Q.By stating sign conventions and assumptions made, derive mirror formula 1/v + 1/u = 1/f for the concave mirror.

Using the Cartesian sign convention (distances measured from the pole, along the principal axis positive in the direction of incident light) and similar triangles from the ray diagram, the mirror formula 1/v + 1/u = 1/f follows directly.

Sign convention (Cartesian, New Cartesian Sign Convention):

• All distances are measured from the pole (P) of the mirror.
• Distances measured in the direction of the incident light (i.e. away from the mirror, to the right if light travels left to right) are taken positive; distances measured against the incident light (to the left) are taken negative.
• Heights measured upward (above the principal axis) are positive; downward, negative.

Assumptions: the object is small and lies close to the principal axis, and only paraxial rays (making small angles with the axis) are considered, so the mirror can be treated as producing a point image via the mirror formula.

Derivation: Consider a concave mirror with pole P, centre of curvature C, and focus F. An object AB is placed on the principal axis beyond C, forming a real, inverted image A'B' between C and F.

A ray from B parallel to the principal axis strikes the mirror at M and reflects through F. Another ray from B passes through C (striking the mirror normally) and reflects back along the same path, meeting the first ray at B' (forming the image).

From the similar triangles $\triangle ABC$ and $\triangle A'B'C$ (formed by the ray through C):

$\dfrac{A'B'}{AB} = \dfrac{A'C}{AC}$

From the similar triangles $\triangle MPF$ and $\triangle A'B'F$ (using the ray through F, and treating M as lying essentially on the mirror's pole line for a small aperture, so $MP \approx AB$):

$\dfrac{A'B'}{AB} = \dfrac{A'F}{PF}$

Equating the two expressions for $A'B'/AB$:

$\dfrac{A'C}{AC} = \dfrac{A'F}{PF}$

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