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Two monochromatic light waves of equal intensities produce an interference pattern. At a point in the pattern, the phase difference between the interfering waves is pi//2 rad. Express the intensity at this point as a fraction of the maximum intensity in the pattern. |
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Answer» Solution :Data: `I_(1) = I_(2) = I_(0),PHI = pi//2` rad The resultant intensity at the point in the pattern is `I= I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cosphi` `=I_(0) + I_(0) = 2sqrt(I_(0)I_(0))cosphi` `= 2I_(0)(1+1) = 4I_(0)`...................(i) Also, `1+ cosphi = 2 cos^(2) phi/2`................(2) Also, `1+cosphi = 2 cos^(2) phi/2`.........(3) From Eqs. (1) and (3), `I= 2I_(0)(2cos^(2)phi/2)`[From Eq. (2)] `I=2I_(0) ( 2cos^(2)phi/2)` `=I_("MAX")(cospi/4)^(2) = I_("max")(1/sqrt(2))^(2)` `THEREFORE I= 1/2I_("max")` |
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