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:

ERmmmmmm28.125 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: ERmm 12.5

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

zERd12.5 mmd   (XP distance)

zmmmmd200  12  (Pupil distance)

Solve for fEYE, f F and t:

fEP 12mm d 25 mm

dmm13

 dtmmEYE 13 EP

dmm13 EYEtd EP 0.52 fmmEP 25

EYEt  0.52

From Condition 3):

zmmmmdmm200  12   225

11  zz EP

zmm  28.125

zERd12.5 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 tffmm0.82EP EP  25 EP

tmm 20.5

EYE 0.52 /tmm 0.02536 /

fmmEYE  39.42

F 0.625 /tmm 0.03049 /

fmmF  32.8