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(A) In a standard Young's Double Slit Experiment $(YDSE)$ with equal slit widths, the amplitudes of the waves from both slits are equal, say $a$. The

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The intensities of both the maxima and the minima increase

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(A) In a standard Young's Double Slit Experiment $(YDSE)$ with equal slit widths, the amplitudes of the waves from both slits are equal, say $a$. The maximum intensity is $I_{max} \propto (a + a)^2 = 4a^2$ and the minimum intensity is $I_{min} \propto (a - a)^2 = 0$.When one slit is made twice as wide as the other, the intensity of the light passing through it increases. Since intensity $I \propto 	ext{width}$, if the width of the first slit is $w$, the width of the second is $2w$. The amplitudes are related by $A_2 = \sqrt{2} A_1$. Let $A_1 = a$, then $A_2 = a\sqrt{2}$.The new maximum intensity is $I'_{max} \propto (a + a\sqrt{2})^2 = a^2(1 + \sqrt{2})^2 \approx 5.83a^2$, which is greater than $4a^2$.The new minimum intensity is $I'_{min} \propto (a\sqrt{2} - a)^2 = a^2(\sqrt{2} - 1)^2 \approx 0.17a^2$, which is greater than $0$.Thus, the intensities of both the maxima and the minima increase.

In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern

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