i don't know the answer yaar.. please tell how u calculated??
A lens of focal length 15 cm is dipped in water. What is the new focal length given that ref index of glass wrt air is 3/2 and that of water wrt air is 4/3.
If the lens is dipped in a liquid under what condition, the rays
are not deviated.
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5 Answers
soumyadeep sanyal
·2010-02-11 09:07:32
the ans is 60 cm.
ref index of glass wrt water is (3/2)/(4/3)=9/8
now use lens makers formula...r1,r2 being the same in both cases would cancel.
f2/f1=(μ1-1)/(μ2-1)
f2=focal length in water.
μ1=3/2
μ2=9/8
f1=15
Manmay kumar Mohanty
·2010-02-14 05:41:20
1f1 = (32 - 1) [1R1 - 1R2] .........(1)
1f2 = (3/24/3 - 1) [1R1 - 1R2] .........(2)
Dividing (1) by (2),
f2f1 = 0.50.128
f2 = 0.5 x 150.128
f2 = 7.50.128 = 58.59 cm.
Is it correct.