Saved Bookmarks
| 1. |
Two light sources with amplitudes 5 units and 3 units respectively interfere with each other. Calculate the ratio of maximum and minimum intensities. |
|
Answer» Solution :Amplitudes, `a_(1) = 5, a_(2) 3` RESULTANT amplitude, `A=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cosphi)` Resultant amplitude is MAXIMUM when, `phi=0,cos0=1,A_(MAX)=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2))` `A_(min)=sqrt((a_(1)-a_(2))^(2))=sqrt((5-3)^(2))=sqrt((2)^(2))=2units` `1 prop A^(2)` `(I_(max))/(I_(min))=((A_(max))^(2))/((A_(min))^(2))` Substituting `(I_(max))/(I_(min))=((8)^(2))/((2)^(2))=(64)/(4)=16(or)I_(max):I_(min)=16:1` |
|