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

Published by Patrick Mutisya · 14 days ago

Cambridge IGCSE Physics 0625 – 3.2.3 Thin Lenses: Real Image Formation by a Converging Lens

3.2.3 Thin Lenses – Real Image Formation by a Converging Lens

Learning Objective

Draw and use ray diagrams to show how a converging (convex) lens forms a real image.

Key Concepts

  • A converging lens is thicker at the centre than at the edges and bends parallel rays toward a focal point on the opposite side.

  • The principal axis is the straight line passing through the centre of the lens and both focal points.

  • The optical centre (also called the centre of the lens) is the point where a ray passes undeviated.

  • The focal length (\$f\$) is the distance from the optical centre to either focal point (F). For a converging lens \$f\$ is positive.

  • A real image is formed when the refracted rays actually converge on the other side of the lens. It can be projected onto a screen.

Ray‑Diagram Construction – Step by Step

  1. Draw the principal axis as a horizontal line.

  2. Mark the optical centre (O) of the lens at the centre of the diagram.

  3. Mark the focal points F (on the image side) and F' (on the object side) at a distance \$f\$ from O.

  4. Place the object (an upright arrow) on the object side of the lens, at a distance \$u\$ from O. For a real image we require \$u > f\$.

  5. Draw the three principal rays from the top of the object:

    • Parallel Ray: From the top of the object travel parallel to the principal axis, then refract through the focal point F on the image side.

    • Focal Ray: From the top of the object pass through the focal point F' on the object side, then emerge parallel to the principal axis.

    • Central Ray: From the top of the object pass straight through the optical centre O; this ray is undeviated.

  6. The point where the refracted rays intersect on the image side gives the position of the real image. Mark its height \$v\$ and note that the image is inverted.

Suggested diagram: Ray diagram showing a converging lens forming a real, inverted image of an object placed beyond the focal length.

Mathematical Relationship

The lens formula relates object distance \$u\$, image distance \$v\$, and focal length \$f\$:

\$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\$

The magnification \$m\$ is given by:

\$m = \frac{v}{u} = \frac{\text{image height } h'}{\text{object height } h}\$

For a real image formed by a converging lens, \$v\$ is positive and \$h'\$ is negative (inverted).

Worked Example

Problem: An object 3 cm tall is placed 30 cm in front of a converging lens of focal length \$f = 10\,\$cm. Determine the position, size, and nature of the image using a ray diagram and the lens formula.

  1. Apply the lens formula:

    \$\frac{1}{10} = \frac{1}{30} + \frac{1}{v} \quad\Rightarrow\quad \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}\$

    Hence \$v = 15\,\$cm on the image side of the lens.

  2. Calculate magnification:

    \$m = \frac{v}{u} = \frac{15}{30} = 0.5\$

    The image height \$h' = m \times h = 0.5 \times 3\,\$cm = \$1.5\,\$cm, and because \$m\$ is positive for a real image formed by a converging lens, the image is inverted (sign convention gives \$h' = -1.5\,\$cm).

  3. Interpretation:

    • The image is real (can be projected on a screen).

    • It is formed 15 cm behind the lens.

    • The image is inverted and half the size of the object.

  4. Draw the ray diagram using the three principal rays described above. The intersection of the refracted rays should occur 15 cm from the lens on the opposite side, confirming the calculation.

Summary Table

ParameterSymbolSign Convention (Converging Lens)Typical \cdot alue for Real Image
Focal length\$f\$PositiveGiven (e.g., 10 cm)
Object distance\$u\$Positive (object on incoming side)\$u > f\$ (e.g., 30 cm)
Image distance\$v\$Positive (image on opposite side)Calculated from lens formula (e.g., 15 cm)
Magnification\$m\$Negative for inverted real image\$m = v/u\$ (e.g., \$-0.5\$)
Image nature—Real, inverted, reduced/ enlarged depending on \$u\$Real, inverted, reduced (since \$u > 2f\$)

Common Mistakes to Avoid

  • Confusing the direction of the focal ray: it must pass through the focal point on the *object* side before emerging parallel.

  • Using the wrong sign for \$v\$ when applying the lens formula; for a real image formed on the opposite side of the lens, \$v\$ is positive.

  • Forgetting that the central ray passes undeviated through the optical centre.

  • Placing the object inside the focal length (\$u < f\$); this produces a virtual, upright image, not a real one.

Practice Questions

  1. A converging lens has a focal length of 8 cm. An object is placed 12 cm from the lens. Determine the image distance, magnification, and describe the image.

  2. Draw a ray diagram for a converging lens with \$f = 5\,\$cm when the object is placed 20 cm from the lens. State the image characteristics.

  3. If the image formed by a converging lens is real, inverted, and the same size as the object, what is the object distance in terms of the focal length?

Further Reading

Students should review the derivation of the thin‑lens equation and practice constructing ray diagrams for both converging and diverging lenses to solidify understanding.