Saturday, November 10, 2018

optics - Why, in order to obtain distinct interference, is a small distance between the two waves essential?


This is quoted from Concepts of Physics by H.C.Verma, chapter "Light Waves", page 370, under the topic "17.9 Coherent and Incoherent Sources":



In order to obtain a fairly distinct interference pattern, the path difference between the two waves originating from coherent sources should be kept small. This is so because the wavetrains are finite in length & hence with large difference in path, the waves do not overlap at the same instant in the same region of space. The second wavetrain arrives well after the first train has already passed & hence, no interference takes place.



Why did the author write the wavetrains are finite in length? What is the cause for the finiteness of the wavetrain? Even if it is finite, why doesn't interference occur when the distance between the two waves is large?


Also, in another stanza, he writes:



Because of the incoherent nature of the be=asic process of light emission in ordinary sources, these sources can't emit highly monochromatic light. ... The light emitted by an ordinary source always has a spread in wavelength. An ordinary sodium vapour lamp emits light of wavelength $589.0~\text{nm} \, \& \, 589.6~\text{nm}$ with a spread of about $\pm 0.01~\text{nm}$ in each line.




Now, what does the author mean by "spread in wavelength"? How does it hamper/disturb interference effect?




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