Light – Reflection and Refraction · Lesson 7 of 15
Sign Convention for Reflection by Spherical Mirrors
“One tiny sign mistake can send the image to the wrong side of the universe.”
• Apply the New Cartesian Sign Convention to spherical mirrors. • Use the mirror formula to find object, image or focal distance. • Calculate magnification and image height. • Interpret signs to determine image nature and orientation. • Solve multi-step mirror problems and check their reasonableness.
A ray diagram predicts an image visually, but calculations require every direction to have an agreed sign. Without a sign convention, the same formula would give ambiguous results for images in front of and behind a mirror. The New Cartesian convention turns the principal axis into a number line.
New Cartesian Sign Convention
- Take the pole P as the origin and the principal axis as the x-axis.
- Place the object on the left, so incident light travels from left to right.
- Measure every axial distance from P.
- Distances to the right of P are positive; distances to the left are negative.
- Heights above the principal axis are positive; heights below it are negative.
| Quantity | Common sign |
|---|---|
| Object distance u | Negative for a real object on the left |
| Concave-mirror focal length f and radius R | Negative because F and C are on the left |
| Convex-mirror focal length f and radius R | Positive because F and C are behind the mirror |
| Real-image distance v | Negative when the image is in front |
| Virtual-image distance v | Positive when the image is behind |
| Erect image height h′ | Positive |
| Inverted image height h′ | Negative |
Mirror Formula
A reliable method is to list the given quantities with signs, identify the unknown, choose the mirror formula, isolate the unknown algebraically, substitute values with units, and then interpret the sign of the result. A numerical value is incomplete until its physical meaning is stated.
Magnification
The magnitude |m| compares sizes. If |m| > 1 the image is enlarged; if |m| = 1 it is the same size; if |m| < 1 it is diminished. Once m is known, image height follows from h′ = mh.
Problem-Solving Strategy
- Sketch the mirror and object region to predict the likely result.
- Convert all lengths to one unit.
- Write u, f and any known v with Cartesian signs.
- Use the mirror formula and simplify fractions carefully.
- Use m = -v/u and h′ = mh if size is required.
- Check whether the signs and magnitude agree with the ray-diagram case.
Problem
A convex mirror has radius of curvature 32 cm. Find its focal length.
- 1.For a convex mirror, R = +32 cm because C is behind the mirror.
- 2.Use R = 2f, so f = R/2.
- 3.f = +32/2 = +16 cm.
- 4.The positive sign correctly places F behind the convex mirror.
Problem
A convex rear-view mirror has R = 3.00 m. A bus is 5.00 m in front. Find image position and magnification.
- 1.Given R = +3.00 m, so f = +1.50 m; u = -5.00 m.
- 2.From 1/v = 1/f - 1/u, 1/v = 1/1.50 - 1/(-5.00) = 0.6667 + 0.2000 = 0.8667 m⁻¹.
- 3.Therefore v = +1.15 m. The image is 1.15 m behind the mirror.
- 4.m = -v/u = -(+1.15)/(-5.00) = +0.23.
- 5.Positive m means erect and virtual; |m| < 1 means diminished to about 0.23 times the object size.
Problem
A 4.0 cm object is 25.0 cm in front of a concave mirror of focal length 15.0 cm. Find screen position and image height.
- 1.Given h = +4.0 cm, u = -25.0 cm and f = -15.0 cm.
- 2.1/v = 1/f - 1/u = -1/15 + 1/25 = (-5 + 3)/75 = -2/75 cm⁻¹.
- 3.Thus v = -37.5 cm. Place the screen 37.5 cm in front of the mirror.
- 4.m = -v/u = -(-37.5)/(-25.0) = -1.5.
- 5.h′ = mh = -1.5 × 4.0 = -6.0 cm.
- 6.The negative height agrees with a real inverted image; its 6.0 cm magnitude shows enlargement.
Do not give a concave mirror a positive focal length merely because length is usually positive. Do not insert unsigned distances and add a sign afterward. Do not confuse the minus sign in m = -v/u with the lens formula, which uses a different relationship.
Quiz
What is the sign of focal length for a concave mirror?
What does negative mirror magnification indicate?
For u = -20 cm and v = -40 cm, what is m?
A computed v is positive for a mirror. Where is the image?
Which check best detects a sign mistake?
Practice Problems
- A concave mirror has R = 24 cm. Find f with sign. Solution: R = -24 cm, so f = R/2 = -12 cm.
- For u = -30 cm and f = -15 cm, find v. Solution: 1/v = -1/15 + 1/30 = -1/30, so v = -30 cm. The image is real at C.
- An image has m = +0.4 and h = 10 cm. Find h′ and describe it. Solution: h′ = mh = +4 cm. Positive m means erect and virtual; |m| < 1 means diminished.
- A concave mirror forms a three-times enlarged real image of an object 10 cm in front. Locate the image. Solution: Real enlarged means m = -3. Using m = -v/u with u = -10 cm: -3 = -v/(-10) = v/10, so v = -30 cm. It is 30 cm in front.
- A convex mirror has f = +20 cm and u = -60 cm. Find v and m. Solution: 1/v = 1/20 + 1/60 = 4/60 = 1/15, so v = +15 cm. m = -(15)/(-60) = +0.25: virtual, erect and diminished.
Key Takeaways
• All mirror distances are measured from P using Cartesian signs. • Concave f is negative and convex f is positive. • The mirror formula relates u, v and f. • Mirror magnification is m = h′/h = -v/u. • The sign of m identifies orientation and, here, image nature. • The magnitude of m identifies enlargement or diminution. • A ray-diagram prediction is an important check on a calculation.