E. Physics: Light and Geometric Optics · Light and Geometric OpticsThe Thin Lens Equation: Predicting and Testing Images
E2.5 E3.5 A1.5 A1.8 A1.10 A1.13About 75 minutes
Learning goals- I can use the thin lens equation and magnification equation for converging and diverging lenses.
- I can use sign conventions to tell whether a lens image is real or virtual, upright or inverted.
- I can predict an image's position and size, then test the prediction with an investigation.
- I can calculate percentage difference and explain sources of error.
Words to know
thin lens equation
1/f = 1/dₒ + 1/dᵢ, relating focal length, object distance and image distance for a lens.
magnification equation
M = hᵢ/hₒ = −dᵢ/dₒ.
percentage difference
|predicted − measured| ÷ predicted × 100%.
prediction
A statement of what you expect to happen, based on a model, made before testing.
sources of error
Reasons why measured values differ from true values, such as limits of equipment or judgement.
Predict, then test
Engineers designing a phone camera need to know exactly how far the lens must be from the sensor. A ray diagram tells you roughly; an equation tells you precisely. Good news: lenses use the same equations as curved mirrors, with a slightly different meaning for the signs.
The thin lens equation and magnification equation
1/f = 1/dₒ + 1/dᵢ M = hᵢ / hₒ = −dᵢ / dₒ
Distances are measured from the optical centre of the lens.
Sign conventions for lenses
- f is positive for a converging lens and negative for a diverging lens.
- dₒ is positive for a real object (on the side the light comes from).
- dᵢ is positive for a real image on the far side of the lens (where light really goes), and negative for a virtual image on the same side as the object.
- hᵢ and M are positive for upright images and negative for inverted images.
Compare with mirrors: for a mirror, real images are in front (the side the light comes from and returns to); for a lens, real images are on the far side (where the light goes through to). In both cases: real = positive dᵢ.
Worked exampleExample 1: A camera
A camera lens has f = 50.0 mm. A person 1.80 m tall stands 4.00 m away. Where must the sensor be, and how tall is the image? Will it fit on a 24 mm tall sensor?
- Use consistent units: dₒ = 4.00 m = 4000 mm, hₒ = 1800 mm, f = +50.0 mm.
- 1/dᵢ = 1/f − 1/dₒ = 1/50.0 − 1/4000 = 0.0200 − 0.00025 = 0.01975 mm⁻¹.
- dᵢ = 1 ÷ 0.01975 = 50.6 mm. Positive: a real image behind the lens, where the sensor goes.
- M = −dᵢ/dₒ = −50.6 ÷ 4000 = −0.0127.
- hᵢ = M × hₒ = −0.0127 × 1800 mm = −22.8 mm.
The sensor must be 50.6 mm behind the lens. The image is 22.8 mm tall and inverted, so it just fits on a 24 mm sensor. To focus on closer objects, a camera moves the lens slightly farther from the sensor.
Worked exampleExample 2: A magnifying glass
A magnifying glass with f = 10.0 cm is held 6.0 cm above a 2.0 cm tall stamp. Find the image distance, magnification and image height.
- Given: f = +10.0 cm, dₒ = 6.0 cm, hₒ = 2.0 cm.
- 1/dᵢ = 1/10.0 − 1/6.0 = 0.100 − 0.1667 = −0.0667, so dᵢ = −15 cm.
- Negative: a virtual image on the same side as the stamp, 15 cm from the lens.
- M = −dᵢ/dₒ = −(−15)/6.0 = +2.5: upright, 2.5 times larger.
- hᵢ = 2.5 × 2.0 cm = 5.0 cm.
dᵢ = −15 cm, M = +2.5, hᵢ = 5.0 cm: a virtual, upright image 2.5 times larger.
Example 2 drawn to scale. The dashed extensions meet 15 cm from the lens on the stamp's side, at a height of 5.0 cm: exactly what the equations predicted.Worked exampleExample 3: Designing a projector
A projector lens has f = 10.0 cm and must form a sharp image on a screen 3.00 m away. How far from the lens must the 2.4 cm tall display be, and how tall is the image?
- Given: f = 10.0 cm, dᵢ = +300 cm (real image on the screen), hₒ = 2.4 cm.
- 1/dₒ = 1/f − 1/dᵢ = 1/10.0 − 1/300 = 0.1000 − 0.00333 = 0.0967, so dₒ = 10.3 cm.
- M = −dᵢ/dₒ = −300 ÷ 10.34 = −29.0.
- hᵢ = −29.0 × 2.4 cm = −70 cm.
The display must be 10.3 cm from the lens (just outside F′), and the image is 70 cm tall and inverted, which is why the display is mounted upside down.
Worked exampleExample 4: A diverging lens
An object is 30 cm from a diverging lens with f = −20 cm. Find dᵢ and M.
- 1/dᵢ = 1/f − 1/dₒ = −1/20 − 1/30 = −3/60 − 2/60 = −5/60.
- dᵢ = −12 cm (virtual, on the object's side).
- M = −(−12)/30 = +0.40.
dᵢ = −12 cm, M = +0.40: virtual, upright, smaller, as always for a diverging lens.
Investigation: Test the thin lens equation
Question: Does the thin lens equation correctly predict where a converging lens forms an image?
Method: Set up a lamp (object), a converging lens and a white screen on a metre stick. For several object distances, move the screen to find the sharpest image and record dₒ and dᵢ. Calculate f for each trial. Then predict dᵢ for a new object distance and test the prediction.
| dₒ (cm) | dᵢ (cm) | f = 1 ÷ (1/dₒ + 1/dᵢ) (cm) |
|---|
| 20.0 | 61.0 | 15.1 |
| 25.0 | 37.0 | 14.9 |
| 30.0 | 30.5 | 15.1 |
| 40.0 | 24.0 | 15.0 |
| 60.0 | 19.8 | 14.9 |
Prediction and test: using f = 15.0 cm, predict dᵢ for dₒ = 22.0 cm: 1/dᵢ = 1/15.0 − 1/22.0 = 0.0212, so dᵢ = 47.1 cm. The student measures 46.5 cm. Percentage difference = |46.5 − 47.1| ÷ 47.1 × 100% = 1.3%, well within the uncertainty of judging the sharpest image (about ±1 cm). The prediction is supported.
Sources of error: judging when the image is sharpest (the image looks sharp over a range of about 1 cm), measuring from the centre of a lens that has thickness, a lamp that is not a single point, and reading the metre stick.
Check every answer three ways
- Signs: does the sign of dᵢ match what you expect (real or virtual)?
- Ray-diagram pattern: for example, an object beyond 2F′ must give an image between F and 2F; an object inside F′ must give a virtual image.
- Magnitude: if |M| > 1, is the object between F′ and 2F′ (or inside F′)?
Just for funWhy did the student's virtual image refuse to come to class?
It said it was attending remotely, from the other side of the lens, with a negative distance.
Hands-on activityLens-equation lab with a magnifying glass
You need: a magnifying glass (a lens with f between about 10 cm and 30 cm), a small bright object (a phone flashlight shining through a paper cut-out of an arrow, or a candle only with a grown-up), a white card screen, a measuring tape, books to prop things up, a dark room.
- Find f roughly by focusing a distant window onto the card.
- Place the object at about 1.5f, 2f, 2.5f and 3f from the lens. Each time, move the screen until the image is sharpest. Record dₒ and dᵢ in a table, and note the image's size and attitude.
- Calculate f for each trial and the average.
- Predict dᵢ for a new dₒ (for example 1.8f) with the thin lens equation. Then test it, and calculate the percentage difference.
- Graph 1/dᵢ against 1/dₒ and find the intercepts. Write a conclusion: is the equation supported? Include sources of error.
What to look for: f values within about 5% of each other; the predicted and measured dᵢ within a few percent.
Safety (grown-up note): If a candle is used, a grown-up must light it and supervise; keep paper away from the flame. Never use the lens with sunlight.
Your turn
1Write the thin lens equation and the magnification equation.
1/f = 1/dₒ + 1/dᵢ and M = hᵢ/hₒ = −dᵢ/dₒ.
2For a lens, what does a positive image distance mean?
The image is real, on the far side of the lens.
3A converging lens has f = 12 cm. An object is 18 cm away. Find dᵢ and M.
1/dᵢ = 1/12 − 1/18 = 3/36 − 2/36 = 1/36 → dᵢ = 36 cm. M = −36/18 = −2.0 (real, inverted, twice as large).
4A converging lens has f = 20 cm. An object is 15 cm away. Find dᵢ and describe the image.
1/dᵢ = 1/20 − 1/15 = 3/60 − 4/60 = −1/60 → dᵢ = −60 cm. M = +4.0: virtual, upright, 4 times larger, on the object's side.
5A diverging lens has f = −15 cm. An object is 10 cm away. Find dᵢ.
1/dᵢ = −1/15 − 1/10 = −2/30 − 3/30 = −5/30 → dᵢ = −6.0 cm (virtual).
6An object 4.0 cm tall forms an image with M = −0.25. Find hᵢ and describe the image.
hᵢ = −0.25 × 4.0 = −1.0 cm: inverted, one-quarter the size (real for a single lens).
More practice: warm-up
1Is f positive or negative for a diverging lens?
2dₒ = 40 cm and dᵢ = 40 cm. Find f.
1/f = 1/40 + 1/40 = 1/20 → f = 20 cm.
3M = +3.0. Is the image upright or inverted?
4A virtual image formed by a lens is on which side?
The same side as the object.
5dᵢ = 30 cm and dₒ = 10 cm. Find M.
More practice: on your own
1A converging lens forms a sharp image on a screen 40 cm from the lens when the object is 8.0 cm from the lens. Find f, and check that the object position makes sense.
1/f = 1/8.0 + 1/40 = 5/40 + 1/40 = 6/40 → f = 6.7 cm. The object is between F′ (6.7 cm) and 2F′ (13 cm), which gives a larger real image beyond 2F, matching an image 40 cm away.
2A phone camera lens has f = 4.0 mm. How far behind the lens is the image of a building 50 m away? (Hint: work in mm.)
dₒ = 50 000 mm. 1/dᵢ = 1/4.0 − 1/50 000 = 0.25 − 0.00002 = 0.24998 → dᵢ ≈ 4.0 mm: for distant objects, the image is essentially at F.
3You want a same-size, inverted image with a converging lens of f = 25 cm. Where should the object be, and where will the image form?
Same size and inverted means M = −1, so dᵢ = dₒ. Then 1/25 = 2/dₒ → dₒ = 50 cm (at 2F′) and dᵢ = 50 cm (at 2F).
4A student predicts dᵢ = 32.0 cm but measures 33.5 cm. Calculate the percentage difference and suggest one reason for it.
|33.5 − 32.0| ÷ 32.0 × 100% = 4.7%. Possible reasons: difficulty judging the sharpest image, measuring to the lens surface instead of its centre, or an inaccurate value of f.
5An object 3.0 cm tall is 8.0 cm from a converging lens with f = 12 cm. Find dᵢ and hᵢ.
1/dᵢ = 1/12 − 1/8.0 = 2/24 − 3/24 = −1/24 → dᵢ = −24 cm. M = 24/8.0 = 3.0 → hᵢ = +9.0 cm (virtual, upright).
Challenge
1A converging lens forms a real image 4.0 times as tall as the object on a screen. The screen is 100 cm from the object. Find dₒ, dᵢ and f.
Real image → inverted: M = −4.0, so dᵢ = 4.0dₒ. dₒ + dᵢ = 100 → 5.0dₒ = 100 → dₒ = 20 cm, dᵢ = 80 cm. 1/f = 1/20 + 1/80 = 5/80 → f = 16 cm.
2Show that a converging lens cannot form a real image on a screen closer to the object than 4f.
For a real image, the object-to-image distance is D = dₒ + dᵢ. The smallest D happens when dₒ = dᵢ = 2f (try values: dₒ = 1.5f gives dᵢ = 3f, D = 4.5f; dₒ = 3f gives dᵢ = 1.5f, D = 4.5f; dₒ = 2f gives D = 4f). So the screen must be at least 4f from the object.
3A camera with an f = 50.0 mm lens is focused on a flower 0.50 m away. By how much must the lens move compared with focusing on a very distant mountain?
Mountain: dᵢ ≈ f = 50.0 mm. Flower: 1/dᵢ = 1/50.0 − 1/500 = 0.0200 − 0.0020 = 0.0180 → dᵢ = 55.6 mm. The lens must move about 5.6 mm farther from the sensor.
Quiz
1. The thin lens equation is:
2. For a lens, a negative image distance means the image is:
3. A converging lens has f = 10 cm and the object is at 30 cm. dᵢ is:
4. A converging lens (f = 15 cm) has an object at 10 cm. The image is:
5. If M = −2.0 for a lens, the image is:
6. The focal length of a diverging lens is:
7. dₒ = 30 cm, dᵢ = 60 cm for a converging lens. f is:
Big idea: Expectation E2.5 asks students to predict the position and characteristics of an image with ray diagrams and algebraic equations, and to test their predictions through inquiry. The thin lens equation has the same form as the mirror equation; only the meaning of the sides changes (real images are on the far side of a lens).
Common difficulties:
- Forgetting the reciprocal at the end (finding 1/dᵢ and stopping).
- Units. Keep everything in the same unit (all cm or all mm).
- Signs. f is negative for diverging lenses; a negative dᵢ means virtual.
How to help: The lab works well on a kitchen table with a magnifying glass and a phone flashlight. Ask your teen to predict before measuring, then compare. If you use a candle, light it yourself and supervise.