What curvature must be given to the bounding surface of `mu = 1.5` for virtual image of an object in the medium of `mu = 1 at 10 cm` to be formed at a distance of `40 cm`. Calculate power of the refracting surface and also two principal focal lengths of the surface.
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`R = ?, mu_(2) = 1.5, mu_(1) = 1` `u = - 10 cm, v = -40 cm` As object is placed in rarer medium, `:. -(mu_(1))/(u) + (mu_(2))/(v) = (mu_(@) - mu_(1))/(R )` `-(1)/(-10) + (1.5)/(-40) = (1.5 - 1)/(R ) = (1)/(2R)` `(1)/(2R) = (1)/(10) - (3)/(80) = (5)/(80) = (1)/(16)` `R = (16)/(2) = 8cm` As `R` is positive, the refracting surface is convex. Power of surface, `P = (mu_(2) - mu_(1))/(R ) = (1.5 - 1)/(8//100) = 6.25 D`
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