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Physics · 08 / 12

Lenses: reconstruct an image

High school · intermediate · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Units: numbers have dimensionsEquations: keep the balance level
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

For a thin converging lens, a ray parallel to the axis exits through the image focus; a ray through the optical centre keeps its direction. Their intersection locates an image point. For a real object beyond the focus and a real image, positive distances u, v, f obey 1/f = 1/u + 1/v. Signed-distance conventions use a different form.

1/v+1/u=1/f
THE PICTURE TO REMEMBERLenses: reconstruct an image
  1. 01Object and axis
  2. 02Two known rays
  3. 03Image intersection
  4. 04Distance and size

WORKED EXAMPLE

From principle to reasoning

For f = 10 cm and u = 30 cm, 1/v = 1/10 − 1/30 = 1/15. The real image is 15 cm away and inverted, with magnification −v/u = −0.5.

Solve a problem in this subject

  1. Draw the system

    Define the studied object, reference frame and positive axis. A force diagram shows external interactions acting on the system.

  2. Sort quantities

    Separate data and unknowns; convert units before substitution. 36 km/h = 10 m/s, so never insert 36 into a formula expecting SI speed.

  3. Write the model

    Choose a law matching assumptions: constant acceleration, thin lens, ohmic resistor. Isolate the unknown symbolically before rounding.

  4. Interpret

    Check dimensions and signs. Negative velocity gives direction on the chosen axis. Discuss scale and model limitations.

YOUR TURN

01 · Try before checking

Where is the image of an object 20 cm from that lens?

One hint

Isolate 1/v, keeping all lengths in the same unit.

Reveal and understand the solution

1/v = 1/20, so v = 20 cm. Magnification −1 means an inverted image of equal size.

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

A converging lens has f′ = 0.20 m. Find optical power.

A hint to get started

Use focal length in metres.

Step-by-step solution
  1. C = 1/f′ = 1/0.20.
  2. C = +5 dioptres.
  3. Positive power denotes a converging lens.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

With a rightward axis, OA = −30 cm and f′ = +10 cm. Find OA′.

A hint to get started

Use signed distances in 1/OA′ − 1/OA = 1/f′.

Step-by-step solution
  1. 1/OA′ + 1/30 = 1/10.
  2. 1/OA′ = 1/15, so OA′ = +15 cm.
  3. The image forms to the right of the lens.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

For OA = −30 cm and OA′ = +15 cm, interpret magnification.

A hint to get started

γ = OA′/OA; sign and magnitude convey different information.

Step-by-step solution
  1. γ = 15/(−30) = −0.5.
  2. The negative sign denotes an inverted image.
  3. Magnitude 0.5 means half the object size.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

Do not mix positive-distance and signed-distance conventions.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

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