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A thin convex lens L (refractive index =...

A thin convex lens L (refractive index =1.5) is placed on a plane mirror M. when a pin is placed at A, such that OA=18 cm, its real inverted image is formed at A itself, as shown in figure. when a liquid of refractive index `mu_(i)` is put between the lens and the mirror, the pin has to be removed to A',. such that OA'=27 cm, to get its inverted real image at A' itself. the value of `mu_(i)` will be
a) `(4)/(3)` b) `(3)/(2)` c) `sqrt(3)` d) `sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
1.33
`(1)/(f_(1)) = (1)/(2 ) xx (2)/(18) = (1)/(18), (1)/(f_(2)) - ((mu_(1) - 1))/(-18)`
When `mu_(1)` is filled between lens and mirror
`P = (2)/(18) - (2)/(18) (mu_(1) - 1) = (2 - 2 mu_(1) + 2)/(18)`
`= F_(m) = - ((18)/(2 - mu_(1))), 2 = 6 - 3 mu_(1)`
`3 mu_(1) = 4, mu_(1) = (4)/(3)`
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