Draw and use ray diagrams for the formation of a virtual image by a converging lens

3.2.3 Thin Lenses – Virtual Images

Key Concepts

A converging lens (convex) can produce a virtual image when the object is placed inside the focal length. The image appears on the same side as the object, is upright, and larger than the object.

Ray Diagram Basics

Think of a magnifying glass: when you look through it at a small object close to you, the light rays diverge after passing through the lens. Your brain traces them back to a point behind the lens, creating a virtual image that looks bigger. 📐

  • Ray 1: Parallel to the principal axis → refracts through the focal point on the opposite side.

  • Ray 2: Through the centre of the lens → continues straight.

  • Ray 3: Through the focal point on the object side → refracts parallel to the axis.

Constructing a Ray Diagram

  1. Draw the lens as a vertical line with a small “+” at the centre.

  2. Mark the principal axis (horizontal line) and the focal points \(F\) on both sides.

  3. Place the object (arrow) between the lens and the near focal point.

  4. Draw Ray 1, Ray 2, and Ray 3 as described above.

  5. Extend the refracted rays backward (dotted lines) to find their intersection – that point is the virtual image.

  6. Note the image is upright, larger, and on the same side as the object.

Lens Formula & Magnification

FormulaMeaning
\$\frac{1}{f} = \frac{1}{do} + \frac{1}{di}\$\(f\) = focal length, \(do\) = object distance, \(di\) = image distance (negative for virtual)
\$m = \frac{hi}{ho} = -\frac{di}{do}\$\(m\) = magnification, \(hi\) = image height, \(ho\) = object height. Negative sign → inverted; positive → upright.

Exam Tips

1. Identify the type of image first: If the object is inside the focal length of a converging lens, the image is virtual, upright, and magnified.

2. Use the lens formula wisely: Remember that for virtual images \(di\) is negative, so the equation becomes \(\frac{1}{f} = \frac{1}{do} - \frac{1}{|d_i|}\).

3. Check your ray diagram: The refracted rays must diverge after the lens; extend them backwards to find the intersection point.

4. Practice quick calculations: If \(f = 10\,\text{cm}\) and \(do = 5\,\text{cm}\), then \(\frac{1}{di} = \frac{1}{f} - \frac{1}{do} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10}\), so \(di = -10\,\text{cm}\) (virtual).

🧠 Remember: “Virtual = behind the lens, same side as object; Real = on the opposite side.”