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(C) Given that the amplitude $A$ is proportional to the slit width $w$,we have $A \propto w$.Let the widths be $w_1$ and $w_2$,where $w_2 = 3w_1$. The

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$4: 1$

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(C) Given that the amplitude $A$ is proportional to the slit width $w$,we have $A \propto w$.Let the widths be $w_1$ and $w_2$,where $w_2 = 3w_1$. Therefore,the amplitudes are $A_1$ and $A_2 = 3A_1$.The intensity $I$ is proportional to the square of the amplitude,$I \propto A^2$.The ratio of maximum to minimum intensity is given by the formula:$\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + A_2}{|A_1 - A_2|}\right)^2$Substituting $A_2 = 3A_1$ into the equation:$\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + 3A_1}{|A_1 - 3A_1|}\right)^2$$= \left(\frac{4A_1}{2A_1}\right)^2$$= (2)^2 = 4$Thus,the ratio is $4: 1$.

In a Young's double slit experiment,the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

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