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Question
State how the focal length of a glass lens (Refractive Index 1.5) changes when it is completely immersed in:
(i) Water (Refractive Index 1.33)
(ii) A liquid (Refractive Index 1.65)
Solution
We know
`1/"f" = (mu - 1) (1/"R"_1 - 1/"R"_2)`
When the lens is submerged into water of R.I. = 1.33, then `1/"f" = (mu_12 - 1) (1/"R"_1 - 1/"R"_2)`
`("f"')/"f" = (mu -1)/(mu_12 - 1) = ((1.56 - 1)/(1.56/1.33 -1))`
`therefore ("f"')/"f" = ((0.56) xx 1.33)/(1.56 - 1.33) = 3.24`
Thus, focal length of the lens increases.
Similarly, for another liquid
`("f"')/"f" = ((1.56 - 1)/(1.56/1.65 - 1)) = - (0.56 xx 1.65)/0.09 = - 9.7`
i.e. the convex lens behaves as a concave lens.
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