Module 3 · Lesson

Ray Optics

Waves and Thermodynamics

Ray Optics

Orientation

Lesson goal: use ray diagrams and thin-lens relationships to predict image location, size, orientation, and type.

The main discipline is to make the diagram and calculation agree. A ray diagram is evidence, not decoration.

Core Content

The ray model approximates light as travelling in straight-line paths through a uniform medium. Thin-lens and mirror equations are useful when the paraxial approximation is reasonable: rays stay close to the principal axis and angles are small.

Key equations:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

$$m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$$

EvidenceMeaning
real imagerays physically converge and image can be projected on a screen
virtual imagerays appear to diverge from a point and cannot be projected
negative magnificationinverted image in the chosen convention
$\lvert m\rvert > 1$image is larger than object

Concept Check

  1. In a thin-lens ray diagram, a ray through the optical centre is usually drawn:

    • A. as strongly curved
    • B. approximately straight
    • C. parallel then stopped
    • D. backwards only

    Answer: B.

  2. A real image can be:

    • A. projected onto a screen
    • B. seen only by extending imaginary rays
    • C. formed without any light rays
    • D. created only by a plane mirror

    Answer: A.

  3. If $\lvert m\rvert > 1$, the image is:

    • A. smaller
    • B. the same size
    • C. magnified
    • D. always virtual

    Answer: C.

  4. Short response: explain why sign convention must be stated before using the thin-lens equation.

Applied Practice

Worked Example

An object is $0.40\ \text{m}$ from a convex lens with focal length $0.15\ \text{m}$. Find the image distance.

  1. State the equation:

    $$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

  2. Substitute:

    $$\frac{1}{0.15} = \frac{1}{0.40} + \frac{1}{d_i}$$

  3. Solve:

    $$6.667 = 2.500 + \frac{1}{d_i}$$

    $$d_i = 0.240\ \text{m}$$

Final answer: the image forms $0.240\ \text{m}$ on the opposite side of the lens. With this convention it is a real image.

Practice Problem

An object is $0.30\ \text{m}$ from a convex lens with focal length $0.10\ \text{m}$. Calculate image distance and magnification, then state whether your ray diagram should show a real or virtual image.

Deep Practice And Writing

Prompt: evaluate whether a thin-lens model is adequate for a classroom lens experiment. Your answer must mention the paraxial approximation and one source of image error.

Strong response pattern:

  1. identify the model,
  2. state the useful assumption,
  3. identify a limitation,
  4. judge whether the model is sufficient for the given purpose.

Tutor Context

Use this lesson context when the student asks about:

Tutor should first check whether the student has a diagram that agrees with the calculation.

Useful tutor diagnostic:

Does your ray diagram show the rays physically meeting, or only appearing to meet when extended backward?

Maintenance Loop

Fast retrieval:

  1. A real image can be projected on a ______.
  2. $\lvert m\rvert > 1$ means the image is ______.
  3. The thin-lens equation should be used only after stating the ______.

Source Trace

This lesson is materialised from the existing textbook section, app lesson YAML, roadmap lesson, roadmap concept questions, and Module 3 notes. It is ready for conversion into a typed content manifest and panelled-course render block.