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Question
The focal length of a convex lens made of glass of refractive index (1.5) is 20 cm.
What will be its new focal length when placed in a medium of refractive index 1.25?
Is focal length positive or negative? What does it signify?
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Solution
Given aμg = 1.5
Focal length of the given convex lens when it is placed in air is f = + 20 cm
Refractive index of the given medium with respect to air is aμm = 1.25
New focal length of the given convex lens when placed in a medium is f'
`1/"f" = ("a"_(mu"g") - 1)[(1/"R"_1) + (1/"R"_2)]` .....(A)
`1/"f'" = ("m"_(mu"g") - 1)[(1/"R"_1) + (1/"R"_2)]` .....(B)
Dividing (A) by (B), we get
`"f'"/"f" = (("a"_(mu"g") - 1))/(("m"_(mu"g") - 1)) = ((1.5 - 1))/((1.2 - 1)) = 0.5/0.2 = 5/2 = 2.5`
f' = 2.5f = (2.5 × 20) cm = +50 cm as mµg = `mu_"g"/mu_"m" = 1.5/1.25 = 1.2`
New focal length is positive.
The significance of the positive sign of the focal length is that the given convex lens is still converging in the given medium.
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