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Two light sources of equal amplitudes interfere with each other. Calculate the ratio of maximum and minimum intensities. |
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Answer» Solution :LET the amplitude be a. The intensity is, `(I_(max))/(I_(MIN))=((8)^(2))/((2)^(2))=(64)/(4)=16(or)I_(max):I_(min)=16:1` Resultant intensity is maximum when, `phi = 0, COS 0 = I, I_(max) PROP 4a^(2)` Resultant amplitude is minimum when, `phi=pi,cos(pi//2)=0,I_(min)=0` `I_(max):I_(min)=4a^(2):0` |
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