Describe the characteristics of an image formed by a converging lens

3.2.3 Thin Lenses – Converging (Convex) Lens

Learning objective

Describe the characteristics of an image formed by a converging (convex) lens and predict its size, orientation and nature for any object position.

1. Key terms

• Converging (convex) lens: thicker at the centre than at the edges; parallel incident rays are brought together (converge) to a point.
• Principal axis: the straight line passing through the optical centre and the two focal points.
• Optical centre (O): geometric centre of the lens; a ray through O is not deviated.
• Principal (focal) point, F: point on the principal axis where a ray initially parallel to the axis meets after refraction.
• Focal length (f): distance between O and F. Positive for a converging lens.
• Real image: formed on the side opposite the object; can be projected on a screen.
• Virtual image: formed on the same side as the object; cannot be projected.
• Inverted / upright: orientation of the image relative to the object.
• Magnified / reduced: |m| > 1 → magnified, |m| < 1 → reduced.
• Optical power (P): P = 1/f (dioptres, D). The larger the power, the stronger the lens.
• Lens‑maker’s equation (for thin lenses):

  \[

  \frac{1}{f}= (n-1)\!\left(\frac{1}{R1}-\frac{1}{R2}\right)

  \]

  where *n* is the refractive index of the lens material and *R₁*, *R₂* are the radii of curvature (positive if the centre of curvature lies on the outgoing‑light side).

2. Sign convention (Cartesian – as used in the Cambridge IGCSE)

3. Fundamental relations

• Lens formula (Cartesian form)

  \[

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

  \]

• Magnification

  \[

  m = \frac{h'}{h}= -\frac{v}{u}

  \]

  • m > 0 → upright image
  • m < 0 → inverted image
  • |m| > 1 → magnified, |m| < 1 → reduced

• Optical power

  \[

  P=\frac{1}{f}\quad(\text{f in metres, }P\text{ in dioptres})

  \]

• Lens‑maker’s equation (for thin lenses) – appears in the supplementary part of the syllabus.

4. Constructing a ray diagram

Use the three principal rays. The point where the refracted rays intersect (or appear to intersect) gives the image position.

1. Parallel ray: from the top of the object, travels parallel to the principal axis, then refracts through the focal point on the far side.
2. Focal ray: from the top of the object, passes through the focal point on the object side, then emerges parallel to the principal axis.
3. Centre ray: passes straight through the optical centre O without deviation.

For a virtual image (object inside f) extend the refracted rays backwards; their apparent intersection gives the virtual image.

5. Image characteristics for the five standard object positions

6. Worked example (AO2 – handling information)

Given: f = +10 cm, object distance u = ‑30 cm (object beyond 2f). Find the image distance v and the magnification m.

1. Lens formula

   \[

   \frac{1}{10}= \frac{1}{v}+ \frac{1}{-30}

   \Longrightarrow \frac{1}{v}= \frac{1}{10}+ \frac{1}{30}= \frac{3+1}{30}= \frac{4}{30}= \frac{2}{15}

   \]

   \[

   v = \frac{15}{2}=7.5\ \text{cm}\;(>0\;\text{→ real image})

   \]

2. Magnification

   \[

   m = -\frac{v}{u}= -\frac{+7.5}{-30}= -0.25

   \]

   • m < 0 → inverted
   • |m| < 1 → reduced

3. Result: a real, inverted, reduced image 7.5 cm behind the lens.

7. Exam‑style question (AO2)

Question: A converging lens of focal length 12 cm forms a real image of a candle that is 4 cm tall. The image is 9 cm tall and inverted. Determine (a) the object distance, (b) the image distance and (c) whether the object is placed inside, between or beyond 2f.

Solution:

1. Magnification from the heights:

   \[

   m = \frac{h'}{h}= \frac{-9}{4}= -2.25

   \]

   (negative → inverted, magnitude = 2.25 → magnified)

2. Using \(m = -\dfrac{v}{u}\) gives

   \[

   -2.25 = -\frac{v}{u}\;\Longrightarrow\; \frac{v}{u}=2.25\;\Longrightarrow\; v = 2.25u

   \]

3. Insert into the lens formula (f = +12 cm):

   \[

   \frac{1}{12}= \frac{1}{v}+ \frac{1}{u}= \frac{1}{2.25u}+ \frac{1}{u}= \frac{1+2.25}{2.25u}= \frac{3.25}{2.25u}

   \]

   \[

   u = \frac{3.25}{2.25}\times12 = 17.33\ \text{cm (negative by sign convention)}\;\Rightarrow\; u = -17.3\ \text{cm}

   \]

4. Image distance: \(v = 2.25u = 2.25\times(-17.33) = -39.0\ \text{cm}\).

   Because v is positive in the Cartesian convention, we write \(v = +39\ \text{cm}\) (real image on the far side).

5. Comparison with f: \(f = 12\ \text{cm},\; 2f = 24\ \text{cm}\).

   Since \(|u| = 17.3\ \text{cm}\) lies between f and 2f, the object is between f and 2f.

8. Practical investigation – measuring the focal length of a converging lens

This activity satisfies the AO3 requirement to plan, carry out and interpret a simple experiment.

1. Apparatus: convex lens, screen, metre ruler, distant object (e.g., a building or a wall clock), clamps.
2. Method (using the “distant‑object” technique)

   1. Place the lens on a stand and set a screen a few centimetres behind it.
   2. Move the screen until a sharp, inverted image of the distant object appears on it.
   3. Measure the distance between the lens and the screen – this is the focal length f (the object is effectively at infinity, so u ≈ ∞ and 1/u ≈ 0, giving 1/f ≈ 1/v).

3. Data table (example)

   Trial	Screen‑lens distance (cm)	f (cm)
   1	10.2	10.2
   2	9.8	9.8
   3	10.1	10.1

4. Analysis

   • Calculate the mean focal length and the percentage uncertainty.
   • Discuss sources of error (parallax when reading the scale, lens not perfectly thin, alignment of the screen, finite distance of the “distant” object).

9. Everyday applications (link to the syllabus)

• Camera lens: object (scene) is well beyond f → real, inverted image on the sensor.
• Magnifying glass: object placed inside f → virtual, upright, magnified image.
• Projector: slide placed beyond f; the lens projects a real, inverted image onto a screen.
• Focusing light into an optical fibre: a converging lens concentrates parallel light onto the fibre end‑face, ensuring total internal reflection.
• Eye correction:

  • Farsightedness (hyperopia) – a converging lens is placed in front of the eye to bring the image of a near object onto the retina.
  • Nearsightedness (myopia) – a diverging lens (supplementary topic) is used to diverge incoming rays so that the image forms further back on the retina.

10. Common misconceptions (and how to address them)

• All images from a converging lens are inverted.

  Correction: Only real images are inverted. When the object is inside the focal length the image is virtual and upright.

• A larger object always gives a larger image.

  Correction: Image size depends on the ratio |v/u|, not on the absolute size of the object.

• The focal length changes with object distance.

  Correction: f is a property of the lens (for a given material and surrounding medium) and is independent of where the object is placed.

• Virtual images must be smaller.

  Correction: For a converging lens, a virtual image (object inside f) is upright and magnified.

11. Summary checklist (AO1 + AO2)

1. Identify the object distance relative to f and 2f.
2. Write the sign‑convention‑consistent lens formula \(\displaystyle\frac{1}{f}= \frac{1}{v}+ \frac{1}{u}\).
3. Calculate the image distance v (real → positive, virtual → negative).
4. Find magnification \(m = -\dfrac{v}{u}\) to decide size and orientation.
5. Classify the image as real/virtual and inverted/upright; note whether it is magnified or reduced.
6. Confirm the result by sketching a ray diagram using the three principal rays.
7. If required, compute optical power \(P = 1/f\) and, for design questions, use the lens‑maker’s equation.