Eyepieces Design three different eyepieces for an optical system. All three eyepieces have a Magnifying Power of 10, and are to be used with a relaxed eye (the image presented to the eye is at infinity). The system objective presents an intermediate image to the eyepiece, and the intermediate pupil of the system is 200 mm to the left of this intermediate image plane. This intermediate pupil is the image of the stop through any optical elements between the stop and the eyepiece. a) A simple eyepiece consisting of just an eye lens. Determine the focal length and the eye relief. b) A compound eyepiece with a field lens located at the intermediate image plane. The field lens has a focal length of 40 mm. Determine the eye relief. c) A Ramsden-style eyepiece with the same eye relief as found with compound eyepiece of part (b). The field lens is located 12 mm to the right of the intermediate image plane. Determine the focal lengths of the two lenses and their separation. Hint: Three conditions must be met by the design -- the eyepiece must have the proper magnifying power, the final image presented to the eye must be at infinity, and the required eye relief must be obtained. Solution: MPmmf 250 / f 25mm MP 10 EP EP -To be used with a relaxed eye, the intermediate image must be at the front focal plane FEP of the eyepiece. -The intermediate pupil is 200mm to the left of the intermediate image plane or the front focal plane FEP of the eyepiece. a) Simple Eyepiece f fmm 25 EYE EP The eye lens is 25mm to the right of the intermediate image plane. Image the pupil through the eye lens for the XP and the ER: Eyerelief z zfmm 200 225 EYE 11 1 zzf EYE ERz 28.125 mm b) Compound Eyepiece f 40mm t 25 mm F Because the field lens is located at FEYE , only the rear principle plane and the ER change from the simple eyepiece. f fmm 25 EP EYE The front focal point of the eyepiece is coincident with the front focal point of the eye lens (and intermediate image plane). The front principal plane of the eyepiece is coincident with the front principal plane of the eye lens (and is at the eye lens). Rear principal plane of the eyepiece is shifted: f dtt FFEYE f f EYE EP EYE F dmm 15.625 This shift reduces the eye relief of the simple eyepiece to that of the compound eyepiece: ERmmmmmm28.125 15.625 12.5 c) Ramsden Eyepiece Design to have the same ER as compound eyepiece. The field lens is located 12 mm to the right of the intermediate image Intermediate Intermediate Image f f XP Pupil t F P P 12 d d′ ER -200 f z z′ Three conditions must be met: 1) fEP 25mm 2) FEP located at the intermediate image 3) The intermediate pupil is imaged to the proper ER: ERmm12.5 Condition 1) EP F EYE F EYEtf 1/ EP dt EYE dt F EP EP Condition 2) fEP 12mm d 25 mm Condition 3) 11 1 zzf EP zERd12.5 mmd (XP distance) zmmmmd200 12 (Pupil distance) Solve for fEYE, f F and t: fEP 12mm d 25 mm dmm13 EYE dtmm13 EP dmm13 EYEtd EP 0.52 fmmEP 25 EYEt 0.52 From Condition 3): zmmmmdmm200 12 225 11 zz EP zmm 28.125 zERd12.5 mmd dmm 15.625 dt F EP d FEPtd fEP Ft 0.625 There are now three equations with three unknowns: EP F EYE F EYEtf 1/ EP EYEt 0.52 F t 0.625 Solving: EP F EYE F EYEt EPtt F EYE t ()() F t EYE t EPt 0.625 0.52 (0.625)(0.52) EPt 0.82 0.82 tffmm0.82EP EP 25 EP tmm 20.5 EYE 0.52 /tmm 0.02536 / fmmEYE 39.42 F 0.625 /tmm 0.03049 / fmmF 32.8 .