a) Design a thin-lens peephole that covers a total field of 90 and has a total length of 25 mm (it has to fit within the thickness of the door).

Transcription

1 Peephole The goal of this problem is to design a door peephole to view your visitors (you see them, but they don t see you). The eye should be included as part of the design, and for this problem we will model the eye as a single refracting surface of radius 5.65 mm. The index of the eye is that of water (1.333), and the pupil or iris diameter is 4 mm and is located at the cornea. The retina is located at the rear focal point of the cornea. The macula is the central portion of the retina and is specialized for resolution. The fovea is the highest resolution part of the macula and has a diameter of about 1.5 mm. The resolution of the retina drops off outside the macula. The macula has a diameter of about 3 mm, and we will somewhat arbitrarily use an area of about 1.5 times the macula diameter to define the visual field of view for this application. The peephole is a non-inverting afocal system used to increase the field of view of the eye through the door. The iris of the eye is the system stop. a) Design a thin-lens peephole that covers a total field of 90 and has a total length of 25 mm (it has to fit within the thickness of the door). b) What are the required element diameters for a totally vignetted field of this size? The separation between the cornea and the rear element of the optical system should also be 25 mm (nose room). Solution: ye: R5.65mm n ( n1) / R / mm f 16.95mm f nf 22.6mm R

2 Macula: 3mm Dia Chief ray angles at cornea: Image size limit = 1.5 Macula = 4.5mm Dia h 4.5 mm/ mm After: uh/ f 2.25 mm/ f R R Before: u nu n2.25 mm/ f 2.25 mm/ f R FOV of the eye: u 1 FOV 2 tan 15.2 This corresponds to the Central High-Resolution Zone on the retina. a) Telescope Requirement: FOV 90 (Full Field) Chief ray angle in object space (paraxial angle) FOV u0 tan Required MP: MP u / u / m1/ MP 7.52 Remember that m and MP are paraxial quantities, and we must use the values of the chief ray, not the FOV, to determine the MP. This application requires a reducing telescope that will make the scene appear farther away to increase the FOV.

3 Telescope Design None inverting Use a reversed Galilean telescope (negative lens in front) t f1 f2 m f2 / f1 f mf 2 1 t f mf 1 m f f1 t /1 m t1 25mm m 7.52 f1 3.83mm / mm f mm / mm b) For vignetting, we need the marginal and chief ray value of both lenses. The marginal ray is easy since the telescope is afocal and the object is at infinity. The marginal ray is parallel to the axis between L2 and the eye. y L2 y Cornea 2 mm Pupil Diameter /2 y L1 y L2 / m 0.266mm The chief ray values can be found by starting at the cornea: y Cornea 0mm u Cornea Use a backward raytrace to get the chief ray heights at the lenses of the telescope: t2 25mm (Second Lens to Cornea) 2 Cornea Cornea 2 y L y u t

4 yl mm u L u L y L L ul y L y L u L t yl Co u L u rnea t1 25mm (Telescope Lens Separation) If extended to object space, these results produce u 1.00 in object space. This raytrace can also be done on a raytrace sheet. Inside the eye (after the cornea) reduced distances ( f R / n f ) and optical angles ( nu, nu ) are shown on the raytrace sheet.

5 b) For a totally vignetted field: L1: L2: a y y and a y a and a a1 3.50mm and a a1 3.50mm D1 7.00mm a and a a and a Here, the second condition must be applied since the second lens cannot become the system stop. The fully vignetted field is therefore not limited by the second lens. a2 2.00mm D2 4.00mm The second lens of the telescope and the pupil of the eye now have exactly the same diameters. The result of this condition is that the unvignetted field of view of this system is zero. It is only unvignetted on axis.