The ratio of maximum and minimum intensities | vedclass

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

Question

(D) Given the ratio of maximum intensity to minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{36}{1}$.We know that $I_{\max} = (a_1 + a_2)^2$ an

Vedclass · Accepted Answer

$7: 5$

Vedclass · Answer

(D) Given the ratio of maximum intensity to minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{36}{1}$.We know that $I_{\max} = (a_1 + a_2)^2$ and $I_{\min} = (a_1 - a_2)^2$,where $a_1$ and $a_2$ are the amplitudes of the two waves.Therefore,$\frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{36}{1}$.Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = \frac{6}{1}$.By cross-multiplying,$a_1 + a_2 = 6(a_1 - a_2) = 6a_1 - 6a_2$.Rearranging the terms,$5a_1 = 7a_2$.Thus,the ratio of the amplitudes is $\frac{a_1}{a_2} = \frac{7}{5}$.

The ratio of maximum and minimum intensities in an interference pattern is $36: 1$. The ratio of the amplitude of the two interfering waves will be

Similar Questions

If the distance between a point source and a screen is doubled,then the intensity of light on the screen will become:

Which phenomena can be explained by the wave nature of light?

Two light waves of intensities $I_0$ and $9I_0$ superpose at a point to produce a resultant intensity of $7I_0$. Calculate the phase difference between the light waves.

Two coherent monochromatic beams of intensities $I$ and $4I$ respectively are superposed. The maximum and minimum intensities in the resulting pattern are

Two light sources have amplitudes $a$ and $2a$ respectively. If they interfere with a phase difference of $\pi$,what is the resultant minimum intensity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module