1 Stops in optical systems (5.3) Hecht 5.3 Monday September 30, 2002.

Presentation on theme: "1 Stops in optical systems (5.3) Hecht 5.3 Monday September 30, 2002."— Presentation transcript:

1 Stops in optical systems (5.3) Hecht 5.3 Monday September 30, 2002

2 Stops in Optical Systems In any optical system, one is concerned with a number of things including: 1.The brightness of the image SS’ Two lenses of the same focal length (f), but diameter (D) differs More light collected from S by larger lens Bundle of rays from S, imaged at S’ is larger for larger lens Image of S formed at the same place by both lenses

3 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

4 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

5 Theory of Stops We wish to develop an understanding of how and where the bundle of rays are limited by a given optical system

6 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

7 Aperture Stop (AS) O

8 Entrance Pupil (E n P) 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) O L1L1L1L1 E E E’ E’ How big does the aperture stop look to someone at O E n P – defines the cone of rays accepted by the system F1’F1’F1’F1’ E’E’ = E n P

9 Exit Pupil (E x P) The exit pupil is the image of the aperture stop in the lenses coming after it (i.e. to the right of the AS) O L1L1L1L1 E E E’’ E’’ F2’F2’F2’F2’ E”E” = E x P

10 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 eyepiece Objective=AS=E n P

11 Example: Eyepiece E E 9 cm Ф 1 = 1 cm Ф 2 = 2 cm f 1 ’ = 6 cm f 2 ’ = 2 cm Ф D = 1 cm 1 cm 3 cm O

12 Example: Eyepiece Find aperture stop for s = 9 cm in front of L 1. To do so, treat each element in turn – find E n P for each (a) Lens 1 – no elements to the left tan µ 1 = 1/9 defines cone of rays accepted 9 cm 1 cm O µ1µ1µ1µ1

13 Example: Eyepiece Find aperture stop for s = 9 cm in front of Diaphragm. Find E n P (b) Diaphragm – lens 1 to the left E E 9 cm 1 cm Ф D ’ = 1.2 cm E’ E’ 1.2 cm Look at the system from behind the slide O

14 Example: Eyepiece Calculate maximum angle of cone of rays accepted by entrance pupil of diaphragm E’ E’ O 0.6 cm 9 + 1.2 cm µ2µ2µ2µ2 tan µ 2 = 0.6/10.2 ≈ 1/17

15 Example: Eyepiece (c) Lens 2 – 4 cm to the left of lens 1 Ф 2 ’ = 6 cm Look at the system from behind the slide 9 cm 4 cm O Ф 2 = 2 cm

16 Example: Eyepiece Calculate maximum angle of cone of rays accepted by entrance pupil of lens 2. O 3 cm 9 + 12 cm µ3µ3µ3µ3 tan µ 3 = 3/21 = 1/7

17 Example: Eyepiece ComponentEntrance pupilAcceptance cone angle (degrees) Lens 1tan µ 1 = 1/96.3 Diaphragmtan µ 2 = 1/173.4 Lens 2tan µ 3 = 1/78.1 Thus µ 2 is the smallest angle The diaphragm is the element that limits the cone of rays from O Diaphragm = Aperture Stop

18 Example: Eyepiece Entrance pupil is the image of the diaphragm in L 1. 9 cm E’ E’ 1.2 cm O µ 2 = tan -1 (1/17) Ф D ’ = 1.2 cm EnPEnPEnPEnP

19 Example: Eyepiece Exit pupil is the image of the aperture stop (diaphragm) in L 2. E E 9 cm f 2 ’ = 2 cm Ф D = 1 cm 1 cm 3 cm O Ф 2 ’ = 2 cm

20 Example: Eyepiece E E f 2 ’ = 2 cm Ф D = 1 cm 3 cm O Ф E x P ’ = 2 cm E’ E’ 6 cm ExPExPExPExP

21 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 E n P and E x P are also conjugate planes of the complete system

22 Marginal Ray Those rays (for a given object point) that pass through the edge of the entrance and exit pupils (and aperture stop).

23 Ray tracing with pupils and stops P’ Q’ O EnPEnPEnPEnP Q’’ P’’ ExPExPExPExP P Q AS T Marginal Rays from T,O Must proceed towards edges of E n PMust proceed towards edges of E n P Refracted at L 1 to pass through edge of ASRefracted at L 1 to pass through edge of AS Refracted at L 2 to pass (exit) through E x P.Refracted at L 2 to pass (exit) through E x P. L1L1L1L1 L2L2L2L2 Chief Ray from T Proceed toward centre of E n PProceed toward centre of E n P Refracted at L 1 to pass though centre of ASRefracted at L 1 to pass though centre of AS Refracted at L 2 to pass (exit) through centre of E x PRefracted at L 2 to pass (exit) through centre of E x PT’ O’

24 Exit Pupil Defines the bundle of rays at the image O Q’’ P’’ T T’ O’ α’α’α’α’

25 Field Stop That component of the optical system that limits the field of view θ A d θ = angular field of view A = field of view at distance d

Download ppt "1 Stops in optical systems (5.3) Hecht 5.3 Monday September 30, 2002."

Similar presentations