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(B) The intensity $I$ at any point on the screen is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.Phase difference $\phi$

Vedclass · Accepted Answer

$K/2$

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(B) The intensity $I$ at any point on the screen is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.For path difference $\Delta x = \lambda$,the phase difference is $\phi_1 = \frac{2\pi}{\lambda} 	imes \lambda = 2\pi$.The intensity at this point is $K = I_{max} \cos^2(2\pi/2) = I_{max} \cos^2(\pi) = I_{max}(1)^2 = I_{max}$.For path difference $\Delta x = \lambda/4$,the phase difference is $\phi_2 = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{4} = \frac{\pi}{2}$.The intensity at this point is $I_2 = I_{max} \cos^2(\phi_2/2) = I_{max} \cos^2(\pi/4)$.Since $I_{max} = K$ and $\cos(\pi/4) = 1/\sqrt{2}$,we have $I_2 = K 	imes (1/\sqrt{2})^2 = K/2$.

Column-$I$	Column-$II$
$(A) x = \frac{D \lambda}{d}$	$(P) I_0$
$(B) x = \frac{D \lambda}{4d}$	$(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$	$(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$	$(S) 4 I_0$

In Young's double slit experiment,the intensity on the screen at a point where path difference is $\lambda$ is $K.$ What will be the intensity at the point where path difference is $\lambda /4$?

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