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Question
A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take L << |v – f|.
Solution
As the mean distance of object from mirror is u
∴ `u_1 = u - L/2` and `u_2 = (u + L/2)`
Let the image of the two ends of object form at distance v1 and v2(v1 > v2). So length of image on principal axis is L' = (v1 – v2)
`1/v + 1/u = 1/f` or `1/v + 1/f = 1/u`
⇒ `1/v = (u - f)/(uf)`
⇒ `v = (uf)/(u - f)`
So L' = `v_1 - v_2 = ((u - L/2)f)/((u - L/2) - f) - ((u + L/2)f)/((u + L/2) - f)`
⇒ L' = `f (u - L/2)/((u - f - L/2)) - (u + 1/2)/((u - f + L/2))`
⇒ L' = `f[((u - L/2)(u - f + L/2) - (u + L/2)(u - f - L/2))/((u - f - L/2)(u - f + 1/2))]`
= `(f[u^2 - uf + (uL)/2 - (uL)/2 + (fL)/2 - L^2/4 - (u^2 - uf - (uL)/2 + (uL)/2 - (fL)/2 - L^2/4)])/((u - f)^2 - L^2/4)`
∴ `L < < (u - f)`
∴ `L^2/4 < < < (u - f)`
So neglecting the terms `L^2/4`
`L^2 = f[((fL)/2 + (fL)/2)/(u - f)^2]`
`L^2 = (ffL)/((u - f)^2` ⇒ `L^2 = (Lf^2)/((u - f)^2`
It is the length of image f.
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