Question

Two coherent waves with their intensities in the ratio i1/i2=a

Answer 1

Answer:

HERE IS YOUR ANSWER

Step-by-step explanation:

According to the question/

Maximum intensity is given by (√i1 + √i2)^2 and minus in between if it is minimum intensity. Hope it helps you. Intensity of the wave is directly proportional to square of amplitude. Therefore, ratio of maximum to minimum intensity is 4:1.

Answer 2

Answer: (2√a)/(a + 1)

Concept : Wave Optics

Given : $\frac{i_{1} }{i_{2} } = a$

To Find : The ratio of  $\frac{I_{max} - I_{min} }{I_{max} + I_{min} }$

Step-by-step explanation:

According to the question ,

Two coherent waves with their intensities in the ratio  $\frac{i_{1} }{i_{2} } = a$.

Intensity,  $I = I_{1} + I_{2} + 2\sqrt{I_{1} I_{2} } cos \alpha$

at Δα = 0, $I = I_{max}$

                   $= I_{1} + I_{2} + 2\sqrt{I_{1} I_{2} } * 1$

                   $= (\sqrt{I_{1} } + \sqrt{I_{2} } )^{2}$

at Δα = $\pi$, $I = I_{min}$

                   $= I_{1} + I_{2} + 2\sqrt{I_{1} I_{2} } * (-1)$

                   $= (\sqrt{I_{1} } - \sqrt{I_{2} } )^{2}$

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

                   $= \frac{4 \sqrt{I_{1} I_{2} } }{2(I_{1} + I_{2} )}$

                  $= \frac{4 \sqrt{I_{1} * \frac{I_{1} }{a} } }{2(I_{1} + \frac{I_{1} }{a} )}$

                 $= \frac{2}{\sqrt{a} } * (\frac{a}{a +1} )$

                 $= \frac{2\sqrt{a} }{(a + 1)}$

                   

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