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Suppose to observe spectral lines of a mercury or sodium lamp with a prism spectrometer.

If there is a slit in front of the source to make the light collimated, and if regulate that slit in width, do I see the spectral lines change in their width?

In other words, do spectral lines width depend on the intensity ($W/m^2$) of light?

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No, the width of the lines does not depend on the intensity of the light, it depends on the width of the slit! You could replace the lamp with a more powerful version and the width of the lines would be unchanged.

When you form a spectrum, what you are (crudely) doing is forming an image of the slit at each wavelength; so the width of the lines corresponds to the width of the slit.

Actually if you think about it, in the experiment you have described, the intensity of the light is not changed at all.

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  • $\begingroup$ It has more to do with the frequency of the photons ( wavelength of light) and the distance to the screen from the edge of the slit. The spread will be wider as the screen moves further away. If the light source is missing photons of certain frequencies then there will be gaps or black lines at those locations along the spectrum. From the edge of the slit you can calculate the position that a color or (frequency/wavelength) will be in the spectrum like this: Y= The square root of (The wavelength times the distance to the screen times 3/4) $\endgroup$ May 17, 2017 at 5:18
  • $\begingroup$ Depending on the detection technology used there is some risk of an intensity dependent detector systematic. Consider a photographic emulsion: there is a certain intensity that will fully expose the film, and for a brighter source that threshold is reached further and further out the tails of the light distribution. Of course, a well designed with a well understood apparatus will by operated to minimize this effect and with good enough data taking it can be removed in analysis if needed. But the potential is there as a confounding factor. $\endgroup$ Jul 2, 2018 at 23:37

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