Imagine a collimated beam of white light falling on one refracting face of a prism. Let the light emerging from the second face be focused by a lens onto a screen. Suppose there is linear dispersion at the screen. Now by introducing a narrow "exit slit" (on the order of 0.01cm) in the screen, one has a type of monochromator that provides a nearly monochromatic beam of light. The white light has a mean wavelength of 500nm.


What is meant by linear dispersion at the screen. Furthermore how does the slit effect the coherence length and coherence time of the light.


The beam of white light is refracted by the prism, but the different wavelengths in the light are refracted by different angles. That means when the light falls on the screen the different wavelengths are spread out over a region of the screen. This is what is meant by dispersion.


The dispersion is linear if the different wavelengths of light are spread out evenly across the screen. So the wavelength at the point I've shown as $d$ would be given by:

$$ \lambda_d = \lambda_{red} - \frac{d \left(\lambda_{red} - \lambda_{blue}\right)}{L} $$

In general the dispersion won't be linear, but for glass across the range of red to blue light it's not a bad approximation. The reason this matters is when you try to calculate the coherence length from the size of the slit. The coherence length is given by:

$$ L = \frac{2\space ln(2)}{\pi n} \frac{\lambda^2}{\Delta \lambda} $$

Because the slit has a finite width the image is casts on the screen has a finite width and therefore covers a range of frequencies. Going back to my diagram above, suppose the image of the slit on the screen has a width of $w$, then if the dispersion can be assumed linear the range of frequencies is simply:

$$ \Delta \lambda = \frac{w}{L} \left(\lambda_{red} - \lambda_{blue} \right) $$

Put this value for $\Delta \lambda$ back into the equation for the coherence length to get the effect of the slit width on coherence length.


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