In Young's double-slit experiment,if one slit | vedclass

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(D) In Young's double-slit experiment,the intensity $I$ is proportional to the width of the slit $w$ $(I \propto w)$.If the width of one slit is $w$ a

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The intensity of bright fringes will decrease and the intensity of dark fringes will become non-zero.

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(D) In Young's double-slit experiment,the intensity $I$ is proportional to the width of the slit $w$ $(I \propto w)$.If the width of one slit is $w$ and the other is $2w$,their amplitudes $a_1$ and $a_2$ are related by $a_2 = \sqrt{2} a_1$.The maximum intensity is $I_{max} \propto (a_1 + a_2)^2$ and the minimum intensity is $I_{min} \propto (a_2 - a_1)^2$.Since $a_1 
eq a_2$,the minimum intensity $I_{min}$ will not be zero; it will be non-zero.Also,the maximum intensity $I_{max}$ will be higher than the case where widths were equal.Therefore,the intensity of bright fringes increases and the intensity of dark fringes becomes non-zero.

In Young's double-slit experiment,if one slit is made twice as wide as the other instead of having equal widths,then in the interference pattern:

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