y-u ray trace example - Montana State University
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Optical Design (S15) Joseph A. Shaw – Montana State University The y-u ray trace method: telescope example The y-u method is just the y-nu method for thin lenses (ignoring n within lenses). This example illustrates the use of paraxial ray tracing for finding cardinal points, identifying aperture and field stops, locating and sizing pupils, etc. Sfc 1: π1 = 10 … π·1 = 2.3 … π‘1 = 16.5 Sfc 4: π4 = 0 … π·4 = 0.7 … π‘4 = 1.00 Sfc 2: π2 = 2 … π·2 = 1.7 … π‘2 = 2.38 Sfc 5: π5 = 1 … π·5 = 1.3 … π‘5 =bfl=? Sfc 3: π3 = 0 … π·3 = 0.25 … π‘3 = 1.62 1 Optical Design (S15) Joseph A. Shaw – Montana State University y-u ray trace: marginal ray to find aperture stop 2 Optical Design (S15) Joseph A. Shaw – Montana State University y-u ray trace: chief ray to find field stop AS 3 Optical Design (S15) Joseph A. Shaw – Montana State University Entrance pupil location We can locate the entrance-pupil plane by finding the distance π₯ππ from surface 1 to where the chief ray appears to cross the axis when viewed from object space. Chief ray Apparent chief ray π¦1 π’1 Chief ray π₯ππ From the triangle in the figure … π’1 = −π¦1 π₯ππ (the minus sign is from the sign convention, as the diagram is drawn for a negative angle) (1) Therefore, the distance from surface 1 to the entrance-pupil plane is Using values from our y-u ray-trace sheet for surface 1 gives π₯ππ = π₯ππ = −(0.163) −0.0247 −π¦1 π’1 = 6.59 (The exit pupil is located 6.59” to the right of the objective lens … inside the telescope) 4 Optical Design (S15) Joseph A. Shaw – Montana State University Exit pupil location Apparent chief ray We can locate the exit-pupil plane by finding the distance π₯π₯π from the last surface (k) to where the chief ray appears to cross the axis when viewed from image space. From the triangle in the figure … π’π ′ = Chief ray π¦π π’π ′ π₯π₯π −π¦π π₯π₯π (k) Therefore, the distance from surface k to the exit-pupil plane is π₯π₯π = −π¦π π’π ′ Using values from our y-u ray-trace sheet for surface k = 5 gives π₯π₯π = −(0.5660) −0.3500 = 1.617 (The exit pupil is located 1.617” to the right of the eye lens) 5 Optical Design (S15) Joseph A. Shaw – Montana State University Entrance pupil diameter Apparent marginal ray The apparent marginal ray height at the pupil plane gives the entrance pupil semi-diameter π¦ππ . π’1 π₯ππ Marginal ray π’1 From the triangle in the figure … π’1 = Therefore, the pupil semi-diameter is π¦ππ −π¦1 π₯ππ π¦1 π¦ππ (1) π¦ππ = π¦1 + π₯ππ π’1 Using values from our y-u ray-trace sheet for surface 1 gives π¦ππ = 0.9645 + 6.59 0.01929 = 1.0916 So the exit pupil diameter is 2π¦ππ = 2.183" 6 Optical Design (S15) Joseph A. Shaw – Montana State University Exit pupil diameter The apparent marginal ray height at the pupil plane gives the exit pupil semi-diameter π¦π₯π . Marginal ray Marginal ray π¦π π¦π₯π π’π ′ π₯π₯π From the triangle in the figure … π’π ′ = (k) − π¦π −π¦π₯π π₯π₯π Therefore, the pupil semi-diameter is π¦π₯π = π¦π + π₯π₯π π’π ′ (k denotes the last surface) Using values from our y-u ray-trace sheet for surface k=5 gives π¦π₯π = 0.07716 + 1.617 0 = 0.07716 So the exit pupil diameter is 2π¦π₯π = 0.154" 7 Optical Design (S15) Joseph A. Shaw – Montana State University Ray trace from object at ∞ Tracing a ray from an object at (-)∞ gives us what we need to find the following: • effective focal length ππ • 2nd focal length π′ • back focal length πππ π′ Ray from −∞ π¦1 2nd principal plane distance π′ From the triangle formed by the 2nd principal plane, the optical axis, and the ray, the “2nd focal length” is π′ = −π¦1 = π′ππ π’π ′ Dividing by the image-space index yields the effective focal length … Replacing the 2nd principal plane in the above figure with the last optical surface gives the back focal length … (distance from the last surface to the focal plane) π’π ′ π ′ = π′ππ (Greivenkamp calls it back focal distance) • π′ πππ = (k = last surface) ππ = −π¦1 ππ’ π ′ −π¦π π’π ′ The distance from the last surface to 2nd principal plane is then π′ = πππ − π ′ = The distance from the last surface to 2nd nodal plane is π ′ = πππ − π π¦1 − π¦π π’π ′ 8 Optical Design (S15) Joseph A. Shaw – Montana State University Ray trace from image at ∞ Tracing a ray from an image at ∞ gives us what we need to find the following: • effective focal length ππ • 1st focal length π • front focal length πππ π π • π = πππ principal plane distance π From the triangle formed by the 1st principal plane, the optical axis, and the ray, the “1st focal length” is π¦π π’1 (Greivenkamp calls it front focal distance) 1st π= π¦π = πππ π’1 Dividing by the image-space index yields the effective focal length … Replacing the 1st principal plane in the above figure with the first optical surface gives the front focal length … (distance from the first surface to the focal plane) Ray from ∞ πππ = (k = last surface) ππ = 1 −π¦1 π’1 The distance from the first surface to 1st principal plane is then π = πππ + π = The distance from the first surface to 1st nodal plane is π¦π ππ’ π = πππ + π ′ π¦π − π¦1 π’1 9 Optical Design (S15) Joseph A. Shaw – Montana State University The optical (or Lagrange) invariant Paraxial marginal and chief ray data also can be used to calculate the optical invariant … a quantity that always has the same value, regardless of where in the system it is calculated. πΌππ = π π¦π’ − π¦π’ = π′ π¦π’′ − π¦π’′ I encourage you to carefully consider this simple equation (represented by all sorts of Symbols, including strange Cyrillic characters that resemble ℵ … it can be very useful! Ex: object plane (π¦ = β, π¦ = 0) … πΌππ = πβπ’ = π¦ππ’ Ex: image plane (π¦ = β′, π¦ = 0) … πΌππ = π′β′π’′ = π¦πππππ π′π’′ Transverse magnification: π = β′ β = ππ’ π′ π’′ = β π π′ β′ π¦πππππ π¦ππππππ‘ Can use invariant to calculate image size without tracing chief ray … 10
