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The following text is from Concepts of Physics by Dr. H.C.Verma, from the chapter "Light Waves", page 370, topic "17.9 Coherent and Incoherent Sources":

Because of the incoherent nature of the basic process of light emission in ordinary sources, these sources cannot emit highly monochromatic light. A strictly monochromatic light, having well-defined single frequency or wavelength, must be a sine wave which has an infinite expansion. A wave train of finite length may be described by the superposition of a number of sine waves of different wavelengths. [...] Shorter the length of wave train, larger is the spread in wavelength.

I have resolved my questions along with the associated text:

Question 1:

Because of the incoherent nature of the basic process of light emission in ordinary sources, these sources cannot emit highly monochromatic light.

How does the incoherent nature of the light source affect the spread in wavelength (inability to produce highly monochromatic light)? I think incoherence/coherence is just related to the phase difference and not wavelength. So this statement is confusing for me.

Question 2:

A wave train of finite length may be described by the superposition of a number of sine waves of different wavelengths.

Why is it not possible to describe a wave train of finite length by a single wave instead of superposition of multiple waves?

Question 3:

Shorter the length of wave train, larger is the spread in wavelength.

In another paragraph, the author says lasers emit very long wave trains of the order of several hundred metres and this is why the spread in wavelength is small. Why should the length of the wave train affect the spread in wavelength? Aren't the length of the train and its wavelength independent of each other?


Related: Why, in order to obtain distinct interference, is a small distance between the two waves essential? (based on text from the same topic)

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How does the incoherent nature of the light source affect the spread in wavelength (inability to produce highly monochromatic light)?

Coherence means constant phase difference between different parts of a wave. If a wave is incoherent, it can be described as a superposition of short wave packets, each with a random shift of phase and position of the amplitude peak with respect to the others. But since these packets are shorter than the whole wave emitted, this broadens the spectrum.

Why is it not possible to describe a wave train of finite length by a single wave instead of superposition of multiple waves?

Note that the author says "a number of sine(!) waves", not simply waves. And sine waves have, by definition, infinite extent. So to be able to get a shorter wave packet, you need to have a superposition of these sines.

Aren't the length of the train and its wavelength independent of each other?

There's a definite uncertainty relation between frequency and time (and similarly, length of wave packet and wavenumber (which is directly related to wavelength)), known as Gabor limit, so they are not independent.

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  • $\begingroup$ Thank you for your answer. I having a doubt regarding this statement - "So to be able to get a shorter wave packet, you need to have a superposition of these sines." If each sine wave is of infinite length, then how can we obtain a wave packet (of finite length) as a result of superposition of infinitely long sine waves? $\endgroup$ – Guru Vishnu Dec 23 '19 at 13:58
  • $\begingroup$ @M.GuruVishnu it's a tricky one. If you add countable infinity of such waves, then indeed, you'll only get (quasi-)periodic beats instead of a single wave packet. But as you make the distance between frequencies of each of the waves in the series smaller, the distance in time between these beats will increase. In the limit you'll go from discrete Fourier series to the continuous Fourier transform, and your waveform will be a single wave packet. $\endgroup$ – Ruslan Dec 23 '19 at 14:01
  • $\begingroup$ Thanks. But, I didn't understand that properly as I'm unaware of some concepts mentioned in your comment (No problem). However, could you please explain this statement from the explanation for the first question - "But since these packets are shorter than the whole wave emitted, this broadens the spectrum."? $\endgroup$ – Guru Vishnu Dec 23 '19 at 14:06
  • $\begingroup$ Suppose you have a long wave packet. It may have a nice envelope shape, but if you look at the details of its waveform, it'll be not as nice, periodically crossing zero, but will have random shifts of these zero crossings. And consider now another wave packet with the same envelope, but which has regular pattern of crossing zeros. The former has a wider spectrum, i.e. has more non-negligible side frequencies, while the latter is a "clearer" wave packet, which has a more well-defined frequency. The former one can be viewed as a superposition of smaller packets, with random shifts between them. $\endgroup$ – Ruslan Dec 23 '19 at 14:15
  • $\begingroup$ In this view you have a set of nice packets, but since they are shorter than the total wave packet, their spectra are wider. Then, since they add up incoherently (with random phase shifts), so will be the spectrum of the total wave packet. $\endgroup$ – Ruslan Dec 23 '19 at 14:16

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