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The following text is from Concepts of Physics by Dr. H.C.Verma, from the chapter "Dispersion and Spectra", page 434, topic "Dispersive Power":

The dispersive power of a material is defined as the ratio of angular dispersion to the average deviation when a light beam is transmitted through a thin prism placed in a position so that the mean ray (ray having the mean wavelength) passes symmetrically through it.

After the definition of dispersive power, the author has derived the following expression based on the assumptions - the angle of prism is small and so the angle of deviation is small:

$$\omega=\frac{\mu_v-\mu_r}{\mu_y-1}\tag{20.1}$$

This equation itself may be taken as the definition of dispersive power.

In the above equation, $\omega$ is the dispersive power, $\mu_v,\mu_r$, and $\mu_y$ are the refractive indices of violet, red and yellow components of light respectively.

The equation for dispersive power was derived for a specific case - a thin prism (a prism with a small refracting angle). Then how could it be "taken as the definition of dispersive power"? This statement from the book seems to imply that it must also be applicable for dispersing elements other than "thin" prisms. Is the equation really valid for other dispersing elements like a prism of large refracting angle, or a glass sphere, or a grating?

Note: I haven't included the complete derivation of the formula as I thought it will increase the size of the post tremendously. However, I hope I have explained the main point clearly. If not, kindly ask your queries in the comments.

The Wikipedia article on Dispersion doesn't offer any explanation regarding dispersive power.

My search results on dispersing power didn't fetch any thing from reliable sources. So I decided to ask it here.

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    $\begingroup$ Dispersive power is used to study dispersion of visible light in optical instruments. The lack of sources indicate that this is a quantity that isn’t used nowadays. That being said, the above derived definition is a good one because it purely depends on the material and not its geometry. And you’d want to find the maximum deviation between the highest and lowest wavelengths that you are interested in normalised by the mean deviation (this makes it more reliable to compare between materials). $\endgroup$
    – Superfast Jellyfish
    Commented Jan 24, 2020 at 9:58
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    $\begingroup$ @user3518839: Thank you for your comment. I agree that the verbal form of the definition is clear. But, I have a doubt regarding the statement "...it purely depends on the material and not its geometry" from your reply regarding the mathematical form. I can see that the angular dispersion depends only on the indices of refraction of reference wavelengths, but it was derived for a specific case where terms referring to geometric parameters were eliminated while doing approximations based on the assumptions. $\endgroup$
    – Vishnu
    Commented Jan 24, 2020 at 10:30
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    $\begingroup$ Yes. The particular form was derived using a prism, but the final form is independent of geometry. Thus this formula is a good candidate for encapsulating the information about dispersion. Remember that we are at a liberty to define what dispersive power is. $\endgroup$
    – Superfast Jellyfish
    Commented Jan 24, 2020 at 10:36
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    $\begingroup$ @user3518839: Thank you again. That certainly helped. I thought "dispersive power" is something like "optical power" which depends on the device for which it's measured. Rather, it seems like "dispersive power" is the property of the material which is independent of the geometry. Why not write your answer? I could accept and vote it :) $\endgroup$
    – Vishnu
    Commented Jan 24, 2020 at 11:36

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Dispersive power is used to study dispersion of visible light in optical instruments. The lack of sources indicate that this is a quantity that isn’t used nowadays.

That being said, the above derived definition is a good one because it purely depends on the material and not its geometry. And you’d want to find the maximum deviation between the highest and lowest wavelengths that you are interested in normalised by the mean deviation (this makes it more reliable to compare between materials).

Improve this answer
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