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Unformatted text preview: Assignment3a
TAs’ Solution
Problem 1
(i) θ3 θ1
θ2 Since θ1> θ2 >θ3, the stop is the aperture stop. The entrance pupil (EnP) and the exit pupil
(ExP) are shown in the figure.
(ii) θ2
θ1 Since θ1> θ2, the L1 is the field stop. The entrance window (EnW) and the exit window
(ExW) are shown in the figure. (iii) Problem 2 Problem 3
5.84
(a) If so=∞, P=ni/si =1.33/0.02=66.5m1; If so=0.5m, P=n/f=1/0.5 +1.33/0.02=68.5m1
(b) Accommodation is 2m1
(c) P=n/f=1/0.25+1.33/0.02=70.5m1
(d) Need to add 2m1 Problem 4
5.87
(a) MP=25/f+1=25cm/25.4mm+1=10.8
(b)size=MP x D=10.8x5.0mm=54mm
(c) tan a = y / d = 5mm / 25cm ⇒ a = 1.15 0
(d) tan a = y ' / d = 54mm / 25cm ⇒ a = 12.19 0 Problem 5
5.88
(a) The intermediate imagedistance is obtained from the lens formula applied to the
objective; 1/27+1/si=1/25 and si=337.5mm. This is the distance from the objective to the
intermediate image, to which must be added the focal length of the eyepiece to get the
lens separation 337.5+25=362.5mm.
(b) Mo=si/so= 337.5/27= 12.5, Me=254/25=10.2. Thus the total magnification is
12.5*10.2= 127.5; the minus sign just means the image is inverted. Problem 6 ...
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This note was uploaded on 10/16/2011 for the course ELEC 308,315,10 taught by Professor Prof.shenghuisong during the Spring '11 term at CUHK.
 Spring '11
 Prof.ShenghuiSong

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