Light – Reflection and Refraction · Lesson 14 of 15
Power of a Lens
“A short focal length means a powerful lens with very little patience.”
• Define lens power and the dioptre. • Convert between focal length and power. • Interpret positive and negative lens prescriptions. • Calculate the net power of lenses in contact. • Review the chapter's formulas and solve mixed problems.
Two lenses may look similar but bend light with very different strengths. Focal length describes where rays meet; power expresses the same ability as a convenient reciprocal. A short focal length corresponds to a large magnitude of power.
The reciprocal of the focal length expressed in metres.
The power of a lens with focal length one metre; 1 D = 1 m⁻¹.
Before calculating power, convert centimetres to metres. For example, +50 cm = +0.50 m and gives +2.0 D. The sign identifies lens type; the magnitude identifies strength.
Lenses in Contact
Lens systems combine powers to improve magnification and image quality. Cameras, microscopes and telescopes commonly use several elements. A positive and negative lens can partially cancel, producing a longer effective focal length than either unopposed positive component.
Problem
Find the power of a convex lens with f = +50 cm.
- 1.Convert f = +0.50 m.
- 2.P = 1/f = 1/0.50 = +2.0 D.
- 3.The positive sign confirms a convex converging lens.
Problem
Find the focal length and type of a lens prescribed as -2.5 D.
- 1.Use f = 1/P.
- 2.f = 1/(-2.5) = -0.40 m = -40 cm.
- 3.Negative f and P identify a concave diverging lens.
Problem
Lenses of +2.0 D and +0.25 D are in contact. Find net power and effective focal length.
- 1.Add algebraically: P = +2.0 + 0.25 = +2.25 D.
- 2.Use f = 1/P = 1/2.25 = 0.444 m approximately.
- 3.Thus f ≈ +44.4 cm.
- 4.Positive net power means the combination is converging.
Chapter Summary And Practice
Light travels in straight lines within a uniform medium in the ray model. Reflection returns light to the same medium and obeys equal incidence and reflection angles. Plane mirrors give virtual, erect, same-sized, laterally inverted images. Spherical mirrors use P, F and C; concave mirrors can form real or virtual images, while convex mirrors always give virtual erect diminished images.
Refraction is the change in direction at a boundary caused by a change in speed. A ray bends toward the normal on entering an optically denser medium and away on entering a rarer one. A parallel-sided slab produces an emergent ray parallel to the incident ray but laterally shifted. Refractive index connects bending with speed and must not be confused with mass density.
Convex lenses converge and concave lenses diverge. A convex lens has six object-position cases parallel to those of a concave mirror, but real images lie on the opposite side of the lens. A concave lens always produces a virtual erect diminished image for a real object. Ray diagrams and signed formulas must tell the same physical story.
Formula Review
| Relationship | Use |
|---|---|
| R = 2f | Connect radius and focal length of a small-aperture spherical mirror |
| 1/v + 1/u = 1/f | Find mirror object, image or focal distance |
| m = h′/h = -v/u | Find mirror size ratio and orientation |
| sin i/sin r = constant | Relate incidence and refraction for fixed media and colour |
| n = c/v | Relate absolute refractive index and light speed |
| 1/v - 1/u = 1/f | Find lens object, image or focal distance |
| m = h′/h = v/u | Find lens size ratio and orientation |
| P = 1/f | Convert focal length in metres to power in dioptres |
| P = P₁ + P₂ + … | Find net power of thin lenses in contact |
Common Conceptual Mistakes
- Measuring angles from a surface instead of the normal.
- Treating virtual images as unreal; they are visible but cannot be projected.
- Giving all focal lengths a positive sign.
- Using the mirror magnification sign in a lens calculation.
- Confusing optical density with mass density.
- Using centimetres directly in P = 1/f when power in dioptres is required.
Quiz
Which lens has positive power?
What is the power of a lens with f = 0.25 m?
What is the net power of +3 D and -1 D in contact?
Which formula belongs to a spherical lens?
Why must centimetres be converted before finding dioptres?
Practice Problems
- Find P for f = -2 m. Solution: P = 1/(-2) = -0.5 D, so the lens is concave.
- Find f for P = +4 D. Solution: f = 1/4 = +0.25 m = +25 cm; the lens is convex.
- Find net power of +5 D, -2 D and +0.5 D. Solution: P = 5 - 2 + 0.5 = +3.5 D. Effective f = 1/3.5 ≈ +0.286 m.
- A concave mirror has f = -10 cm and u = -15 cm. Find v. Solution: 1/v = -1/10 + 1/15 = -1/30, so v = -30 cm: real and inverted.
- A convex lens has f = +15 cm and u = -10 cm. Find v, m and image type. Solution: 1/v = 1/15 - 1/10 = -1/30, so v = -30 cm. m = (-30)/(-10) = +3. The image is virtual, erect and three times enlarged.
Key Takeaways
• Power is the reciprocal of focal length in metres. • One dioptre equals one inverse metre. • Convex power is positive and concave power is negative. • Powers of thin lenses in contact add algebraically. • Ray diagrams, sign conventions and formulas must agree. • Reflection and refraction are linked by precise geometric relationships. • Unit conversion and sign interpretation are essential checks in optical calculations.