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(A) The intensity $I$ at any point in an interference pattern is given by $I = I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference.Given

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$\frac{I_0}{4}$

Vedclass · Answer

(A) The intensity $I$ at any point in an interference pattern is given by $I = I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference.Given the path difference $\Delta x = \frac{\lambda}{3}$.The phase difference $\phi$ is related to the path difference by $\phi = \frac{2\pi}{\lambda} \Delta x$.Substituting the value of $\Delta x$: $\phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{3} = \frac{2\pi}{3}$.Now, calculate the intensity: $I = I_0 \cos^2(\frac{2\pi/3}{2}) = I_0 \cos^2(\frac{\pi}{3})$.Since $\cos(\frac{\pi}{3}) = \cos(60^{\circ}) = 0.5 = \frac{1}{2}$, we have $I = I_0 (\frac{1}{2})^2 = \frac{I_0}{4}$.

In a biprism experiment, the maximum intensity is $I_0$. If the path difference between the two interfering waves is $\frac{\lambda}{3}$, then the intensity at the point on the screen is [$\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \sin 60^{\circ} = \cos 30^{\circ} = \sqrt{3}/2$].

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