If one of the two slits of a Young's double slit experiment is painted over so that it transmits half the light intensity of the other, then:

A. the fringe system would disappear

B. the bright fringes would be brighter & dark fringes would be darker

C. the dark fringes would be brighter and bright fringes would be darker

D. bright as well as dark fringes would be darker

Answers (1)

Case 1: When slit are of equal width, if $I_{0}$ is intensity of light then,
$\mathrm{I}_{\max }=\mathrm{I}_{\mathrm{o}}+\mathrm{I}_{\mathrm{o}}+2 \sqrt{\mathrm{I}_{\mathrm{o}} \mathrm{I}_{\mathrm{o}}}=4 \mathrm{I}_{\mathrm{o}}$ \\ $I_{\min }=I_{o}+I_{o}-2 \sqrt{I_{o} I_{o}}=0$ 
Case 2: When one of the slit is painted over, its intensity would reduce to half then,

$\mathrm{I}_{\mathrm{newmax}}=\mathrm{I}_{\mathrm{o}} / 2+\mathrm{I}_{\mathrm{o}}+2 \sqrt{\mathrm{I}_{\mathrm{o}} \mathrm{I}_{\mathrm{o}} / 2}=2.932 \mathrm{I}_{\mathrm{o}}$ 
$\mathrm{I}_{\mathrm{newmin}}=\mathrm{I}_{\mathrm{o}} / 2+\mathrm{I}_{\mathrm{o}}-2 \sqrt{\mathrm{I}_{\mathrm{o}} \mathrm{I}_{\mathrm{o}} / 2}=0.068 \mathrm{I}_{\mathrm{o}}$
Therefore, dark fringes get brighter and bright fringes get darker

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