Question

The variation of refractive index, n, with wavelength for fused silica is given by

$$\begin{gathered} {\mathbf{n}}^{\mathbf{2}}\mathbf{‐}\mathbf{1}\mathbf{=}\frac{\left(0.6961663\right){\mathbf{\lambda }}^{\mathbf{2}}}{{\mathbf{\lambda }}^{\mathbf{2}}\mathbf{‐}{\left(0.0684043\right)}^{\mathbf{2}}}\mathbf{+}\frac{\left(0.4079426\right){\mathbf{\lambda }}^{\mathbf{2}}}{{\mathbf{\lambda }}^{\mathbf{2}}\mathbf{‐}{\left(0.1162414\right)}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\frac{\left(0.8974794\right){\mathbf{\lambda }}^{\mathbf{2}}}{{\mathbf{\lambda }}^{\mathbf{2}}\mathbf{‐}{\left(9.896161\right)}^{\mathbf{2}}} \end{gathered}$$

where $\mathbf{\lambda }$ is expressed in $\mathbf{\mu m}$.

(a) Make a graph of n versus $\mathbf{\lambda }$ with points at the following wavelengths: 0.2,0.4,0.6,0.8,1,2,3,4,5, and $\mathbf{6}\mathbf{\mu m}$.

(b) The ability of a prism to spread apart (disperse) neighboring wavelengths increases as the slope $\mathbf{dn}\mathbf{/}\mathbf{d\lambda }$increases. Is the dispersion of fused silica greater for blue light or red light?

Short answer

(a) The wavelengths of 0.2,0.4,0.6,0.8,1,2,3,4,5, and $6\mathrm{\mu m}$ is

(b) The dispersion of fused silica is greater for blue light ( 400 nm ) than red light ( 600 nm ).

Step-by-step solution

Definition of wavelenght

• The wavelength of a waveform signal conveyed in space or down a wire is the distance between identical points (adjacent crests) in adjacent cycles.

Determine the following wavelengths: 0.2,0.4,0.6,0.8,1,2,3,4,5, and 6μm.

A graph depicting the relationship between n and $\lambda$It must be marked.

The graph was created using the formula provided.

Wavelength$=\frac{{\mathrm{\lambda }}_0}{\mathrm{n}}$

${\mathrm{\lambda }}_0$is the wavelength in vacuum

n is the refraction inden

(a)

For blue light or red light, the dispersion of fused silica is larger.

Determine the dispersion of fused silica greater for blue light or red light

(b)

For blue light or red light, the dispersion of fused silica is larger.

$\frac{\mathrm{dn}}{\mathrm{d\lambda }}$a wavelength of blue light$\left(~400\mathrm{nm}\right)$ then the red light comes on $\left(~600\mathrm{nm}\right)$.