Undergrad Width of Spectrum produced by a prism?

[itsnaresh]

How to calculate the width of spectrum produced by a dispersive prism of a given apex angle and refractive index, of white light of given width incident on first face of prism?

 

[blue_leaf77]

If by spectrum you mean the frequency content of the light, then the spectrum width after passing through the prism is the same as that before entering it, assuming the propagation inside the prism is of linear nature. On the other hand, if you mean the width of the visible color spread, which is not the same as the spectrum defined previously, then you also need to specify the distance from prism back surface to the observation plane and the tilt of this plane. Rather than going through the trouble of specifying these parameters, it's much simpler to calculate the angular spread instead.

 

[itsnaresh]

Thank you very much for your reply. I am attaching below two pictures. The top picture shows the spectrum produced on a screen by a prism when white light is incident on its first face. Spectrum produced on the screen has distinct widths of various colors from red to blue. I want to know how we can calculate these widths of various colors when a collimated beam of white light is incident. The second pictures shows the layout of the system. I can work out the dispersion for various colours having different colours for a single ray. But I can not work out for the beam. Actually I want to design a wavelength scale for this system. The parameters Fo, D and X can be fixed based on the calculations.

 

[Khashishi]

You list a single refractive index, but of course the refractive index depends on the wavelength. The diffraction angle is just given by Snell's law. $n_1 \sin \theta_1 = n_2 \sin \theta_2$.

The Abbe number approximately characterizes the dispersion with just a single number.
Looking at this https://en.wikipedia.org/wiki/Abbe_number#/media/File:Abbe-diagram_2.svg , it looks like the Abbe number for SF1 is around 29.5.
Wikipedia says
$V_D = \frac{n_D-1}{n_F-n_C}$
where nD, nF and nC are the refractive indices of the material at the wavelengths of the Fraunhofer D-, F- and C- spectral lines (589.3 nm, 486.1 nm and 656.3 nm respectively)
So, that means
$29.5 \approx \frac{1.717295-1}{n(486.1)-n(656.3)}$
$n(486.1)-n(656.3) \approx 0.024315$
$\frac{dn}{d\lambda} \approx -1.4286 \times 10^{-4} \mathrm{cm}^{-1}$
Just plug that into Snell's law based on the geometry of your prism.