In Young's double slit experiment,the amplitu | vedclass

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(C) The resultant intensity $I$ when two waves with amplitudes $A_{1}$ and $A_{2}$ interfere with a phase difference $\phi$ is given by $I = I_{1} + I

Vedclass · Accepted Answer

$\frac{I_{0}}{9}(5+4 \cos \phi)$

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(C) The resultant intensity $I$ when two waves with amplitudes $A_{1}$ and $A_{2}$ interfere with a phase difference $\phi$ is given by $I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} \cos \phi$.Since intensity $I \propto A^{2}$,we can write $I = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2} \cos \phi$.Given $A_{1} = A$ and $A_{2} = 2A$,the resultant intensity is:$I = A^{2} + (2A)^{2} + 2(A)(2A) \cos \phi = A^{2}(1 + 4 + 4 \cos \phi) = A^{2}(5 + 4 \cos \phi) \quad \dots (i)$.The maximum intensity $I_{0}$ occurs when $\cos \phi = 1$:$I_{0} = A^{2}(5 + 4(1)) = 9A^{2} \implies A^{2} = \frac{I_{0}}{9} \quad \dots (ii)$.Substituting $(ii)$ into $(i)$,we get:$I = \frac{I_{0}}{9}(5 + 4 \cos \phi)$.

In Young's double slit experiment,the amplitudes of the two waves incident on the two slits are $A$ and $2A$. If $I_{0}$ is the maximum intensity,then the intensity at a spot on the screen,where the phase difference between the two interfering waves is $\phi$,is:

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