Two yellow flowers are separated by 60 cm along a line perpendicular to your line of sight to the flowers. How far are you from a flower when they are at the limit of resolution according to the Rayleigh criterion? Assume the light from the flowers has a single wavelength of 550 nm and that your pupil has a diameter of 5.5 mm.

The given data can be listed below,

• The distance between flowers is,$D=60\text{cm}$
• The wavelength of the light from flower is,$\lambda =500\text{nm}$
• The diameter of the pupil is, $d=5.5\text{mm}$

The capacity to recognise two closely spaced points as distinct is referred to as resolution power. Resolution is the capacity to recognise detail.

The angle of diffraction is given by,

$$\begin{gathered} \sin \theta =\frac{1.22\lambda }{d} \\ \theta ={\sin }^{-1}\left(\frac{1.22\lambda }{d}\right) \end{gathered}$$ …(i)

The distance between eye and length of the resolution is given by,

$$\begin{gathered} L\theta =D \\ \theta =\frac{D}{L} \end{gathered}$$ …(ii)

Compare equation (i) and (ii) the distance of the flowers from the observation point is given by,

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

Here,$\lambda$ is the wavelength of the light, dis the diameter of the pupil, D is the smallest resolution distance.

Substitute all the values in the above,

$$\begin{gathered} L=\frac{\left(0.60\text{m}\right)\left(5,5\times {10}^{-3}\right)}{1.22\left(550\times {10}^{-9}\text{m}\right)} \\ =4918\text{m} \end{gathered}$$

Thus, the distance of flowers from the observation point is 4918 m.