Question

Suppose that two points are separated by 2.0 cm. If they are viewed by an eye with a pupil opening of 5.0 mm, what distance from the viewer puts them at the Rayleigh limit of resolution? Assume a light wavelength of 500 nm.

Short answer

The required distance is 164 m.

Step-by-step solution

Describe the diffraction by a circular aperture or a lens

The expression to calculate the distance between the eye and two objects is given by,

$\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}}$

Here, $\mathit{\lambda }$ is the wavelength, $\mathit{d}$ is diameter, and $\mathit{\theta }$ is angle.

Let D be the distance between two objects, and L be the distance between the eye and two objects. Then,

$$\begin{gathered} \mathit{L}\mathit{\theta }\mathbf{=}\mathit{D} \\ \mathit{\theta }\mathbf{=}\frac{\mathbf{D}}{\mathbf{L}} \end{gathered}$$

From the above equation,

$$\begin{gathered} \mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}} \\ \frac{\mathbf{D}}{\mathbf{L}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}} \\ \mathit{L}\mathbf{=}\frac{\mathbf{D}\mathbf{d}}{\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{\lambda }} \end{gathered}$$ ….. (1)

Determine the distance between the eye and two objects

Substitute $500\times {10}^{-9}\text{m}$ for $\lambda$, $5.0\times {10}^{-3}\text{m}$ for $d$, and $2.0\times {10}^{-2}\text{m}$ for $D$ in equation (1).

$$\begin{gathered} L=\frac{\left(2.0\times {10}^{-2}\text{m}\right)\left(5.0\times {10}^{-3}\text{m}\right)}{\left(1.22\right)\left(500\times {10}^{-9}\text{m}\right)} \\ =164\text{m} \end{gathered}$$

Therefore, the required distance is 164 m.