Light – Reflection and Refraction · Lesson 13 of 15
Image Formation in Lenses Using Ray Diagrams
“Three clever rays reveal where every lens image is hiding.”
• Apply the three standard lens-ray rules. • Construct convex- and concave-lens ray diagrams. • Apply Cartesian signs to lens quantities. • Use the lens formula and magnification relationship. • Interpret calculated signs, sizes and image positions.
Lens tables become easier to understand when every result can be rebuilt from three ray rules. Calculations then express the same geometry numerically, provided signs are assigned from the optical centre rather than a mirror pole.
Standard Ray Rules
- A ray parallel to the axis passes through F₂ after a convex lens, or appears to come from F₁ after a concave lens.
- A ray through F₁ of a convex lens emerges parallel; a ray directed toward F₂ of a concave lens emerges parallel.
- A ray through optical centre O passes without appreciable deviation.
Use solid lines for actual rays and dashed backward extensions for a virtual image. For a convex lens, two refracted rays meet on the opposite side when the object is outside F₁. Inside F₁ they diverge and extensions meet on the object side. For a concave lens, extensions always meet between F₁ and O.
Sign Convention for Spherical Lenses
- Take O as origin and incident light from left to right.
- Axial distances to the right are positive and to the left negative.
- Heights above the axis are positive and below negative.
- A convex lens has positive f; a concave lens has negative f.
Lens Formula and Magnification
Solve by sketching the expected case, listing signed data, using 1/v = 1/f + 1/u, then calculating m = v/u and h′ = mh. Check that a convex lens outside F₁ gives positive v and negative m, while a concave lens gives negative v and positive m for a real object on the left.
Problem
A concave lens has f = -15 cm and forms an image at v = -10 cm. Find u and m.
- 1.Use 1/u = 1/v - 1/f.
- 2.1/u = -1/10 - (-1/15) = (-3 + 2)/30 = -1/30 cm⁻¹.
- 3.u = -30 cm.
- 4.m = v/u = (-10)/(-30) = +0.33.
- 5.The image is virtual, erect and one-third the object size.
Problem
A 2.0 cm object is 15 cm from a convex lens of f = 10 cm. Find v, h′ and m.
- 1.Given u = -15 cm, f = +10 cm, h = +2.0 cm.
- 2.1/v = 1/f + 1/u = 1/10 - 1/15 = 1/30, so v = +30 cm.
- 3.m = v/u = 30/(-15) = -2.
- 4.h′ = mh = -2 × 2.0 = -4.0 cm.
- 5.The image is real, inverted and twice enlarged, 30 cm on the other side.
Problem
A convex lens of f = 12 cm forms a virtual image twice the object size. Find the object and image distances.
- 1.Virtual erect enlargement gives m = +2 and u < 0, so v/u = 2 and v = 2u.
- 2.Substitute into 1/v - 1/u = 1/f: 1/(2u) - 1/u = 1/12.
- 3.-1/(2u) = 1/12, so u = -6 cm.
- 4.Then v = 2u = -12 cm.
- 5.Both object and virtual image are on the incident side; the object lies within f, as expected.
Mirror magnification is -v/u, while lens magnification is v/u. The mirror formula adds 1/v and 1/u; the lens formula subtracts 1/u. Write the appropriate formula before substituting.
Quiz
What does a ray through O do in the thin-lens model?
What is the sign of f for a concave lens?
Which is the lens magnification formula?
For u = -20 cm and v = +40 cm, what is m?
A negative v for a lens indicates an image where?
Practice Problems
- A convex lens has u = -30 cm and f = +15 cm. Find v. Solution: 1/v = 1/15 - 1/30 = 1/30, so v = +30 cm.
- For u = -24 cm and v = +12 cm, find m. Solution: m = v/u = 12/(-24) = -0.5, so the image is real, inverted and half-sized.
- A concave lens has u = -40 cm and f = -20 cm. Find v. Solution: 1/v = -1/20 - 1/40 = -3/40, so v = -13.3 cm approximately; virtual and on the object side.
- A 3 cm object has m = -1.5. Find image height. Solution: h′ = mh = -1.5 × 3 = -4.5 cm, so it is inverted and 4.5 cm tall.
- A convex lens has f = 20 cm and object at 10 cm. Find v and m. Solution: u = -10 cm. 1/v = 1/20 - 1/10 = -1/20, so v = -20 cm. m = (-20)/(-10) = +2: virtual, erect and enlarged.
Key Takeaways
• Three standard rays are sufficient for reliable lens constructions. • Lens distances are measured from O using Cartesian signs. • Convex f is positive and concave f is negative. • The lens formula is 1/v - 1/u = 1/f. • Lens magnification is m = h′/h = v/u. • Signs reveal image side and orientation; magnitude reveals size change. • A predicted ray-diagram case should agree with every calculated result.