A convex lens has a focal length of 0.1 m in air. Calculate its power. If the lens is completely dipped in CS_(2) of refractive index 1.66, then what will be the change in power of the lens? Given R.I of the convex lens = 1.50.

A convex lens has a focal length of 0.1 m in air. Calculate its power.

1.	A convex lens has a focal length of 0.1 m in air. Calculate its power. If the lens is completely dipped in CS_(2) of refractive index 1.66, then what will be the change in power of the lens? Given R.I of the convex lens = 1.50.
Answer» Solution :`f=0.1m` `THEREFORE "POWER"=(1)/(f)=(1)/(0.1m)` `P=+10D` w.K.t`(1)/(f)=P=(n-1)(k)` `therefore k=(P)/(n-1)=(10)/(1.5-1)=(10)/(0.5)=20` Also,`" "_(1)n_(2)=(n_(2))/(n_(1))=(n_(g))/(n_(cs_(2))) =(1.5)/(1.66)=0.9036` `therefore(1)/(f_(e))=k(""_(cs_(2))n_(g)-1) =20(0.9036-1)` `therefore P_(e) = 20xx(-0.0964)= -1.928D` LENS behaves as a diverging lens of power = -1.93 D Change in power `DeltaP=P_(f)-P_(1)= -1.93-(+10)= -11.93D`

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A convex lens has a focal length of 0.1 m in air. Calculate its power. If the lens is completely dipped in CS_(2) of refractive index 1.66, then what will be the change in power of the lens? Given R.I of the convex lens = 1.50.

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