Two waves having the intensities in the ratio | vedclass

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

Question

(B) Let the intensities of the two waves be $I_1$ and $I_2$.Given that the ratio of intensities is $I_1 : I_2 = 9 : 1$. Let $I_1 = 9x$ and $I_2 = x$.T

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(B) Let the intensities of the two waves be $I_1$ and $I_2$.Given that the ratio of intensities is $I_1 : I_2 = 9 : 1$. Let $I_1 = 9x$ and $I_2 = x$.The formula for the ratio of maximum to minimum intensity in an interference pattern is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$.Substituting the values,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{9x} + \sqrt{x}}{\sqrt{9x} - \sqrt{x}} \right)^2$.This simplifies to $\frac{I_{\max}}{I_{\min}} = \left( \frac{3\sqrt{x} + \sqrt{x}}{3\sqrt{x} - \sqrt{x}} \right)^2 = \left( \frac{4\sqrt{x}}{2\sqrt{x}} \right)^2$.$\frac{I_{\max}}{I_{\min}} = (2)^2 = 4$.Thus,the ratio of maximum to minimum intensity is $4 : 1$.

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

Similar Questions

In Young's double slit experiment,the intensity of light coming from the first slit is double the intensity from the second slit. The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is

In Young's double-slit experiment using monochromatic light of wavelength $\lambda$,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where the path difference is $\lambda / 3$?

Two coherent light sources having intensity in the ratio $2x$ produce an interference pattern. Then the value of $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be

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 $\pi / 2$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module