Consider an YDSE that has different slit widt | vedclass

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Question

(C) The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.Given amplitudes are $A$ and $2A$,the intensities are $I_1 = I$

Vedclass · Accepted Answer

$\frac{I_0}{9} [5 + 4 \cos \phi]$

Vedclass · Answer

(C) The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.Given amplitudes are $A$ and $2A$,the intensities are $I_1 = I$ and $I_2 = 4I$.The maximum intensity $I_0$ is given by $I_0 = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.Thus,$I = \frac{I_0}{9}$.The resultant intensity $I'$ at a point with phase difference $\phi$ is given by $I' = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Substituting the values: $I' = I + 4I + 2\sqrt{I \cdot 4I} \cos \phi = 5I + 4I \cos \phi$.Substituting $I = \frac{I_0}{9}$,we get $I' = \frac{I_0}{9} (5 + 4 \cos \phi)$.

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