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Stops in optical systems (5.3) Hecht 5.3 Monday September 30, 2002 1 Stops in Optical Systems In any optical system, one is concerned with a number of things including: 1. The brightness of the image Image of S formed at the same place by both lenses S Bundle of rays from S, imaged at S’ is larger for larger lens Two lenses of the same focal length (f), but diameter (D) differs S’ More light collected from S by larger lens 2 Stops in Optical Systems Brightness of the image is determined primarily by the size of the bundle of rays collected by the system (from each object point) Stops can be used to reduce aberrations 3 Stops in Optical Systems How much of the object we see is determined by: (b) The field of View Q Q’ (not seen) Rays from Q do not pass through system We can only see object points closer to the axis of the system Field of view is limited by the system 4 Theory of Stops We wish to develop an understanding of how and where the bundle of rays are limited by a given optical system 5 Aperture Stop A stop is an opening (despite its name) in a series of lenses, mirrors, diaphragms, etc. The stop itself is the boundary of the lens or diaphragm Aperture stop: that element of the optical system that limits the cone of light from any particular object point on the axis of the system 6 Aperture Stop (AS) O 7 Entrance Pupil (EnP) is defined to be the image of the aperture stop in all the lenses preceding it (i.e. to the left of AS - if light travels left to right) E’ How big does the aperture stop look E to someone at O L1 E’E’ = EnP O F1’ EnP – defines the cone of rays accepted by the system E E’ 8 Exit Pupil (ExP) The exit pupil is the image of the aperture stop in the lenses coming after it (i.e. to the right of the AS) E’’ E L1 E”E” = ExP F2’ O E E’’ 9 Location of Aperture Stop (AS) In a complex system, the AS can be found by considering each element in the system The element which gives the entrance pupil subtending the smallest angle at the object point O is the AS Example, Telescope Objective=AS=EnP eyepiece 10 Example: Eyepiece f1’ = 6 cm f2’ = 2 cm E O E 9 cm 1 cm Ф1 = 1 cm 3 cm ФD = 1 cm Ф2 = 2 cm 11 Example: Eyepiece Find aperture stop for s = 9 cm in front of L1. To do so, treat each element in turn – find EnP for each (a) Lens 1 – no elements to the left tan µ1 = 1/9 defines cone of rays accepted O µ1 1 cm 9 cm 12 Example: Eyepiece Find aperture stop for s = 9 cm in front of Diaphragm. Find EnP (b) Diaphragm – lens 1 to the left Look at the system E E’ from behind the slide 1 1 1 1 s' 6 s ' 1.2cm O M E 9 cm s' 1.2 s E’ 1 cm 1.2 cm ФD’ = 1.2 cm 13 Example: Eyepiece Calculate maximum angle of cone of rays accepted by entrance pupil of diaphragm E’ O µ2 0.6 cm 9 + 1.2 cm tan µ2 = 0.6/10.2 ≈ 1/17 E’ 14 Example: Eyepiece (c) Lens 2 – 4 cm to the left of lens 1 Look at the system from behind the slide 1 1 1 4 s' 6 s ' 12cm O M s ' 12 3 s 4 Ф2’ = 6 cm 9 cm 4 cm Ф2 = 2 cm 15 Example: Eyepiece Calculate maximum angle of cone of rays accepted by entrance pupil of lens 2. 3 cm O µ3 9 + 12 cm tan µ3 = 3/21 = 1/7 16 Example: Eyepiece Component Entrance pupil Acceptance cone angle (degrees) Lens 1 tan µ1 = 1/9 6.3 Diaphragm tan µ2 = 1/17 3.4 Lens 2 tan µ3 = 1/7 8.1 Thus µ2 is the smallest angle The diaphragm is the element that limits the cone of rays from O Diaphragm = Aperture Stop 17 Example: Eyepiece Entrance pupil is the image of the diaphragm in L1. E’ µ2 = tan-1 (1/17) O E’ EnP 9 cm 1.2 cm ФD’ = 1.2 cm 18 Example: Eyepiece Exit pupil is the image of the aperture stop (diaphragm) in L2. f2’ = 2 cm E 1 1 1 3 s' 2 s ' 6cm O M E 9 cm 1 cm s' 6 2 s 3 Ф2’ = 2 cm 3 cm ФD = 1 cm 19 Example: Eyepiece E’ f2’ = 2 cm E ФExP’ = 2 cm O ExP E 3 cm ФD = 1 cm 6 cm E’ 20 Chief Ray for each bundle of rays, the light ray which passes through the centre of the aperture stop is the chief ray after refraction, the chief ray must also pass through the centre of the exit and entrance pupils since they are conjugate to the aperture stop EnP and ExP are also conjugate planes of the complete system 21 Marginal Ray Those rays (for a given object point) that pass through the edge of the entrance and exit pupils (and aperture stop). 22 Chief Ray from T •Proceed toward centre of EnP Ray tracing with pupils and stops •Refracted at L1 to pass though centre of AS •Refracted at L2 to pass (exit) through centre of ExP T L2 L1 Q’’ P’ P O’ O Marginal Rays from T,O •Must proceed towards edges of EnP •Refracted at L1 to pass through edge of AS •Refracted at L2 to pass (exit) through ExP. T’ Q Q’ P’’ AS EnP ExP 23 Exit Pupil Defines the bundle of rays at the image Q’’ T O’ α’ O T’ P’’ 24 Field Stop That component of the optical system that limits the field of view A θ d A = field of view at distance d θ = angular field of view 25