To understand Rayleigh's criterion as applied to the pupil of the eye, notice that rays do not pass straight through the center of the lens system (cornea + lens) of the eye except at normal incidence because the indices of refraction on the two sides of the lens system are different. In a simplified model, suppose light from two point sources travels through air and passes through the pupil (diameter a). On the other side of the pupil, light travels through the vitreous fluid (index of refraction n). The figure shows two rays, one from each source, that pass through the center of the pupil.
(a) What is the relationship between Δθ, the angular separation of the two sources, and b, the angular separation of the two images?
(b) The first diffraction minimum for light from source 1 occurs at angle Ï•, where a sin Ï• = 1.22λ [Eq. (25-13)]. Here, l is the wavelength in the vitreous fluid. According to Rayleigh's criterion, the sources can be resolved if the center of image 2 occurs no closer than the first diffraction minimum for image 1; that is, if β ‰¥ Ï• or, equivalently, sin β ‰¥ sin Ï• . Show that this is equivalent to Eq. (25-14), where λ0 is the wavelength in air.

Transcribed Image Text:

Retina Center of image 1 Pupil Vitreous fluid (n) Edge of image 1 Center of image 2