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Blazing goIITian

Joined:	5 Nov 2007
Post:	507

29 Apr 2008 17:01:20 IST

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451

Another assignment question-lenses

None

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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SPARTIAN

Blazing goIITian

Joined: 19 Mar 2007

Posts: 585

30 Apr 2008 00:32:17 IST

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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

Harshil

Blazing goIITian

Joined: 16 Nov 2007

Posts: 325

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

P_net= 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..

Learner .

Blazing goIITian

Joined: 5 Nov 2007

Posts: 507

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!:-)

Harshil

Blazing goIITian

Joined: 16 Nov 2007

Posts: 325

30 Apr 2008 17:21:36 IST

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netime frnd!!

Celestine Preetham

Hot goIITian

Joined: 24 Apr 2008

Posts: 163

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

Harshil

Blazing goIITian

Joined: 16 Nov 2007

Posts: 325

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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