A thin glass plate of thickness t = frac{2500 | vedclass

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(D) Before the glass plate is inserted,the point equidistant from the slits has a path difference of $0$. Thus,the phase difference is $0$,and the int

Vedclass · Accepted Answer

$4:3$

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(D) Before the glass plate is inserted,the point equidistant from the slits has a path difference of $0$. Thus,the phase difference is $0$,and the intensity is $I_{max} = 4I_0$ (where $I_0$ is the intensity of each slit).After the glass plate is inserted,the path difference introduced is $\Delta x = (\mu - 1)t$.Substituting the given values: $\Delta x = (1.5 - 1) 	imes \frac{2500}{3} \lambda = 0.5 	imes \frac{2500}{3} \lambda = \frac{1}{2} 	imes \frac{2500}{3} \lambda = \frac{1250}{3} \lambda$.The phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} 	imes \frac{1250}{3} \lambda = \frac{2500\pi}{3} = 833\pi + \frac{\pi}{3} = 833\pi + 60^\circ$.Since $833\pi$ is an odd multiple of $\pi$,the effective phase difference is $\pi + 60^\circ$ or simply $60^\circ$ relative to the interference condition.The intensity is given by $I = 4I_0 \cos^2(\phi/2)$.$I = 4I_0 \cos^2(60^\circ / 2) = 4I_0 \cos^2(30^\circ) = 4I_0 (\sqrt{3}/2)^2 = 4I_0 	imes \frac{3}{4} = 3I_0$.The ratio of intensities before and after is $\frac{4I_0}{3I_0} = 4:3$.

Similar Questions

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