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If one of two identical slits producing interference in Young's experiment is covered with glass, so that the light intensity passing through it is reduced 50 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern. (b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light ? |
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Answer» Solution :(a) Let INTENSITY of light passing through first slit be `I_(1)=I`. So the intensity of light passing through second slit covered with glass `I_(2)=50%` of I=0.5I. `therefore`MAXIMUM intensity of interference PATTERN `I_(max)=I_(1)+I_(2)+2sqrt(I_(1)I_(2))=I+0.5I+2sqrt(I*(0.5I))=(1.5+sqrt(2))I` and minimum intensity of interference pattern `I_("min")=I_(1)+I_(2)-2sqrt(I_(1)I_(2))=I+0.5I-2sqrt(I*(0.5I))=(1.5-sqrt(2))I` `implies(I_(max))/(I_("min"))=(1.5+sqrt(2))/(1.5-sqrt(2))=(1.5+1.414)/(1.5-1.414)=(2.914)/(0.086)=33.9:1` (B) N/A. |
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