In the lens formula, `(1)/(f)=(1)/(v)-(1)/(u)`
we find that `u` and `v` are reversible. Therefore, there are two positions of the object, for which there will be an image on the screen.
Let the first position be when lens is at `C`, Fig.
`:. -u + v = D` = distance between source (O) and screen `I`.
or `u = -(D - v)`
Put in (i)
`(1)/(v)+(1)/(D - v)=(1)/(f)` or `(D - v + v)/(v (D - v)) = (1)/(f)` or `D v - v^2 = Df` or `v^2 - Dv + Df = 0`
`:. v = (D +- sqrt(D^2 - 4 Df))/(2) = (D)/(2) +- (sqrt(D^2 - 4 Df))/(2)`
and `u = -(D - v) = -[(D)/(2) +- sqrt(D^2 - 4 Df)/(2)]`.
Thus if object distance is `u = (D)/(2) + (sqrt(D^2 - 4 Df))/(2)`, then image distance is `v = (D)/(2) - (sqrt(D^2 - 4 Df))/(2)`
Again, if object distance is `u = (D)/(2) - (sqrt(D^2 - 4 Df))/(2)` then image distance is `v = (D)/(2) + (sqrt(D^2 - 4 Df))/(2)`
The distance between two positions of the lens for these two object distance is
`d = (D)/(2) + (sqrt(D^2 - 4 Df))/(2) - (D)/(2) + (sqrt(D^2 - 4 Df))/(2) = sqrt(D^2 - 4 Df)`
If `u = (D)/(2)+ (d)/(2)`, then `v = (D)/(2) - (d)/(2) :.` Magification, `m_1 = (v)/(u) = (D - d)/(D + d)`
If `u = (D)/(2) - (d)/(2)`, then `v = (D)/(2)+(d)/(2) :.` Magnification, `m_2 = (v)/(u) = (D + d)/(D - d)`.
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