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Question
In many 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.
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Solution
From the reversibility of u and v, as seen from the formula for lens,
`1/f = 1/v - 1/u`
It is clear that there are two positions for which there shall be an image on the screen.
Let the first position be when the lens is at O.
Given `-u + v = D`
⇒ `u = -(D - v)`
Placing it in the lens formula
`1/(D - v) + 1/v = 1/f`
⇒ `(v + D - v)/((D - v)v) = 1/f`
⇒ `v^2 - Dv + Df` = 0
⇒ `v^2 = D/2 +- sqrt(D^2 - 4Df)/2`
`u = -(D - v) = - (D/2 + sqrt(D^2 - 4Df)/2)`
Thus, if the object distance is `D/2 - sqrt(D^2 - 4Df)/2` then the image is at `D/2 + sqrt(D^2 - 4Df)/2`
If the object distance is `D/2 + sqrt(D^2 - 4Df)/2`, then the image is at `D/2 - sqrt(D^2 - 4Df)/2`.
The distance between the poles for these two object distances is
`D/2 - sqrt(D^2 - 4Df)/2 - (D/2 - sqrt(D^2 - 4Df)/2) = sqrt(D^2 - 4Df)`
Let `d = sqrt(D^2 - 4Df)`
If `u = D/2 + d/2` then the image is at `v = D/2 - d/2`
∴ The magnification `m_1 = (D - d)/(D + d)`
If `u = (D - d)/2` then `v = (D + d)/2`
∴ The magnification `m_2 = (D + d)/(D - d)` Thus `m_2/m_1 = ((D + d)/(D - d))^2`.
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