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:
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.