A convex lens made up of glass of refractive index 1.5 is dipped, in turn in (i) medium of refractive index 1.6 (ii) a medium of refractive index 1.33.
Will it behave as converging or diverging lens in two cases? justify your answer with necessary equations. How will its focal length change in two media?

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Solution

$(μ_{l} / μ_{s} - 1) (1 / R_{1} - 1 / R_{2}) = 1 / f = p o w e r J u s t s t i c k t o t h e a b o v e f o r m u l a, h e r e μ_{l} = r e f r a c t i v e i n d e x o f l e n s . μ_{s} = r e f r a c t i f e i n d e x o f s u r r o u n d i n g m a t e r i a l . R 1 & R 2 a r e t h e r a d u s o f c u r v a t u r e o f t h e t w o f a c e s o f t h e l e n s . μ_{l} > μ_{s} c o n v e r g i n g e l s e d i v e r g i n g .$

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μl/μs -1(1/R1 - 1/R2) = 1/f = powerJust stick to the above formula,hereμl = refractive index of lens.μs= refractife index of surrounding material.R1 & R2 are the radus of curvature of the two faces of the lens.μl>μs converging else diverging.