Entoptic Halos Images produced by structures within the eye (like lens fibers or cell fragments) are referred to as entoptic images. These images can sometimes take the form of \"halos\" around a bright light seen against a dark background. The halo in such a case is actually the bright outer rings of a circular diffraction pattern, with the central bright spot not visible because it overlaps the direct image of the light. Part A Find the diameter of the eye structure that causes a circular diffraction pattern with the first dark ring at an angle of 2.7 ? when viewed with monochromatic light of wavelength 625 nm. (Typical eye structures of this type have diameters on the order of 10 ?m. Also, the index of refraction of the vitreous humor is 1.36.) Solution The angular resolution of the eye is expressed as follows :- sin(Theta) = 1.22(Lemda / D) For any medium other than air, the above equation becomes :- Sin(Theta) = 1.22(Lemda / nD) D = 1.22(Lemda / (n*sin(Theta)) D = 1.22((625 nm) / (1.36*sin(2.7))) D = 1.22 * ((6.25*10^-7)m / (1.36*0.0471)) D = 11.9 * 10^-6 m D = 11.9 um .

1. Entoptic Halos Images produced by structures within the eye (like lens fibers or cell fragments)
are referred to as entoptic images. These images can sometimes take the form of "halos" around
a bright light seen against a dark background. The halo in such a case is actually the bright outer
rings of a circular diffraction pattern, with the central bright spot not visible because it overlaps
the direct image of the light.
Part A Find the diameter of the eye structure that causes a circular diffraction pattern with the
first dark ring at an angle of 2.7 ? when viewed with monochromatic light of wavelength 625
nm. (Typical eye structures of this type have diameters on the order of 10 ?m. Also, the index of
refraction of the vitreous humor is 1.36.)
Solution
The angular resolution of the eye is expressed as follows :-
sin(Theta) = 1.22(Lemda / D)
For any medium other than air, the above equation becomes :-
Sin(Theta) = 1.22(Lemda / nD)
D = 1.22(Lemda / (n*sin(Theta))
D = 1.22((625 nm) / (1.36*sin(2.7)))
D = 1.22 * ((6.25*10^-7)m / (1.36*0.0471))
D = 11.9 * 10^-6 m
D = 11.9 um