Chapter 24, Problem 74GP

To determine

Index of refraction of a clear material.

Answer to Problem 74GP

Index of refraction of a clear material is $n=2$

Explanation of Solution

Given:

Thickness, $t=150\text{ nm}$

Wavelength, $\lambda =600\text{ nm}$

Formula used:

The thickness of a material is given by: $t=\frac{\lambda }{4n}$

Calculation:

Substituting the given values in the above formula,

  $\begin{array}{l}150\text{ nm}=\frac{600\text{ nm}}{4n}\\ n=1\end{array}$

Here, $n<n_{glass}$ this is not possible.

Hence, the thickness is given by:

  $\begin{array}{l}t=\frac{\lambda }{2n}\\ \left(150\text{ nm}\right)=\frac{\left(600\text{ nm}\right)}{2n}\\ n=2\end{array}$

Conclusion:

Thus, the index of refraction of a clear material is $n=2$