The intensity ratio of the two interfering be | vedclass

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

Question

(B) Let the intensities of the two beams be $I_1$ and $I_2$. The ratio is given as $\frac{I_1}{I_2} = \beta$.Since intensity $I \propto A^2$,we have $

Vedclass · Accepted Answer

$\frac{2\sqrt{\beta}}{1 + \beta}$

Vedclass · Answer

(B) Let the intensities of the two beams be $I_1$ and $I_2$. The ratio is given as $\frac{I_1}{I_2} = \beta$.Since intensity $I \propto A^2$,we have $\frac{A_1}{A_2} = \sqrt{\beta}$.The maximum and minimum intensities are given by $I_{\max} = (A_1 + A_2)^2$ and $I_{\min} = (A_1 - A_2)^2$.We need to calculate the value of $X = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$.Substituting the expressions for $I_{\max}$ and $I_{\min}$:$X = \frac{(A_1 + A_2)^2 - (A_1 - A_2)^2}{(A_1 + A_2)^2 + (A_1 - A_2)^2} = \frac{4A_1 A_2}{2(A_1^2 + A_2^2)} = \frac{2A_1 A_2}{A_1^2 + A_2^2}$.Dividing the numerator and denominator by $A_2^2$:$X = \frac{2(A_1/A_2)}{(A_1/A_2)^2 + 1} = \frac{2\sqrt{\beta}}{\beta + 1}$.

The intensity ratio of the two interfering beams of light is $\beta$. What is the value of $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$?

Similar Questions

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

Two waves having the intensities in the ratio of $9 : 1$ produce interference. The ratio of maximum to the minimum intensity is equal to

In Young's double slit experiment,if the maximum intensity is $I$,then the angular position where the intensity becomes $\frac{I}{4}$ is:

Two coherent sources with intensity ratio $\beta : 1$ produce interference. Fringe visibility will be

The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the fringe width from the centre is

Vedclass Test Series

Exam Paper Generator

Online Exam Module