Reflection and spherical mirrors
Reflection means light bouncing back from a surface. It follows two laws. (1) The angle of incidence equals the angle of reflection, both measured from the normal (the line at right angles to the surface). (2) The incident ray, the reflected ray and the normal lie in one flat plane.
A spherical mirror is a part of a hollow shiny sphere. If the inside shines, it is concave (a bowl). If the outside shines, it is convex (a bulge).
- Pole P: the middle point of the mirror.
- Centre of curvature C: the centre of the sphere.
- Radius of curvature R: the distance PC.
- Principal axis: the line through P and C.
- Focus F: where rays parallel to the axis meet (concave) or seem to spread from (convex).
- Focal length f: the distance PF.
Why f = R/2
Take a ray parallel to the axis hitting the mirror at point M. The line CM is the normal there (a radius is always at right angles to a sphere). Call the angle of incidence θ. The reflected ray also makes θ with CM, and crosses the axis at F.
Because the incident ray is parallel to the axis, angle MCP is also θ. So in triangle MCF, two angles are θ, which means FM = FC. For a small mirror (rays close to the axis, called paraxial rays), M is very close to P, so FM ≈ FP. Hence FP = FC, so F is the midpoint of PC:
f = R / 2
This is exact only for paraxial rays. Rays far from the axis meet a little closer to the mirror. This blur is called spherical aberration.
Cartesian sign convention
- All distances are measured from the pole P.
- Light is drawn coming from the left. Distances in the direction of the light (to the right) are +, against it (to the left) are −.
- Heights above the axis are +, below are −.
So for a mirror, the object distance u is always negative. A concave mirror has f negative; a convex mirror has f positive. A real image (in front) has v negative; a virtual image (behind) has v positive.
Mirror formula: derivation
Put an object AB (B on the axis) in front of a concave mirror. Draw two rays from A: one parallel to the axis (it reflects through F) and one to the pole P (it reflects at the same angle). They meet at A′, and the image A′B′ stands at B′.
Step 1: two pairs of similar triangles
Triangles A′B′P and ABP are similar (equal angles at P), so B′A′ / BA = B′P / BP.
For a small mirror, the mirror is almost flat near P. Triangles A′B′F and MPF are similar, and MP = AB, so B′A′ / BA = B′F / FP.
Step 2: equate
B′P / BP = B′F / FP = (B′P − FP) / FP.
Step 3: put signs in
B′P = −v, BP = −u, FP = −f. So −v / −u = (−v + f) / −f, which gives v/u = (v − f)/f. Divide by v: 1/u = 1/f − 1/v, or
1/v + 1/u = 1/f
The same formula works for convex mirrors and for virtual images, as long as the signs are used correctly.
Magnification
Magnification m = image height ÷ object height = h′/h. From the similar triangles, m = h′/h = −v/u.
- m negative → image is inverted and real.
- m positive → image is erect and virtual.
- |m| > 1 → bigger; |m| < 1 → smaller.
You can also write m = f/(f − u) = (f − v)/f, which is handy when v or u is not given.
Images by a concave and a convex mirror
| Object | Image (concave) |
|---|---|
| At infinity | At F, real, inverted, point-sized |
| Beyond C | Between F and C, real, inverted, smaller |
| At C | At C, real, inverted, same size |
| Between C and F | Beyond C, real, inverted, bigger |
| At F | At infinity |
| Between F and P | Behind the mirror, virtual, erect, bigger |
A convex mirror always gives a virtual, erect, smaller image between P and F behind the mirror.
Try it: find f of a steel spoon
Hold a shiny steel spoon (inside facing you) at arm's length. Your face looks upside down: the image is real. Now bring it very close to one eye: your eye looks huge and upright. The switch happens when your eye crosses the focus. Flip the spoon: the back (convex) always shows you small and upright. In the 3D above, predict the image first, then drag the object to check.
Key formulas and definitions
- f = R / 2
- 1/v + 1/u = 1/f
- m = h′/h = −v/u
- m = f / (f − u) = (f − v) / f
- Concave: f < 0; Convex: f > 0 (Cartesian sign convention)
Worked examples
1. A concave mirror has radius of curvature 40 cm. Find its focal length.
Step 1: f = R/2. Step 2: R = −40 cm (C is in front). Step 3: f = −40/2 = −20 cm. Answer: focal length 20 cm (f = −20 cm).
2. An object is 30 cm in front of a concave mirror of focal length 15 cm. Find the image position and nature.
Step 1: u = −30 cm, f = −15 cm. Step 2: 1/v = 1/f − 1/u = −1/15 + 1/30 = −1/30. Step 3: v = −30 cm. Step 4: m = −v/u = −(−30)/(−30) = −1. Image at C, 30 cm in front, real, inverted, same size.
3. An object 2 cm tall is 10 cm in front of a concave mirror of focal length 15 cm. Find the image and its height.
Step 1: u = −10, f = −15. Step 2: 1/v = −1/15 + 1/10 = 1/30, so v = +30 cm (behind the mirror). Step 3: m = −v/u = −30/−10 = +3. Step 4: h′ = 3 × 2 = 6 cm. Virtual, erect, 3 times bigger.
4. A convex mirror has f = 20 cm. A car is 60 cm away from it. Where is the image and how big is it compared with the car?
Step 1: u = −60, f = +20. Step 2: 1/v = 1/20 + 1/60 = 4/60, v = +15 cm. Step 3: m = −15/−60 = +0.25. The image is 15 cm behind the mirror, virtual, erect and one quarter the size.
5. A concave mirror forms a real image 3 times the size of the object on a screen 60 cm from the mirror. Find the object distance and focal length.
Step 1: real image → v = −60 cm, m = −3. Step 2: m = −v/u → −3 = 60/u → u = −20 cm. Step 3: 1/f = 1/v + 1/u = −1/60 − 1/20 = −4/60. Step 4: f = −15 cm. Object 20 cm in front; focal length 15 cm.
6. Where should an object be placed in front of a concave mirror of focal length 12 cm to get a virtual image twice its size?
Step 1: virtual → m = +2, so v = −2u. Step 2: 1/(−2u) + 1/u = 1/f → 1/(2u) = 1/f. Step 3: u = f/2 = −12/2 = −6 cm. Place it 6 cm in front of the mirror (between P and F).
7. A 5 cm tall candle is 25 cm from a concave mirror of R = 30 cm. The candle is moved 10 cm closer. By how much does the image move?
Step 1: f = −15 cm. Step 2: first u = −25: 1/v = −1/15 + 1/25 = −2/75, v = −37.5 cm. Step 3: new u = −15 (at F): 1/v = −1/15 + 1/15 = 0, image goes to infinity. The image runs from 37.5 cm to very far away; at F no image forms on a screen.
8. Show that for a concave mirror, a real image is formed only when the object is beyond F, using the formula.
Step 1: write v = uf/(u − f) with u < 0, f < 0. Step 2: real image needs v < 0. Step 3: uf is positive (both negative), so we need u − f < 0, that is u < f, meaning |u| > |f|. So the object must be farther than F. If |u| < |f|, v > 0 and the image is virtual.
Common mistakes
- Forgetting signs: putting f = +15 cm for a concave mirror. For a concave mirror f is negative.
- Writing m = v/u for a mirror. For mirrors m = −v/u (for lenses it is v/u).
- Taking 1/v = 1/f + 1/u (the lens form) for a mirror. The mirror formula is 1/v + 1/u = 1/f.
- Thinking f = R/2 is exact for any size of mirror. It holds only for rays close to the axis (paraxial rays).