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If one of the slits of a standard YDSE apparatus is covered by a thin glass slab so that it allows only one half of the light intensity of the other slit, then

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the resultant intensity at dark fringe will increase

the resultant intensity at central bright will decrease

the fringe width will remain unchanged.

the fringe pattern will shift toward the covered slit.

answer is A, B, C, D.

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Detailed Solution

$I_{\max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2},$ so on decreasing $I_{1}, I_{\max}$ decreases $I_{\min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} .$ So on decreasing $I_{1}, I_{\min}$ increases

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