The amplitude of the light waves emerging fro | vedclass

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

Question

(C) Given the ratio of amplitudes $A_1 : A_2 = 2 : 3$.Since intensity $I \propto A^2$,the ratio of intensities is $\frac{I_1}{I_2} = \left(\frac{A_1}{

Vedclass · Accepted Answer

$1 : 25$

Vedclass · Answer

(C) Given the ratio of amplitudes $A_1 : A_2 = 2 : 3$.Since intensity $I \propto A^2$,the ratio of intensities is $\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.Let $I_1 = 4k$ and $I_2 = 9k$.The intensity of the minimum is $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{4k} - \sqrt{9k})^2 = (2\sqrt{k} - 3\sqrt{k})^2 = (-\sqrt{k})^2 = k$.The intensity of the maximum is $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{4k} + \sqrt{9k})^2 = (2\sqrt{k} + 3\sqrt{k})^2 = (5\sqrt{k})^2 = 25k$.Therefore,the ratio $\frac{I_{\min}}{I_{\max}} = \frac{k}{25k} = \frac{1}{25}$.

The amplitude of the light waves emerging from the two slits in Young's experiment is in the ratio of $2 : 3$. The ratio of the intensity of the minimum to that of the consecutive maximum will be:

Similar Questions

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda / 6$ is $I'$. If $I_0$ denotes the maximum intensity,then $I'/I_0$ is equal to

What is the time-averaged value of the intensity of a light wave,and at what phase difference is the intensity equal to $\frac{1}{2}$ of its maximum value?

If $\frac{I_1}{I_2} = \frac{16}{1},$ then $\frac{I_{\max}}{I_{\min}} = ?$

Obtain the formula for the intensity of light if the phase difference at a point from two coherent sources is $\phi$.

Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module