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Question

(A) In a Young's double slit experiment,the resultant intensity $I$ at any point on the screen is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$

Vedclass · Accepted Answer

$0, 4 I_0, 2 I_0$

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(A) In a Young's double slit experiment,the resultant intensity $I$ at any point on the screen is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $I_1 = I_2 = I_0$.Thus,$I = 2 I_0 + 2 I_0 \cos \phi = 4 I_0 \cos^2(\phi / 2)$.$1$. The minimum intensity occurs when $\cos^2(\phi / 2) = 0$,so $I_{\min} = 0$.$2$. The maximum intensity occurs when $\cos^2(\phi / 2) = 1$,so $I_{\max} = 4 I_0$.$3$. The average intensity $I_{	ext{avg}}$ over the entire screen is the average of $I_{\max}$ and $I_{\min}$ for a sinusoidal interference pattern,which is $I_{	ext{avg}} = \frac{I_{\max} + I_{\min}}{2} = \frac{4 I_0 + 0}{2} = 2 I_0$.Therefore,the values are $0, 4 I_0, 2 I_0$.

In a Young's double slit experiment,the intensity of light at each slit is $I_0$. An interference pattern is observed along a direction parallel to the line $S_1 S_2$ on screen $S$. Find the minimum,maximum,and the intensity averaged over the entire screen,respectively.

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