A Young's double-slit experiment is performed | vedclass

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(A) The intensity at any point in a Young's double-slit experiment is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase d

Vedclass · Accepted Answer

$9$

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(A) The intensity at any point in a Young's double-slit experiment is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.Given that the intensity at a path difference of $\lambda$ is $K$,and since the path difference $\lambda$ corresponds to a phase difference of $2\pi$,the intensity is $I_{max} = 4I_0 = K$,where $I_0$ is the intensity of each slit.Thus,$I_0 = K/4$.For a path difference $\Delta x = \frac{\lambda}{6}$,the phase difference $\phi$ is $\frac{2\pi}{\lambda} 	imes \Delta x = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$.The intensity at this point is $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(\frac{\pi}{3}) = 2I_0 + 2I_0 \cos(\frac{\pi}{3})$.Substituting $I_0 = K/4$ and $\cos(\frac{\pi}{3}) = 1/2$,we get $I = 2(K/4) + 2(K/4)(1/2) = K/2 + K/4 = 3K/4$.We are given $I = \frac{nK}{12}$,so $\frac{3K}{4} = \frac{nK}{12}$.Solving for $n$,we get $n = \frac{3 	imes 12}{4} = 9$.

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