Reflection and spherical mirrors: the mirror formula
Every spherical mirror obeys the laws of reflection. With the New Cartesian sign convention, all mirrors follow one equation, 1/v + 1/u = 1/f, with f = R/2.
The words you need
Light reflects from a spherical mirror by the two laws of reflection. (1) The angle of reflection equals the angle of incidence. (2) The incident ray, the reflected ray and the normal at the point of incidence lie in one plane. For a spherical mirror, the normal at any point lies along the radius: the line joining that point to the centre of curvature C. The pole P is the centre of the mirror's surface. The principal axis is the line through P and C. Paraxial rays are rays close to the principal axis that make small angles with it. Paraxial rays parallel to the axis meet at the principal focus F (concave mirror), or seem to spread out from it (convex mirror). The distance PF is the focal length f. Simple geometry with paraxial rays gives f = R/2, where R = PC is the radius of curvature.
The New Cartesian sign convention
- Your textbook calls it the Cartesian sign convention. Use it for every mirror, surface and lens in this chapter.
- Measure all distances from the pole of a mirror (or the optical centre of a lens).
- Draw the object on the left, so the incident light travels from left to right.
- Distances measured in the direction of the incident light are positive. Distances measured against it are negative.
- Heights measured upwards from the principal axis are positive. Heights measured downwards are negative.
- So a real object always has u negative. A concave mirror has f and R negative; a convex mirror has f and R positive.
- For a mirror, a real image forms in front of it (v negative). A virtual image forms behind it (v positive).
- Put the signs in only once, when you substitute numbers. Never change the formula itself.
Derivation: the mirror equation and magnification
Take a concave mirror forming a real image A′B′ of an object AB placed beyond C. Use paraxial rays, so the mirror is nearly flat near P.
- 1A ray from the top A of the object, parallel to the axis, strikes the mirror at M and reflects through F. A second ray from A strikes the pole P and reflects at an equal angle.
- 2Triangles A′B′P and ABP are similar (equal angles at P): B′A′/BA = B′P/BP. ...(1)
- 3Triangles A′B′F and MPF are similar, and PM = BA for paraxial rays: B′A′/BA = B′F/FP. ...(2)
- 4From (1) and (2): B′F/FP = B′P/BP. Since B′F = B′P − FP, this gives (B′P − FP)/FP = B′P/BP. ...(3)
- 5Apply the sign convention: B′P = −v, FP = −f and BP = −u.
- 6Then (−v + f)/(−f) = (−v)/(−u), which gives (v − f)/f = v/u.
- 7Divide both sides by v and rearrange: 1/f − 1/v = 1/u.
- 8★So, the mirror equation is 1/v + 1/u = 1/f.
- 9★In (1), put B′A′ = −h′ (the image is below the axis) and BA = h: −h′/h = (−v)/(−u). So, m = h′/h = −v/u.
★ key step
How to draw the ray diagram for this derivation
- Draw a horizontal principal axis. Draw a concave mirror on the right as an arc, with its hollow side facing left towards the object.
- Mark P where the mirror meets the axis. Mark F and C on the axis to the left of P, with PC = 2 × PF.
- Draw the object AB as an upright arrow beyond C, with B on the axis.
- From A, draw a ray parallel to the axis to the mirror at M. Reflect it back through F.
- From A, draw a ray to P. Reflect it below the axis, making the same angle with the axis.
- Where the two reflected rays meet, mark A′. Drop A′B′ to the axis: a real, inverted, smaller image between F and C.
- Draw a dotted line from M down to the axis. Label A, B, A′, B′, M, P, F, C, u, v and f, and put arrows on every ray.
Worked example: concave and convex mirrors
(a) A 2 cm tall object stands 30 cm in front of a concave mirror of focal length 20 cm. Find the position, size and nature of the image. (b) Repeat with the object 10 cm from the same mirror. (c) A scooter's convex rear-view mirror has a radius of curvature of 2 m. A bus is 9 m behind it. Find the position and magnification of the bus's image.
- (a) Signs: u = −30 cm, f = −20 cm (concave), h = +2 cm.
- 1/v = 1/f − 1/u = −1/20 + 1/30 = (−3 + 2)/60 = −1/60, so v = −60 cm.
- m = −v/u = −(−60)/(−30) = −2, so h′ = m × h = −2 × 2 cm = −4 cm.
- v and m are negative: the image is real, inverted and 4 cm tall, 60 cm in front of the mirror.
- (b) u = −10 cm, f = −20 cm: 1/v = −1/20 + 1/10 = (−1 + 2)/20 = 1/20, so v = +20 cm.
- m = −v/u = −(+20)/(−10) = +2: a virtual, erect image, twice as tall, 20 cm behind the mirror.
- (c) Convex mirror: f = R/2 = +1 m. The bus is in front of the mirror, so u = −9 m.
- 1/v = 1/f − 1/u = 1/1 + 1/9 = 10/9, so v = +0.9 m.
- m = −v/u = −(0.9)/(−9) = +0.1: virtual, erect, one-tenth the size, 0.9 m behind the mirror.
(a) v = −60 cm; real, inverted, 4 cm tall. (b) v = +20 cm; virtual, erect, m = +2. (c) v = +0.9 m, m = +0.1; virtual and erect
Activity: focal length of a concave mirror and of a convex mirror
Two practical experiments on an optical bench, using the no-parallax method. A needle and an image coincide when they stay together as you move your eye sideways.
- 1Concave mirror (u and v): Fix the mirror at one end of the optical bench. Place the object needle beyond F.
- 2Look along the axis at the inverted image. Move the image needle until its tip touches the image tip with no parallax.
- 3Record u and v (both negative) for five or six object positions. Work out f = uv/(u + v) for each, and take the mean.
- 4Result: the mean f is negative, as for any concave mirror. With signs, a graph of 1/v against 1/u is a straight line that cuts both axes at 1/f.
- 5Convex mirror (using a convex lens): With a convex lens L alone, form a real image I of the object needle O. Mark the position of I.
- 6Place the convex mirror between L and I. Slide it until the image of O, after reflection, coincides with O itself with no parallax.
- 7The rays now strike the mirror normally, so I is its centre of curvature. R = distance from the mirror to I, and f = +R/2.
- 8Result: the focal length of the convex mirror is positive and equal to half the measured R.
A dentist holds a small concave mirror close to a tooth to see an enlarged, erect image. Scooters and buses use convex rear-view mirrors, which show a wide view of the road behind, though vehicles look smaller and farther away.
2 marks: “Show that the focal length of a spherical mirror is half its radius of curvature.” A paraxial ray parallel to the axis meets the mirror at M; CM is the normal and the angle of incidence is θ. The reflected ray passes through F, so ∠MFP = 2θ. With D the foot of the perpendicular from M: tan θ = MD/CD and tan 2θ = MD/FD. For small θ, tan θ ≈ θ, so MD/FD = 2 MD/CD and FD = CD/2. D is close to P, so f = R/2.
3 marks: “Derive the mirror equation 1/v + 1/u = 1/f for a concave mirror. State the sign convention you use.” State the New Cartesian sign convention; ray diagram with the object beyond C; similar triangles give B′A′/BA = B′P/BP and B′A′/BA = B′F/FP; put B′P = −v, FP = −f, BP = −u; simplify to 1/v + 1/u = 1/f.
2 marks: “An object is 15 cm in front of a convex mirror of radius of curvature 30 cm. Find the position and nature of the image.” f = +15 cm, u = −15 cm. 1/v = 1/f − 1/u = 1/15 + 1/15 = 2/15, so v = +7.5 cm. m = −v/u = +0.5. The image is virtual, erect and half the size, 7.5 cm behind the mirror.
Mirror: plus in the formula (1/v + 1/u = 1/f), minus in the magnification (m = −v/u). A lens does the opposite.