Two periodic waves of intensities I_{1} and I | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The resultant intensity of two periodic waves is given by the formula:$I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\delta$where $\delta$ is the phas

Vedclass · Accepted Answer

$2(I_{1}+I_{2})$

Vedclass · Answer

(D) The resultant intensity of two periodic waves is given by the formula:$I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\delta$where $\delta$ is the phase difference between the waves.For maximum intensity,$\cos\delta = 1$,so:$I_{\max} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$For minimum intensity,$\cos\delta = -1$,so:$I_{\min} = I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$The sum of the maximum and minimum intensities is:$I_{\max} + I_{\min} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2} + (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$Expanding the squares:$I_{\max} + I_{\min} = (I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}) + (I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}})$$I_{\max} + I_{\min} = 2(I_{1} + I_{2})$

Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

Similar Questions

The phenomenon of interference is shown by

As a result of interference of two coherent sources of light,energy is

In the interference of two coherent sources of different intensities,the ratio of maximum and minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{144}{81}$. What is the ratio of their amplitudes?

Two coherent monochromatic beams of intensities $I$ and $4I$ respectively are superposed. The maximum and minimum intensities in the resulting pattern are

Which of the following is not a correct statement?

Vedclass Test Series

Exam Paper Generator

Online Exam Module