Two coherent sources of intensity ratio alpha | vedclass

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

Question

(A) Let the intensities of the two sources be $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \alpha$. Since $I \propto A^2$,we have $\frac{A_1}{A_2} = \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the intensities of the two sources be $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \alpha$. Since $I \propto A^2$,we have $\frac{A_1}{A_2} = \sqrt{\alpha}$.The maximum and minimum intensities in an interference pattern 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 $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$.Substituting the expressions for $I_{max}$ and $I_{min}$:$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{(A_1 + A_2)^2 - (A_1 - A_2)^2}{(A_1 + A_2)^2 + (A_1 - A_2)^2}$Expanding the squares:$= \frac{(A_1^2 + A_2^2 + 2A_1A_2) - (A_1^2 + A_2^2 - 2A_1A_2)}{(A_1^2 + A_2^2 + 2A_1A_2) + (A_1^2 + A_2^2 - 2A_1A_2)}$$= \frac{4A_1A_2}{2(A_1^2 + A_2^2)} = \frac{2A_1A_2}{A_1^2 + A_2^2}$Dividing the numerator and denominator by $A_2^2$:$= \frac{2(A_1/A_2)}{(A_1/A_2)^2 + 1}$Since $\frac{A_1}{A_2} = \sqrt{\alpha}$,we substitute this into the expression:$= \frac{2\sqrt{\alpha}}{(\sqrt{\alpha})^2 + 1} = \frac{2\sqrt{\alpha}}{\alpha + 1}$.

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

Similar Questions

The ratio of intensities of two interfering waves is $9:1$. The ratio of the maximum amplitude to the minimum amplitude of the resultant wave is ........

In the Young's double slit experiment,the intensities at two points $P_1$ and $P_2$ on the screen are respectively $I_1$ and $I_2$. If $P_1$ is located at the centre of a bright fringe and $P_2$ is located at a distance equal to a quarter of fringe width from $P_1$,then $\frac{I_1}{I_2}$ is

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $k$. What will be the intensity at a point where the path difference is $\lambda/4$? ($\lambda$ = wavelength of light)

In a Young's double slit experiment,the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is . . . . . . . . $\mu m$.
(Given: $\lambda = 600 \ nm, d = 1.0 \ mm, D = 1.0 \ m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module