Two coherent monochromatic light beams of int | vedclass

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

Question

(C) The maximum intensity $I_{\max}$ for the superposition of two coherent beams is given by the formula:$I_{\max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^2$G

Vedclass · Accepted Answer

$9I$ and $I$

Vedclass · Answer

(C) The maximum intensity $I_{\max}$ for the superposition of two coherent beams is given by the formula:$I_{\max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^2$Given $I_{1} = I$ and $I_{2} = 4I$,we have:$I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$The minimum intensity $I_{\min}$ is given by the formula:$I_{\min} = (\sqrt{I_{1}} - \sqrt{I_{2}})^2$Substituting the values:$I_{\min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$Thus,the maximum and minimum intensities are $9I$ and $I$ respectively.

Two coherent monochromatic light beams of intensities $I$ and $4I$ are superposed. The maximum and minimum possible intensities in the resulting beam are:

Similar Questions

The major product of the following reaction is

The sulphate of a metal $M$ contains $9.87 \%$ of $M$. This sulphate is isomorphous with $ZnSO_4 \cdot 7H_2O$. The atomic weight of $M$ is: (in $.3$)

$A$ student is to answer $10$ out of $13$ questions in an examination such that he must choose at least $4$ from the first $5$ questions. The number of choices available to him is

The number of lone pairs of electrons present on $Xe$ in $XeF_4$ is

If $\int \frac{\sqrt{1-x^2}}{x^4} ~d x=A(x)\left(\sqrt{1-x^2}\right)^m+C$ for a suitable chosen integer $m$ and a function $A(x)$,where $C$ is a constant of integration,then $(A(x))^m$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module