Mirrors — plane, concave, convex; mirror equation 1 - v + 1 - u = 1 - f

1. The three mirrors (WHAT)

2. WHY focus is halfway: deriving $f = R/2$ (Derivation from scratch)

Let the incoming ray be parallel to the axis. It hits $M$ and reflects to cross the axis at $F$.

• The incident ray is parallel to axis, so it makes angle $\theta$ with normal $MC$.
• By law of reflection, reflected ray also makes $\theta$ with $MC$.
• The line $MC$ cuts the axis at $C$, making angle $\theta$ (alternate angles, since ray ∥ axis).

So in triangle $MFC$, angles $\angle MCF = \theta$ and $\angle FMC = \theta$ ⇒ isosceles ⇒ $FM = FC$.

For a paraxial ray (close to axis, small $\theta$), $M$ is near $P$, so $FM \approx FP$. Therefore: $FP = FC \;\Rightarrow\; F \text{ is the midpoint of } PC \;\Rightarrow\; PF = \tfrac{1}{2}PC$

3. Sign convention (HOW we keep signs honest)

4. Deriving the mirror equation $\tfrac1v+\tfrac1u=\tfrac1f$ (from scratch)

We use a concave mirror with object beyond C, forming a real, inverted image. Object height $h$, image height $h'$.

Step 1 — Magnification from the pole ray. A ray from the top of the object to the pole $P$ reflects symmetrically about the axis (axis is the normal at $P$). Two right triangles (object & image with the axis) are similar: $\frac{h'}{h} = \frac{\text{image distance}}{\text{object distance}}$ Why this step? The pole ray makes equal angles with the axis, so the triangles share equal angles ⇒ similar ⇒ ratios of sides equal. With signs, $h'$ and $h$ have opposite sign (image inverted), giving magnification: $m = \frac{h'}{h} = -\frac{v}{u}$

Step 2 — Use the focus ray and similar triangles. Take the ray parallel to the axis (height $h$); it reflects through $F$. Consider triangles formed at the pole and at $F$. Comparing the parallel ray triangle (from $P$, height $h$) and the image triangle (from $F$, height $h'$): $\frac{h'}{h} = \frac{PF - \text{(image side)}}{PF}\cdots$ Doing this with proper signed lengths $u, v, f$ (all algebra, no new physics) yields: $\frac{v - f}{f} = -\frac{h'}{h} = \frac{v}{u}$

Step 3 — Solve. $\frac{v-f}{f} = \frac{v}{u} \;\Rightarrow\; uv - uf = vf \;\Rightarrow\; uv = vf + uf$ Divide by $uvf$: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

5. Worked examples

6. Common mistakes (Steel-man + fix)

7. Active recall

Connections

• Reflection of Light — Laws & Normal
• Refraction & Lenses — Lens Maker's Equation (same algebra, different signs)
• Spherical Aberration (why $f=R/2$ is only paraxial)
• Magnification & Image Formation
• Optical Instruments — Telescopes & Microscopes
• Total Internal Reflection

Concept Map

Hinglish (regional understanding)