A diverging lens of refractive index `1.5` and focal length `15 cm` in air has same radii of curvature for both sides. If it is immersed in a liquid of refractive index `1.7`, calculate focal length of the lens in liquid.
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`mu = 1.5, f = -15 cm`, Let `R_(1) = -R, R_(2) = + R` `(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))` `(1)/(-15) = (1.5 - 1)(-(1)/R - (1)/(R )) = -(1)/(R )` `R = 15 cm` Let `f'` be focal length of lens in liquid `(1)/(f') = ((mu_(g))/(mu_(l)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` `= ((1.5)/(1.7) - 1)((1)/(-15) - (1)/(15))` `= -(02)/(1.7)(-(2)/(15)) = (0.4)/(1.7 xx 15)` `f' = (1.7 xx 15)/(0.4) = 63.75cm` The diverging lens will behave as converging lens in the liquid.
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