InterviewSolution
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InterviewSolution
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A parallel beam of light of intensity I_0 is incident an a glass plate, 25% of light is reflected by upper surface and 50% of light is reflected by lower surface. The ratio of maximum to minimum intensity is interference region of reflected rays is (Assume that cohernt light reflected from lower surface I refects from upper surface it entirely come out). |
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Answer» Solution :`I_(1)= IXX (25)/(100)= (I)/(4)` The intensity of transmitted LIGHT is `I.= I-(I)/(4)= (3I)/(4)` The intensity of light reflected fro lower surface is `I_(2)= (3I)/(4)xx(50)/(100)= (3I)/(8)` `(I"MAX")/(I"min")= ((SQRT(I_1)+sqrt(I_2))/(sqrt(I_1)+sqrt(I_2)))^(2)= ((sqrt(I/4)+sqrt((3I)/(8)))^2)/((sqrt(I/4)-sqrt((3I)/(8)))^2)=(((1)/(2)+sqrt((3)/(8)))^2)/((1)/(2)-sqrt((3)/(8)))^(2)`. 
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