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(A) Case $1$: When slits are of equal width,if $I_0$ is the intensity of light from each slit,then:$I_{\max} = I_0 + I_0 + 2\sqrt{I_0 I_0} = 4I_0$$I_{

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the intensities of both the maxima and minima increase.

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(A) Case $1$: When slits are of equal width,if $I_0$ is the intensity of light from each slit,then:$I_{\max} = I_0 + I_0 + 2\sqrt{I_0 I_0} = 4I_0$$I_{\min} = I_0 + I_0 - 2\sqrt{I_0 I_0} = 0$Case $2$: When one slit is twice as wide as the other,the intensity from that slit becomes $2I_0$. The intensities are now $I_1 = 2I_0$ and $I_2 = I_0$.$I_{	ext{newmax}} = I_1 + I_2 + 2\sqrt{I_1 I_2} = 2I_0 + I_0 + 2\sqrt{2I_0^2} = 3I_0 + 2\sqrt{2}I_0 \approx 5.83I_0$$I_{	ext{newmin}} = I_1 + I_2 - 2\sqrt{I_1 I_2} = 2I_0 + I_0 - 2\sqrt{2I_0^2} = 3I_0 - 2\sqrt{2}I_0 \approx 0.17I_0$Comparing the two cases,the intensity of the maxima increases (from $4I_0$ to $5.83I_0$) and the intensity of the minima also increases (from $0$ to $0.17I_0$).

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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