Two coherent sources whose intensity ratio is 81:1produces interfernces fringes .calculate the ratio of intensity of maxima and minima in the fringe system ?

Question

Answer

I1/I2= a1*2/a2*2...a1/a2=9:1..Imax/I min = (a1+a2)/(a1-a2)*2..=(9+1)*2/(9-1)*2 = 100/64 =25/16

Answer

I1=81i2 =1so a1= 9a2=1imax=(9+1)2imin=(9-1)2100/64=25/16100percent👍👍👍👍👍

Answer

I max by i min ratio is given by under root I1 plus under root I2 ka square divided by under root I1 minus under root I2 ka square...thus imax by i min is 100 by 64...

Answer

Ans 100/64.
given, i1/i2= 81/1

eqn of imax/ imin = (√i1+√i2) all square/(√i1-√i2) square.
= (9+1) square/(9-1) square
= 10square/8square
= 100/64

Answer

With solution

Answer

Introduction:
The phenomenon of interference occurs when two or more coherent sources of light superpose and create regions of constructive and destructive interference. The resulting interference pattern consists of bright and dark fringes. The intensity of these fringes depends on the relative amplitudes of the interfering waves.

Given:
The intensity ratio of the two coherent sources is 81:1.

Calculating the ratio of intensity of maxima and minima:
To determine the ratio of intensity of maxima and minima in the fringe system, we need to consider the superposition of the waves from the two sources at different points in the interference pattern.

Intensity of maxima:
At the maxima of the interference pattern, the two waves are in phase and add constructively. The resultant amplitude is the sum of the amplitudes of the individual waves. Since intensity is proportional to the square of the amplitude, the intensity at the maxima is the square of the sum of the amplitudes.

Let the amplitudes of the two sources be A and B, respectively. The intensity at the maxima (Imax) is given by:
Imax = (A + B)^2

Intensity of minima:
At the minima of the interference pattern, the two waves are in phase opposition and add destructively. The resultant amplitude is the difference between the amplitudes of the individual waves. Again, intensity is proportional to the square of the amplitude, so the intensity at the minima is the square of the difference of the amplitudes.

The difference in amplitudes between the two sources is given by 9A. Therefore, the intensity at the minima (Imin) is given by:
Imin = (A - B)^2 = (9A)^2 = 81A^2

Calculating the ratio:
To find the ratio of intensity at the maxima and minima, we divide the intensity at the maxima by the intensity at the minima:
Ratio = Imax / Imin = [(A + B)^2] / [(A - B)^2]

Since the intensity ratio of the two sources is 81:1, we can express B in terms of A:
B = sqrt(81A^2) = 9A

Substituting this value of B into the ratio equation:
Ratio = [(A + 9A)^2] / [(A - 9A)^2] = (10A)^2 / (-8A)^2 = (100A^2) / (64A^2) = 100/64 = 25/16

Therefore, the ratio of intensity of maxima to minima in the fringe system is 25:16.

Top Courses for Class 12

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams