The width of one of the two slits in a Young' | vedclass

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(A) Given that the amplitude $A$ is proportional to the slit width $w$,so $A \propto w$.Since intensity $I$ is proportional to the square of the ampli

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(A) Given that the amplitude $A$ is proportional to the slit width $w$,so $A \propto w$.Since intensity $I$ is proportional to the square of the amplitude,$I \propto A^2 \propto w^2$.Let the widths be $w_1 = 3w_0$ and $w_2 = w_0$. Then the intensities are $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \left(\frac{w_1}{w_2}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$.Thus,$I_1 = 9I_2$. Let $I_2 = I$,then $I_1 = 9I$.The ratio of minimum to maximum intensity is given by $\frac{I_{\min}}{I_{\max}} = \left(\frac{\sqrt{I_1} - \sqrt{I_2}}{\sqrt{I_1} + \sqrt{I_2}}\right)^2$.Substituting the values: $\frac{I_{\min}}{I_{\max}} = \left(\frac{\sqrt{9I} - \sqrt{I}}{\sqrt{9I} + \sqrt{I}}\right)^2 = \left(\frac{3\sqrt{I} - \sqrt{I}}{3\sqrt{I} + \sqrt{I}}\right)^2 = \left(\frac{2\sqrt{I}}{4\sqrt{I}}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.Given the ratio is $x:4$,we have $\frac{x}{4} = \frac{1}{4}$,which implies $x = 1$.

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width,the ratio of minimum to maximum intensity in the interference pattern is $x: 4$ where $x$ is ..... .

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