Question

Analysis of an interference effect in clear solid shows that the wavelength of light in the solid is${}^{392}\mathrm{nm}$. Knowing this light comes from a He-Ne laser and has a wavelength ofrole="math" localid="1654057129736" ${}^{633}$nm in air, is the substance zircon or diamond?

Short answer

The refractive index of the material is${}^{1.92}$ equal to the refractive index of zircon.

Step-by-step solution

Refractive index and wavelength

The refractive index of any medium reduces the speed of light as compared to the vacuum.

Given data and formula used

The wavelength of light in the solid is ${}^{329\mathrm{nm}}$.

This light comes from a He-Ne laser and has a wavelength of ${}^{633\mathrm{nm}}$in the air.

We can use the formula to define the refractive index of the material,

$\mathrm{n}=\frac{\mathrm{\lambda }}{{\mathrm{\lambda }}_{\mathrm{n}}}$

Here,${}^{{\mathrm{\lambda and\lambda }}_{\mathrm{n}}}$are the wavelengths of light in the air and medium, respectively.

Determine the refractive index of solid medium

Substituting the value, we get

$$\begin{gathered} \mathrm{n}=\frac{633\mathrm{nm}}{329\mathrm{nm}} \\ =1.92 \end{gathered}$$

Thus the refractive index of the solid medium is $1.92$.

Since this value of the refractive index is the same as for the zircon, so medium is of zircon.