5L}{4}-L=\frac{L}{4}\end{array}$$ Thus, maximum and minimum intensities will be $\left(\frac{9L}{4},\frac{L}{4}\right)$