CIE iGCSE Co-ordinated Sciences-P3.2.3 Thin converging lens- Study Notes- New Syllabus
CIE iGCSE Co-ordinated Sciences-P3.2.3 Thin converging lens – Study Notes
CIE iGCSE Co-ordinated Sciences-P3.2.3 Thin converging lens – Study Notes -CIE iGCSE Co-ordinated Sciences – per latest Syllabus.
Key Concepts:
Core
1. Describe the action of a thin converging lens on a parallel beam of light and know that rays of light from an object at distance can be assumed to be parallel
2. Define and use the terms principal axis, principal focus (focal point) and focal length
3. Draw and use ray diagrams for the formation of an image by a thin converging lens, limited to real images
4. Describe the characteristics of an image using the terms enlarged / same size / diminished and upright / inverted
Supplement
5. Draw and use ray diagrams for the formation of a virtual image by a thin converging lens
6. Describe the characteristics of an image using the terms real / virtual
7. Describe the use of a single lens as a magnifying glass
CIE iGCSE Co-Ordinated Sciences-Concise Summary Notes- All Topics
Thin Converging Lens
(a) Converging Lens and Parallel Rays
- A thin converging lens (convex lens) bends parallel rays of light so that they converge to a single point on the other side of the lens.
- This point is called the principal focus (F).
- The distance from the centre of the lens to the principal focus is the focal length (\( f \)).
(b) Rays from a Distant Object
- When an object is at a very large distance from the lens (e.g. the Sun), the rays reaching the lens are almost parallel.
- The lens focuses these parallel rays to a sharp point at the focal plane.
- Thus, distant objects form an image at the focus of the lens.
(c) Key Assumptions
- Rays from infinity (very distant objects) are treated as parallel rays.
- The focal length \( f \) is fixed for a given lens, depending on its curvature and refractive index.
Example
A thin converging lens has a focal length of \( 15~\text{cm} \). A parallel beam of light from a distant lamp falls on the lens. Find the position of the image formed.
▶️Answer/Explanation
Step (1) – Distant Object Approximation:
For a very distant object, the incoming rays are parallel, so the image is formed at the focal point.
Step (2) – Position of Image:
The image is at a distance equal to the focal length on the other side of the lens.
Final Answer:
The image is formed \( 15~\text{cm} \) from the lens at the principal focus.
Lens Terminology
(a) Principal Axis
- The principal axis is the straight line that passes through the centre of the lens and is perpendicular to its surfaces.
- It is the reference line about which the lens is symmetrical.
- All definitions of focus and focal length are made relative to the principal axis.
(b) Principal Focus (Focal Point)
- The principal focus (F) of a converging lens is the point on the principal axis where rays parallel to the axis converge after passing through the lens.
- For a diverging lens, it is the point on the axis from which rays appear to diverge after refraction.
- This point is unique to a lens and depends on its curvature and material.
(c) Focal Length (f)
- The focal length (\( f \)) is the distance between the optical centre of the lens and the principal focus.
- It determines the converging power of the lens: shorter focal length → stronger convergence.
- Focal length is related to the refractive index (\( n \)) and radii of curvature of the lens surfaces by the lens maker’s formula:
\( \dfrac{1}{f} = (n – 1)\left(\dfrac{1}{R_1} – \dfrac{1}{R_2}\right) \).
Example
A lens has a principal focus \( 12~\text{cm} \) from its centre. A beam of rays parallel to the principal axis falls on the lens. Where will the rays meet?
▶️Answer/Explanation
Step (1) – Use of Principal Focus:
By definition, parallel rays to the axis converge at the principal focus.
Step (2) – Focal Length:
The focal length is given as \( f = 12~\text{cm} \).
Final Answer:
The rays will meet at the principal focus, \( 12~\text{cm} \) from the centre of the lens.
Image Formation by a Thin Converging Lens
(a) Rays Used in Lens Diagrams
- To construct ray diagrams for a thin converging lens, we usually draw at least two of these principal rays:
- Ray 1: A ray parallel to the principal axis refracts through the lens and passes through the principal focus (F) on the far side.
- Ray 2: A ray passing through the centre of the lens (optical centre) goes straight through without deviation.
- Ray 3 (optional): A ray passing through the principal focus on the near side of the lens emerges parallel to the axis.
- The point where the rays converge gives the position of the image.
(b) Real Image Formation
- When an object is placed at a distance greater than the focal length (\( u > f \)), the refracted rays actually converge to form a real image on the opposite side of the lens.
- A real image can be projected onto a screen because the light rays physically meet.
(c) Cases of Real Images (Converging Lens)
- Object beyond 2F: Image is real, inverted, diminished, and formed between F and 2F.
- Object at 2F: Image is real, inverted, same size, and formed at 2F.
- Object between F and 2F: Image is real, inverted, magnified, and formed beyond 2F.
- Object at infinity: Image is real, inverted, highly diminished (point image), formed at F.
(d) Lens Formula
The relation between object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) is given by:
\( \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} \)
Example
An object is placed \( 30~\text{cm} \) from a thin converging lens of focal length \( 10~\text{cm} \). Find the position and nature of the image.
▶️Answer/Explanation
Step (1) – Lens Formula:
\( \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} \)
Here, \( f = 10~\text{cm} \), \( u = 30~\text{cm} \).
Step (2) – Substitution:
\( \dfrac{1}{10} = \dfrac{1}{30} + \dfrac{1}{v} \)
Step (3) – Rearrangement:
\( \dfrac{1}{v} = \dfrac{1}{10} – \dfrac{1}{30} = \dfrac{3 – 1}{30} = \dfrac{2}{30} = \dfrac{1}{15} \).
Step (4) – Image Distance:
\( v = 15~\text{cm} \).
Step (5) – Nature of Image:
Since the rays converge:
- Image is real (can be projected).
- Image is inverted.
- Image is diminished (smaller than the object).
Final Answer:
The image is formed \( 15~\text{cm} \) from the lens on the opposite side, real, inverted, and diminished.
Virtual Image Formation by a Thin Converging Lens
(a) Virtual Image Formation
- A virtual image is formed when the refracted rays do not actually meet, but appear to diverge from a point when extended backward.
- In the case of a converging lens, a virtual image occurs when the object is placed between the focal point (F) and the lens.
- The image cannot be projected onto a screen because the rays do not physically converge.
(b) Using Ray Diagrams
- Ray 1: A ray drawn from the top of the object parallel to the principal axis refracts through the lens and passes through the far-side principal focus.
- Ray 2: A ray drawn through the optical centre passes straight through without deviation.
- The two rays diverge after refraction. When extended backward, they appear to meet on the same side of the lens as the object, forming a virtual image.
(c) Characteristics of the Virtual Image (Converging Lens)
- The image is upright (same orientation as object).
- The image is magnified (larger than object).
- The image is virtual (cannot be projected onto a screen).
- The image appears on the same side of the lens as the object.
(d) Lens Formula for Virtual Image
- The lens formula still applies: \( \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} \).
- However, for a virtual image, the image distance (\( v \)) comes out as negative (indicating the image is on the same side as the object).
Example
An object is placed \( 5~\text{cm} \) from a thin converging lens of focal length \( 10~\text{cm} \). Find the position and nature of the image.
▶️Answer/Explanation
Step (1) – Lens Formula:
\( \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} \)
Here, \( f = 10~\text{cm} \), \( u = 5~\text{cm} \).
Step (2) – Substitution:
\( \dfrac{1}{10} = \dfrac{1}{5} + \dfrac{1}{v} \)
Step (3) – Rearrangement:
\( \dfrac{1}{v} = \dfrac{1}{10} – \dfrac{1}{5} = \dfrac{1}{10} – \dfrac{2}{10} = -\dfrac{1}{10} \).
Step (4) – Image Distance:
\( v = -10~\text{cm} \).
Step (5) – Interpretation:
The negative sign means the image is on the same side as the object, i.e. it is virtual.
Final Answer:
The image is formed \( 10~\text{cm} \) from the lens on the same side as the object, virtual, upright, and magnified.
Characteristics of Images (Converging Lens)
| Object Position | Image Position | Size | Orientation | Nature |
|---|---|---|---|---|
| Beyond 2F | Between F and 2F (opposite side) | Diminished | Inverted | Real |
| At 2F | At 2F (opposite side) | Same size | Inverted | Real |
| Between F and 2F | Beyond 2F (opposite side) | Enlarged | Inverted | Real |
| At F | At infinity | Highly enlarged | Inverted | Real |
| Between F and Lens | Same side as object | Enlarged | Upright | Virtual |
Example:
An object is placed at different positions relative to a converging lens of focal length \( 10 \,\text{cm} \). Use the table of image characteristics to determine the nature of the image in each case.
- Object at \( 25 \,\text{cm} \) from the lens.
- Object at \( 20 \,\text{cm} \) from the lens.
- Object at \( 15 \,\text{cm} \) from the lens.
- Object at \( 10 \,\text{cm} \) from the lens.
- Object at \( 5 \,\text{cm} \) from the lens.
▶️Answer/Explanation
Step (1) – Recall positions:
Focal length \( f = 10 \,\text{cm} \). So:
\( 2F = 20 \,\text{cm} \).
Step (2) – Use Table:
| Object Position | Image Position | Size | Orientation | Nature |
|---|---|---|---|---|
| Object at 25 cm (> 2F) | Between F and 2F | Diminished | Inverted | Real |
| Object at 20 cm (at 2F) | At 2F | Same size | Inverted | Real |
| Object at 15 cm (between F and 2F) | Beyond 2F | Enlarged | Inverted | Real |
| Object at 10 cm (at F) | At infinity | Highly enlarged | Inverted | Real |
| Object at 5 cm (< F) | Same side as object | Enlarged | Upright | Virtual |
Final Answers:
- 25 cm → Diminished, inverted, real (between F and 2F).
- 20 cm → Same size, inverted, real (at 2F).
- 15 cm → Enlarged, inverted, real (beyond 2F).
- 10 cm → Highly enlarged, inverted, real (at infinity).
- 5 cm → Enlarged, upright, virtual (same side as object).
Use of a Single Lens as a Magnifying Glass
A single thin converging lens can be used as a magnifying glass to view small objects more clearly.
- The object must be placed closer to the lens than the focal length (\( u < f \)).
- In this case, the lens produces a virtual, upright, and enlarged image of the object.
- The image appears on the same side of the lens as the object.
- The lens allows the eye to see details of the object at a comfortable viewing distance, instead of bringing the eye extremely close.
Key points:
- Image: virtual, upright, magnified
- Condition: object inside focal length (\( u < f \))
Example
A lens has a focal length of \( f = 10 \,\text{cm} \). An object is placed \( u = 5 \,\text{cm} \) from the lens. Find the image distance and describe the nature of the image.
▶️Answer/Explanation
Lens Formula:
\( \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \)
Substitute values:
\( \dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{5} \)
\( \dfrac{1}{v} = \dfrac{1}{10} – \dfrac{1}{5} = \dfrac{1}{10} – \dfrac{2}{10} = -\dfrac{1}{10} \)
\( v = -10 \,\text{cm} \)
Interpret result:
Negative \( v \) means the image forms on the same side of the lens as the object.
Image is virtual, upright, and enlarged.