Answer to Question #195290 in Optics for Maeve

Question

Our eyes have to balance between focus and the amount of light. When we’re in a dark space, the slit in our eyes will open up. Using the concept of diffraction, how does this affect the image that we see?

Answer 1

The diffraction limit of a telescope is given approximately by the formula

$\theta=1.22\dfrac{\lambda}{D}$

$\theta$ is the angular size of the blurring due to the wave nature of light

$λ$ is the wavelength of the light (about 0.5 µm for visible light)

D is the diameter of the objective lens

Pupil diameter is about 3 mm

$\theta=1.22\bigg(\dfrac{0.5\times10^{-6}}{3\times10^{-3}}\bigg)=0.203\times10^{-3}\space rad$

There are 206,000 arcseconds in a radian

$\therefore0.203\times10^{-3}\space rad$ have = $(0.203\times10^{-3})\times(206000)=41.88\space arcseconds$

Therefore, our eye is working at its diffraction limit, corresponding to 41.88 arc seconds of smearing in the real image formed at focal plane in the eye