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Discuss the pattern of interference fringes obtained on the screen away from the two point sources. |
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Answer» Solution :For TWO waves, ratio of their intensities, `(I_(1))/(I_(2))=alpha` (given) But we know that `IpropA^(2)`, where A is an AMPLITUDE. `:.(I_(1))/(I_(2))=(A_(1)^(2))/(A_(2)^(2))=alpha` `:.(A_(1))/(A_(2))=(SQRT(alpha))/(l)` Taking componendo and dividendo, `:.(A_(1)+A_(2))/(A_(1)-A_(2))=(A_(max))/(A_(min))=(sqrtalpha+1)/(sqrtalpha-1)` `:.(I_(max))/(I_(min))=(A_(max)^(2))/(A_(min))=(sqrtalpha+1)^(2)/((sqrtalpha-1)^(2))=(alpha+2sqrtalpha+1)/(alpha-2sqrt+1)` `(I_(max)+I_(min))/(I_(max)-I_(min))=((alpha+2sqrtalpha+1)+(alpha-2sqrtalpha+1))/((alpha+2sqrtalpha+1)-(alpha-2sqrtalpha+1))` `=(alpha+1)/(2sqrtalpha)` Note : Reciprocal of the above term, i.e. `(I_(max)-I_(min))/(I_(max)+I_(min))` is KNOWN as visibility of fringes. |
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