Question

A 10.0-cm interference wedge is to be built that has a linear dispersion from 400 to 700 nm. Describe details of its construction. Assume that a dielectric with a refractive index of 1.32 is to be used.

Short answer

The thickness of interference wedge at $400\mathrm{nm}=0.152\mathrm{\mu m}$.

The thickness of interference wedge at $700\mathrm{nm}=0.265\mathrm{\mu m}$.

Step-by-step solution

Step 1:  Given information

$10.0\mathrm{cm}$interference wedge with linear dispersion from $400-700\mathrm{nm}$.

The refractive index of $1.32$.

Relationship between the wavelength and the thickness of the dielectric constant 

An interference wedge is described by the calculation of the thickness of the wedge at both ends. And this calculation is done by using the formula of the relationship between the wavelength of the absorption band and the thickness of the dielectric constant.

This is given as follows:

$$\begin{gathered} \lambda =\frac{2dn}{\mathrm{n}} \\ d=\frac{\lambda \mathrm{n}}{2n}...\left(1\right) \\ \mathrm{Where}, \\ \lambda =\mathrm{wavelength} \\ d=\mathrm{thickness} \\ n=\mathrm{interference}\mathrm{order} \\ \mathrm{n}=\mathrm{refractive}\mathrm{index}\mathrm{of}\mathrm{dielectric}\mathrm{medium} \end{gathered}$$

Calculate thickness at 400 nm

Put the values as $\lambda =400\times {10}^{-9}\mathrm{m},n=1.32\mathrm{and}\mathrm{n}=1$in equation (1) as follows:

$\begin{array}{rcl}d& =& \frac{\lambda \mathrm{n}}{2n}\\ & =& \frac{400\times {10}^{-9}\mathrm{m}\times 1}{2\times 1.32}\\ & =& 0.152\times {10}^{-6}\mathrm{m}\\ & =& 0.152\mathrm{\mu m}\end{array}$

Calculate thickness at 700 nm

Put the values as$\lambda =700\times {10}^{-9}\mathrm{m},n=1.32\mathrm{and}\mathrm{n}=1$ in equation (1) as follows:

$\begin{array}{rcl}d& =& \frac{\lambda \mathrm{n}}{2n}\\ & =& \frac{700\times {10}^{-9}\mathrm{m}\times 1}{2\times 1.32}\\ & =& 0.265\times {10}^{-6}\mathrm{m}\\ & =& 0.265\mathrm{\mu m}\end{array}$

Conclusion

For the given dispersion range $400\mathrm{nm}$and $700\mathrm{nm}$of $10\mathrm{cm}$interference wedge thickness is calculated. For $400\mathrm{nm}$thickness of one and of the wedge is $0.152\mathrm{\mu m}$and for $700\mathrm{nm}$thickness of the other end of the wedge is $0.265\mathrm{\mu m}$. The thickness of both ends of the wedge describes the construction of an interference wedge.