In may experimental set-ups the source and screen are fixed at a distance say `D` and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Explanation

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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