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Question

(A) The intensity $I$ at a point on the screen in a Young's double slit experiment is given by the formula $I = I_0 \cos^2 \left( \frac{\phi}{2} \righ

Vedclass · Accepted Answer

$\frac{3}{4}$

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(A) The intensity $I$ at a point on the screen in a Young's double slit experiment is given by the formula $I = I_0 \cos^2 \left( \frac{\phi}{2} ight)$,where $I_0$ is the maximum intensity and $\phi$ is the phase difference.The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.Given the path difference $\Delta x = \frac{\lambda}{6}$,we calculate the phase difference:$\phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$.Now,substitute $\phi$ into the intensity formula:$\frac{I}{I_0} = \cos^2 \left( \frac{\pi/3}{2} ight) = \cos^2 \left( \frac{\pi}{6} ight)$.Since $\cos \left( \frac{\pi}{6} ight) = \frac{\sqrt{3}}{2}$,we have:$\frac{I}{I_0} = \left( \frac{\sqrt{3}}{2} ight)^2 = \frac{3}{4}$.

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of light used) is $I$. If $I_0$ denotes the maximum intensity,then $\frac{I}{I_0} = $ . . . . . .

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