If you look at something 40 m from you, what is the smallest length (perpendicular to your line of sight) that you can resolve, according to Rayleigh’s criterion? Assume the pupil of your eye has a diameter of 4.00 mm, and use 500 nm as the wavelength of the light reaching you.

The given data can be listed below,

• The diameter of the pupil of the eye is,$d=4\text{mm}$
• The wavelength of the light is,$\lambda =500\text{nm}$
• The distance of the object from eye is,$L=40\text{m}$

The smallest distance at which two points can be separated while still being recognized as two separate points is known as the "resolving power," or resolution, of an image.

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 smallest length of resolution is given by,

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

Here, $\lambda$is the wavelength of the light, dis the diameter of the pupil, L is the distance of the object from the eye

Substitute all the values in the above,

$$\begin{gathered} D=\frac{1.22\left(500\times {10}^{-9}\text{m}\right)\left(40\text{m}\right)}{4.00\times {10}^{-3}\text{m}} \\ =0.0061\text{m} \end{gathered}$$

Thus, the smallest length of resolution is 0.0061 m.