Question

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.

Short answer

The distance of flowers from the observation point is 4918 m.

Step-by-step solution

Identification of given data

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, $d=5.5\text{mm}$

Concept/Significance of resolution power

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

Determination of the distance of flowers from the observation point.

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.