The wings of tiger beetles (Fig. 36-41) are coloured by interference due to thin cuticle-like layers. In addition, these layers are arranged in patches that are $\mathbf{60}\mathbf{}\mathit{\mu }\mathit{m}$ across and produce different colours. The colour you see is a pointillistic mixture of thin-film interference colours that varies with perspective. Approximately what viewing distance from a wing puts you at the limit of resolving the different coloured patches according to Rayleigh’s criterion? Use $\mathbf{550}\mathbf{nm}$ as the wavelength of light and $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{nm}$ as the diameter of your pupil.

Diameter of pupil $d=3\text{mm}$

Wave length of light $\left(\lambda \right)=550\text{nm}$

The Rayleigh criteria specify the minimal distance between two light sources that must exist in order to resolve them into separate objects.

Raleigh criterion for resolving with angular separation ${\theta }_R$is $\sin {\theta }_R=\frac{1.22\lambda }{d}$

Diameter of pupil $d=3\text{mm}$

$$\begin{gathered} d=\text{3}\text{mm}\left(\frac{{10}^{-3}\text{m}}{\text{mm}}\right) \\ =3\times {10}^{-3}\text{m} \end{gathered}$$

Wave length of light $\left(\lambda \right)=550\text{nm}$

$$\begin{gathered} d=550\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{m}}\right) \\ =550\times {10}^{-9}\text{m} \end{gathered}$$

Also if D is a size of the object that eye resolve and L is distance between eye and object

Then we have the condition $\tan {\theta }_R=\frac{D}{L}$

Here size of the object that eye resolved

$$\begin{gathered} D=60\mu m \\ =60\times {10}^{-6}m \end{gathered}$$

As i${\theta }_R$s small $\text{tan}{\theta }_R\approx \sin {\theta }_R\approx {\theta }_R$

Therefore from above equations

$$\begin{gathered} \frac{D}{L}=\frac{1.22\lambda }{d} \\ L=\frac{Dd}{1.22\lambda } \end{gathered}$$

Substitute given values

$$\begin{gathered} L=\frac{\left(60\times {10}^{-6} \text{m}\right)\left(3\times {10}^{-3}\text{m}\right)}{1.22\left(550\times {10}^{-9}\text{m}\right)} \\ =0.27 \text{m}\backslash \mathrm{hfill} \\ \text{= 0.27} \text{m}\left(\frac{100\text{cm}}{\text{m}}\right) \\ =27 \text{cm} \end{gathered}$$

Hence, the value is $27cm$.