The ratio of amplitudes of two coherent waves in Young's double-slit experiment is A 1 A 2 = 1 3 . What is the ratio of maximum and minimum intensities of fringes?

I max I min = I 1 + I 2 2 ( I 1 - I 2 ) 2 = A 1 + 3 A 1 A 1 - 3 A 1 2 I max I min = A 1 + 3 A 1 A 1 - 3 A 1 2 = 4 2 2 = 4

Start Free Mock LoginAlready registered? Login here

Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsYDSEEasy2 minQB

PhysicsEasynumerical

The ratio of amplitudes of two coherent waves in Young's double-slit experiment is $\frac{A_{1}}{A_{2}} = \frac{1}{3}$ . What is the ratio of maximum and minimum intensities of fringes?

Answer:

4.00

Solution:

$\frac{I_{\max}}{I_{\min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{(\sqrt{I_{1}} - \sqrt{I_{2}})^{2}} = {(\frac{A_{1} + 3 A_{1}}{A_{1} - 3 A_{1}})}^{2}$

$\frac{I_{\max}}{I_{\min}} = {(\frac{A_{1} + 3 A_{1}}{A_{1} - 3 A_{1}})}^{2} = {(\frac{4}{2})}^{2} = 4$

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:YDSE

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...