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What change is observed in interference pattern of Young's double slit experiment , if one of the two slits is painted. So that it transmits half the light intensity of the other? |
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Answer» Solution :When the slit is not painted `a_1^2=a_2^2=1` `therefore I_max=(sqrtI+sqrtI)^2 =4I`and ` I_min=0`. When the slit is painted `a_1^2=2,a^2_2=I/2` `therefore I_max=(sqrtI+sqrtI/2)^2=2.915 ` I I_min =(sqrtI-sqrtI/2)^2=0.086 I. THUS , upon painting the maximum intensity DECREASES and minimum intensity increases. HENCE , there is decrease in CONTRAST. |
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