E. Physics: Light and Geometric Optics · Light and Geometric Optics

The Mirror and Magnification Equations

E3.3 E2.3 A1.8 A1.12 A1.13About 75 minutes

Learning goals

I can use the mirror equation 1/f = 1/dₒ + 1/dᵢ to find an image or focal length.
I can use M = hᵢ/hₒ = −dᵢ/dₒ to find magnification and image height.
I can use the sign convention to tell whether an image is real or virtual, upright or inverted.
I can check equation answers against ray diagrams and analyse measured data.

Words to know

mirror equation
1/f = 1/dₒ + 1/dᵢ, which relates focal length, object distance and image distance.

magnification (M)
The ratio of image height to object height: M = hᵢ/hₒ = −dᵢ/dₒ.

object distance (dₒ)
The distance from the object to the mirror's vertex.

image distance (dᵢ)
The distance from the image to the mirror's vertex; negative for a virtual image.

sign convention
Rules for positive and negative values that show whether images are real or virtual and upright or inverted.

reciprocal
1 divided by a number; the reciprocal of 20 is 1/20 = 0.050.

From drawings to equations

Ray diagrams are great for seeing what kind of image forms. But if an engineer needs to know exactly where to put a sensor in a telescope, a drawing is not precise enough. Physicists use two equations that give the exact location and size of the image. They are the algebraic version of the ray rules you already know.

The mirror equation

1/f = 1/dₒ + 1/dᵢ

f = focal length
dₒ = object distance (from the object to the vertex)
dᵢ = image distance (from the image to the vertex)

The magnification equation

M = hᵢ / hₒ = −dᵢ / dₒ

M = magnification (no units)
hᵢ = image height, hₒ = object height

If |M| > 1 the image is larger than the object; if |M| < 1 it is smaller; if |M| = 1 it is the same size.

Sign conventions

Because images can be in front of or behind the mirror, and upright or inverted, we use signs to carry that information:

Distances to real objects and images (in front of the mirror) are positive.
Distances to virtual images (behind the mirror) are negative.
f is positive for concave mirrors and negative for convex mirrors.
Heights above the principal axis (upright) are positive; heights below (inverted) are negative. So a positive M means upright, and a negative M means inverted.

The sign convention. Everything on the side the light comes from (in front of the mirror) is positive; image distances behind the mirror are negative. Upright heights are positive and inverted heights are negative.

Worked example

Example 1: An object beyond C

A 4.0 cm tall candle is 30.0 cm in front of a concave mirror with f = 10.0 cm. Find the image distance, the magnification and the image height. Describe the image.

Given: f = +10.0 cm (concave), dₒ = 30.0 cm, hₒ = 4.0 cm. Required: dᵢ, M, hᵢ.
Mirror equation: 1/dᵢ = 1/f − 1/dₒ = 1/10.0 − 1/30.0 = 3/30.0 − 1/30.0 = 2/30.0.
dᵢ = 30.0 ÷ 2 = 15.0 cm. Positive, so the image is real, in front of the mirror.
Magnification: M = −dᵢ/dₒ = −15.0/30.0 = −0.500. Negative, so the image is inverted; |M| < 1, so it is smaller.
Image height: hᵢ = M × hₒ = −0.500 × 4.0 cm = −2.0 cm (2.0 cm tall, pointing down).

dᵢ = 15.0 cm, M = −0.500, hᵢ = −2.0 cm. The image is real, inverted, half the size, 15.0 cm in front of the mirror (between C at 20 cm and F at 10 cm), which matches the ray diagram rules.

Example 1 drawn to scale (f = 10.0 cm, C at 20.0 cm). The rays cross 15.0 cm from the mirror, and the image is half as tall as the object and inverted, exactly as the equations predicted.

Worked example

Example 2: A magnifying makeup mirror

A concave mirror has f = 15 cm. Your face is 10 cm away. Where is the image, and what is its magnification?

Given: f = +15 cm, dₒ = 10 cm.
1/dᵢ = 1/15 − 1/10 = 2/30 − 3/30 = −1/30, so dᵢ = −30 cm.
Negative dᵢ: the image is virtual, 30 cm behind the mirror.
M = −dᵢ/dₒ = −(−30)/10 = +3.0. Positive: upright; 3.0: three times larger.

dᵢ = −30 cm, M = +3.0. The image is virtual, upright, 3 times larger, behind the mirror (the face is inside F).

Worked example

Example 3: A convex security mirror

A convex security mirror in a store has a focal length of −0.12 m (negative because it is convex). A shopper 1.6 m tall stands 2.4 m from the mirror. Find the image distance and the image height.

Given: f = −0.12 m, dₒ = 2.4 m, hₒ = 1.6 m.
1/dᵢ = 1/f − 1/dₒ = 1/(−0.12) − 1/2.4 = −8.333 − 0.417 = −8.750 m⁻¹.
dᵢ = 1 ÷ (−8.750) = −0.114 m: about 11 cm behind the mirror (virtual), between V and F.
M = −dᵢ/dₒ = −(−0.114)/2.4 = +0.0476.
hᵢ = M × hₒ = 0.0476 × 1.6 m = 0.076 m = 7.6 cm.

dᵢ ≈ −0.11 m and hᵢ ≈ 7.6 cm: a tiny, upright, virtual image just behind the mirror, which is why a whole aisle fits in one security mirror.

Worked example

Example 4: Finding the focal length from measurements

In a lab, a lamp 30.0 cm from a concave mirror makes a sharp image on a screen 60.0 cm from the mirror. Find f and R.

The image is on a screen, so it is real: dᵢ = +60.0 cm. dₒ = 30.0 cm.
1/f = 1/dₒ + 1/dᵢ = 1/30.0 + 1/60.0 = 2/60.0 + 1/60.0 = 3/60.0.
f = 60.0 ÷ 3 = 20.0 cm, and R = 2f = 40.0 cm.

f = 20.0 cm, R = 40.0 cm.

A problem-solving routine (GRASS)

Given: list the values with signs (concave f +, convex f −).
Required: what you need to find.
Analysis: choose the equation(s).
Solution: substitute and solve. On a calculator, use the reciprocal key (x⁻¹ or 1/x): for 1/dᵢ = 1/15 − 1/10, type 15 x⁻¹ − 10 x⁻¹ = then x⁻¹ again.
Statement: answer with units and the right number of significant digits, then interpret the signs (real or virtual? upright or inverted?) and check against a ray diagram.

Testing the equation with data

A student measures object and image distances for a concave mirror (f is unknown) and calculates f for each trial:

dₒ (cm)	dᵢ (cm)	f = 1 ÷ (1/dₒ + 1/dᵢ) (cm)
30.0	61.0	20.1
40.0	39.5	19.9
50.0	33.0	19.9
60.0	30.5	20.2

Sample data. The four values of f agree within 0.3 cm; the average is 20.0 cm.

Another way to analyse the data: graph 1/dᵢ against 1/dₒ. The mirror equation says 1/dᵢ = 1/f − 1/dₒ, a straight line with slope −1 whose intercepts both equal 1/f = 0.050 cm⁻¹, so f = 20 cm.

Just for fun

Why was the virtual image always so negative?

It lived behind the mirror, where every distance gets a minus sign.

Hands-on activity

Measure the focal length of a mirror two ways

You need: the concave makeup mirror from earlier lessons, a small bright object (a phone flashlight covered with a piece of paper with an arrow cut out, or a small LED light), a white card screen, a measuring tape, a calculator, a dim room.

Method 1 (distant object): form an image of a distant window on the card. Record the distance as f₁.
Method 2 (mirror equation): place the object at four distances between about 1.5f₁ and 3f₁. For each, move the screen to find a sharp image. Record dₒ and dᵢ in a table.
Calculate f for each trial with 1/f = 1/dₒ + 1/dᵢ, and find the average. Round to the right number of significant digits.
Compare f₁ with your average. Calculate the percentage difference: |f₁ − f_avg| ÷ f_avg × 100%.
List two sources of error (for example, judging when the image is sharpest; measuring from the vertex).

What to look for: the values of f from method 2 should agree within a few percent; method 1 usually gives a slightly larger value because the window is not infinitely far away.

Safety (grown-up note): Keep the mirror out of direct sunlight.

Your turn

Write the mirror equation and the magnification equation.

What does a negative image distance tell you?

A concave mirror has f = 12 cm. An object is 36 cm away. Find dᵢ and M.

A concave mirror has f = 8.0 cm. An object is 4.0 cm away. Find dᵢ and M, and describe the image.

A convex mirror has f = −20 cm. An object is 20 cm away. Find dᵢ and M.

An image has M = −3.0. Describe it.

More practice: warm-up

Is f positive or negative for a convex mirror?

M = +0.25. Is the image upright or inverted, larger or smaller?

dₒ = 20 cm, dᵢ = 20 cm. Find f.

hₒ = 5.0 cm and M = −2.0. Find hᵢ.

What does a positive image distance tell you about a mirror image?

More practice: on your own

A concave mirror with R = 50 cm forms an image of an object placed 75 cm away. Find dᵢ, M and the image height if the object is 6.0 cm tall.

An object 2.0 cm tall is 15 cm in front of a convex mirror with f = −10 cm. Find dᵢ and hᵢ.

A concave mirror forms a real image 3.0 times as tall as the object, 45 cm from the mirror. Find dₒ and f.

Use the mirror equation to show that an object at C (dₒ = 2f) gives an image at C.

What happens in the mirror equation when dₒ = f? Explain what this means physically.

Challenge

A concave mirror with f = 12 cm forms an upright image 3.0 times the size of the object. Where is the object?

Show, using the mirror equation, that a convex mirror always gives a virtual image for any real object (dₒ > 0).

A dentist wants a mirror that gives an upright image 2.0 times larger when held 1.5 cm from a tooth. What focal length (and type) of mirror is needed?

Check your understanding

Quiz

The mirror equation is:

A concave mirror has f = 10 cm and an object is at 15 cm. The image distance is:

A magnification of −2.0 means the image is:

In the sign convention used here, the focal length of a convex mirror is:

A mirror forms an image with dᵢ = −12 cm. The image is:

dₒ = 40 cm and dᵢ = 40 cm for a concave mirror. The focal length is:

A convex mirror (f = −15 cm) has an object 30 cm away. The magnification is:

For parents and teachers

Big idea: Expectation E3.3 asks students to describe image characteristics using ray diagrams and algebraic equations where appropriate. The mirror equation and magnification equation turn ray diagrams into numbers. A1.12 and A1.13 (symbols, units, significant digits) are practised here too.

Common difficulties:

Reciprocals. Students often add 1/dₒ + 1/dᵢ and forget to flip the result. Encourage writing "1/dᵢ = ..." then "dᵢ = ..." on separate lines.
Signs. Concave f is +, convex f is −; a negative dᵢ means virtual; negative M means inverted.
Checking. Every answer should be checked against the ray-diagram rules (e.g., an object inside F must give a virtual image).

How to help: Ask your teen to explain what each sign in their answer means in words. If they have a scientific calculator, help them find the reciprocal (x⁻¹) key.