GCSE — physicsHard5 min read

Lenses and Images

Converging and diverging lenses; real and virtual images; ray diagrams

Tutor AI Team
Tutor AI Team

# Lenses and Images — GCSE Physics

Lenses are used in cameras, glasses, microscopes, telescopes, and projectors. Understanding how lenses form images — and whether those images are real or virtual — is an important GCSE topic.

1. Types of Lenses

Converging (Convex) Lens

  • Thicker in the middle, thinner at the edges
  • Brings parallel rays of light to a focus at the focal point (FF)
  • Can form both real and virtual images

Diverging (Concave) Lens

  • Thinner in the middle, thicker at the edges
  • Spreads parallel rays of light apart — they appear to come from the focal point behind the lens
  • Always forms virtual, upright, diminished images

2. Key Terms

Term Definition
Principal axis Horizontal line through the centre of the lens
Optical centre (C) Centre of the lens — rays pass through without bending
Focal point (F) Point where parallel rays converge (or appear to diverge from)
Focal length (f) Distance from the centre of the lens to the focal point
Real image Formed where rays actually converge — can be projected on a screen
Virtual image Formed where rays appear to come from — cannot be projected

3. Ray Diagrams for Converging Lenses

To draw a ray diagram, use two of these three standard rays:

  1. Ray parallel to the principal axis → refracts through the focal point FF on the other side
  2. Ray through the optical centre → passes straight through without bending
  3. Ray through the focal point FF → refracts parallel to the principal axis

Where these rays meet (or appear to meet) is where the image forms.

Image Characteristics

Object Position Image Position Image Properties
Beyond 2F Between F and 2F Real, inverted, diminished
At 2F At 2F Real, inverted, same size
Between F and 2F Beyond 2F Real, inverted, magnified
At F At infinity No image (rays parallel)
Between F and lens Same side as object Virtual, upright, magnified

4. Magnification

magnification=image heightobject height\boxed{\text{magnification} = \frac{\text{image height}}{\text{object height}}}

Also: magnification=image distanceobject distance\text{magnification} = \frac{\text{image distance}}{\text{object distance}}

  • Magnification > 1 → image is larger than the object
  • Magnification < 1 → image is smaller (diminished)
  • Magnification = 1 → same size

5. Uses of Lenses

Converging Lens Uses

  • Camera: Object beyond 2F → small, real, inverted image on sensor
  • Magnifying glass: Object between F and lens → large, virtual, upright image
  • Eye: The lens focuses light onto the retina
  • Projector: Object between F and 2F → large, real image on screen

Diverging Lens Uses

  • Correcting short-sightedness (myopia): diverging lens spreads light before it enters the eye
  • Peepholes in doors
  • Combined with converging lenses in camera systems

6. The Eye and Vision Defects

How the Eye Focuses

  • The cornea does most of the refraction
  • The lens fine-tunes the focus by changing shape (accommodation)
  • Light focuses on the retina

Short-Sightedness (Myopia)

  • Can see near objects clearly but distant objects are blurry
  • Light focuses in front of the retina
  • Corrected with: diverging (concave) lens

Long-Sightedness (Hypermetropia)

  • Can see distant objects clearly but near objects are blurry
  • Light focuses behind the retina
  • Corrected with: converging (convex) lens

Worked Example: Magnification

Problem

Question: An object 3 cm tall produces an image 9 cm tall. Calculate the magnification.

magnification=93=3\text{magnification} = \frac{9}{3} = 3

The image is 3 times larger than the object.

Solution

Worked Example: Image Height

Problem

Question: A lens has a magnification of 5. An object is 2 cm tall. Find the image height.

image height=magnification×object height=5×2=10 cm\text{image height} = \text{magnification} \times \text{object height} = 5 \times 2 = 10 \text{ cm}

Solution

Worked Example: Describing an Image

Problem

Question: An object is placed between F and 2F of a converging lens. Describe the image.

Answer: The image is real (forms on the opposite side), inverted (upside down), and magnified (larger than the object). It forms beyond 2F.

Solution

8. Practice Questions

    1. State the difference between a converging and a diverging lens. (2 marks)
    1. An object is placed 25 cm from a converging lens. The image forms 50 cm from the lens on the other side. Calculate the magnification. (2 marks)
    1. Describe the image formed when an object is placed between F and a converging lens. (3 marks)
    1. A person has myopia (short-sightedness). (a) Explain why distant objects appear blurry. (2 marks) (b) What type of lens is used to correct this? (1 mark)
    1. A camera lens forms a real image of a distant tree. Is the image upright or inverted? Is it magnified or diminished? (2 marks)

    Answers

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Summary

  • Converging lens: convex, brings rays to a focus; can form real or virtual images
  • Diverging lens: concave, spreads rays; always forms virtual, upright, diminished images
  • Ray diagrams use standard rays through F, C, and parallel to axis
  • Magnification = image height / object height
  • Myopia corrected by diverging lens; hypermetropia by converging lens
Tags:
#gcse#physics#waves#lenses#images#optics

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