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(B) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.For a path differen

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$k/2$

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(B) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.For a path difference $\Delta x_1 = \lambda$,the phase difference is $\phi_1 = \frac{2\pi}{\lambda} 	imes \lambda = 2\pi$.The intensity is given by $I = I_{max} \cos^2(\phi/2)$.For $\phi_1 = 2\pi$,$I_1 = I_{max} \cos^2(\pi) = I_{max} = k$.For a path difference $\Delta x_2 = \lambda/4$,the phase difference is $\phi_2 = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{4} = \frac{\pi}{2}$.The intensity at this point is $I_2 = I_{max} \cos^2(\phi_2/2) = k \cos^2(\pi/4)$.Since $\cos(\pi/4) = 1/\sqrt{2}$,we have $I_2 = k 	imes (1/\sqrt{2})^2 = k/2$.

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $k$. What will be the intensity at a point where the path difference is $\lambda/4$? ($\lambda$ = wavelength of light)

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