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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Apr 2008 17:01:20 IST
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Two lenses,when in contact produce a focal length of power +10D.When they are 0.25m apart,power reduces to +6D.Find focal lengths of the two lenses.
Is it 0.16m &0.26m?
If not,please show the method
Rates assured!!
Thanks!:-)
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when they r placed in contact P1+P2=+10D
when at a distance of .25 mthe formula we used here is P1+P2 -dP1P2 (where d is distance b/w them)
now form the eq.
we get here P1+P2-dP1P2 = +6D
now solve this eq .
AND USING (P1-P2)^2 =(P1+P2)^2-4dP1P2
P1-P2=6
now solve the above eq
P1=8
P2=2
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Apr 2008 10:19:06 IST
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Case 1:
P1+P2=10
and formula when two lenses are separated by a distance d
Pnet= P1+P2-dP1P2 = +6 D
On solving the equations u get two values for the powers of both the lenses
P1=8 ...f=0.125 ...P2=2 ....f=0.5
and vice versa..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Apr 2008 17:04:02 IST
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thanks..i didnt know this formula...
did some really long stuff....it took around two pages...n still wrong!!!!lol!!!
Thanks!:-)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Apr 2008 17:21:36 IST
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netime frnd!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Apr 2008 19:17:15 IST
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Sorry adi the formula you used is applicable only if d is negligible when compared to focal length
The correct ans are .308 , .148 -.058, .0367
Check it for urself !!!! cheers
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IM NO BABY
I HAVE THE SEVENTH SENSE
NONSENSE!!!!!!!!!!! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Apr 2008 19:27:51 IST
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Sorry learner...
what celestine is saying is true...
d shud be negligible...
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