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The wavelenght of a light is 450 nm. Howphaseit will differ for a path of 3 mm? |
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Answer» Solution :The wavelenght is, `lambda = 450 NM- 450 XX 10^(2)m` PATH difference is,`delta = 3 nm = 3 xx 10^(-3)m` Relation between phase difference and path difference is, `phi = (2pi)/(lambda) xx delta` Substituting,`phi=(2pi)/(450xx10^(-9))xx3xx10^(-3)=(pi)/(75)xx10^(6)` `phi=(pi)/(75)xx10^(6)rad` |
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