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|.

Answer 1

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.