??1s?1s?????1s1???1f?1s1(?40)188?402f?110540?即:s??8(cm)? 9.证:
x?s??(?s)2??24(cm)
?y??nn2y1第一次折射:n?n,2n?111y?DP??y??nDP?n?1,211
第二次折射:n?n1y?y??d,y??2111n(y??d)?EP?11?PP??(?EP?)?(?EP?)?EP??EP?
1??1n1n(y??d)?(Dp??d)11(nDp??d)?(Dp??d)11?Dp??11nd?Dp??d?d(1?11n)?dn?1n
由图可知,若 使凹透镜向物体移动d10.解:
n?1n的距离,亦可得到同样的结果。
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?ns??ns?n??n
?
而:s??
s??2?n?1?n??1n??n?2?1P′
?n?2??n??1?,n?2故n??211.解: (1)由P208L3?7经导知:
nR2(n?1)1.5?42(1.5?1)?6(cm)
f???
按题意,物离物方主点H的距离为? 于是由
(6?4),
1s?1s???1s1s??1f?1f??16得?1?10?5?330?115?s??15(cm) (2)
??s?s?156?4?1.5
12.解:
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??ns?n??nsn?s???n??nrn??nr
s
(1)
?s??r?ns1?1n?r?n??nr?nr
仍在原处(球心),物像重合 (2)
即s?r?s??1r2n??2
?ns2??n??nr?2n?r
?n??nr?n??nrs?2nrn??n?nD2(n??n)?1.57?202?(1.53?1)?6.05(cm)
13.解:
(1)
??n?s?ns2??ns?n??nr?nsn?s??n??nr?nr?n?r
n??nr又?s?r?s??r?15cm即鱼在原处(2)
???y?y?s?s?nn??1515?1.331?1.33
14解: (1)
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?f?n?n??n1.50?1.33n?1.50f??r??2?17.647cmn??n1.50?1.33r??1.33?2??15.647cm
而f?s??fs?1即?f?s??1-fs?s?fs??141.1767.647??18.46??18.5cm
?s??(2)
sf?s?f?8?17.647?8?(?15.647)??y?y?s?s?nn???18.5?8?1.331.50?2.046?2
(3)
光路图如右: 15解: (1)
??f??nf?11(n?n?r1?n2?n)r2,n?n?n,r?r?r,n?n?1212?nn?n?r?f?n?n??r?1.33?102?(1.5?1.33)f???f11??39.12??f?1又??f?s?s?1s?凸?s?1s?1fs?1?f??11f?1?20?1?39.12??0.0244
??s??s???40.92cm(2)
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??f?nf??21(n?n?r1?n2?n)r2,n?n?n,r?r?r,n?n?1212n?n?n??r?n?n?r??nr2(n??n)2??1.33?102?(1.5?1.33)??39.12??f2又??f?s?f?s?1s?22???fs?12f???f2?fs1s?2?11f?22?1?20?1?39.12??0.0756
?s??s???13.23cm凹(3)
16.解:(1)透镜在空气中和在水中的焦距分别为:
1f??f?21?(n?1)(?1r1?1r2)1f?2?n?n?11(?)n?rr12n?(n?1)n?n?f?2f?n?(n?1)?(n?n?)?n?)?1.33??140?(1.54?1)f?n(n??f?f?21f?21n?n?n??nn?(1??n?n??1r12f?1f?1f?)?n?(1?f?21f?)1f?2f?f?21.33?(1?136.8140136.840)??3.22?2.091?1.54f?11?1r2?f?(n?1)1?21.6
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