Q40E - University Physics with Modern Physics

Question

If you can read the bottom row of your doctor’s eye chart, your eye has a resolving power of 1 arcminute, equal to $\frac{\mathbf{1}}{\mathbf{60}}$ degree. If this resolving power is diffraction-limited, to what effective diameter of your eye’s optical system does this correspond? Use Rayleigh’s criterion and assume $\mathbf{\lambda }\mathbf{=}\mathbf{550}\mathbf{ }\mathbf{nm}$.

Step-by-Step Solution

Answer

The diameter of the eye optical system is $2.3\times {10}^{-3}\mathrm{m}$. 

Step 1: Given Data

The resolving power of the eye is $={\left(\frac{1}{60}\right)}^{°}=\left(2.91\times {10}^{-4}\mathrm{rad}\right)$

Wavelength of light is 550 nm  

Step 2: Resolving Power

The ability of an optical device to distinguish minute details of an object is known as the resolving power of the lens. The higher the resolving power of an instrument, the smaller the details it can distinguish. The minimum separation angle  that can be resolved by an instrument is given as-

$\mathbf{Sin\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{D}}$ 

Here, $\mathit{\lambda }$ is the wavelength of light used and is the diameter of the aperture of the instrument. The above relation is known as the Rayleigh Criterion for the resolution of an optical system.

Step 3: Diameter

Using the Rayleigh’s criterion, the diameter of the aperture is given as-

 $$\begin{gathered} \mathrm{D}=\frac{1.22\mathrm{\lambda }}{\mathrm{sin\theta }} \\ =\frac{1.22\times \left(550\times {10}^{-9}\mathrm{m}\right)}{\sin \left(2.91\times {10}^{-4}\right)} \\ =2.3\times {10}^{-3}\mathrm{m} \end{gathered}$$

Hence, the diameter of the aperture is $2.3\times {10}^{-3}\mathrm{m}$ .