Thin lenses

1. Overview

Thin lenses are optical components that refract light to form images by changing the direction of light rays. Understanding how light travels through converging and diverging lenses is essential for explaining how the human eye, cameras, telescopes, and corrective eyewear function.

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Key Definitions

• Principal Axis: A horizontal line passing through the exact center of the lens, perpendicular to its surface.
• Principal Focus (Focal Point, $F$): The point on the principal axis where light rays that were originally parallel to the axis converge (meet) or appear to diverge from.
• Focal Length ($f$): The distance from the center of the lens to the principal focus.
• Real Image: An image formed where light rays actually meet; it can be projected onto a screen.
• Virtual Image: An image formed where light rays only appear to come from; it cannot be projected onto a screen and is formed by extrapolating diverging rays backward.

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Core Content

Action of Lenses on Parallel Beams

• Converging (Convex) Lens: Thicker in the middle than at the edges. It brings parallel light rays together to meet at the principal focus.
• Diverging (Concave) Lens: Thinner in the middle than at the edges. It spreads parallel light rays out so they appear to come from the principal focus in front of the lens.

Ray Diagrams for Real Images (Converging Lens)

To find the position and size of an image, draw two "standard" rays from the top of the object:

1. Ray 1: Draw a line parallel to the principal axis; after the lens, it passes through the principal focus ($F$).
2. Ray 2: Draw a line passing through the exact center of the lens; it continues in a straight line without bending.
3. The Image: The point where these two rays cross is the top of the real image.

Describing Images

Every image must be described using three characteristics:

1. Size: Enlarged (bigger), Same size, or Diminished (smaller).
2. Orientation: Upright (same way up as object) or Inverted (upside down).
3. Nature: Real (rays actually cross) or Virtual (rays must be traced back).

Note: Real images are always inverted. Virtual images are always upright.

How to Draw a Ray Diagram (Step-by-Step)

This is a common exam question worth 3–4 marks. Follow this exact process:

1. Draw the lens as a vertical line. Mark the principal axis (horizontal line through the centre) and the two focal points ($F$) on either side.
2. Place the object (an upright arrow) on the left side of the lens.
3. Ray 1: From the tip of the object, draw a line parallel to the principal axis. After the lens, bend it so it passes through $F$ on the far side.
4. Ray 2: From the tip of the object, draw a line straight through the centre of the lens. It does not bend.
5. Where the two rays meet is the top of the image. Draw the image as an arrow from the axis to that point.

Exam tip: If the object is between $F$ and the lens, the rays diverge — extend them backwards with dashed lines to find the virtual image on the same side as the object.

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Extended Content (Extended Only)

The Magnifying Glass

When an object is placed closer to a converging lens than the focal length ($u < f$), the rays emerging from the lens never meet—they diverge.

• To the eye, these rays appear to come from a point behind the object.
• Image Characteristics: Enlarged, Upright, and Virtual.

Correcting Vision

• Short-sightedness (Myopia): The eye cannot focus on distant objects because the light converges in front of the retina.
  • Correction: Use a Diverging lens to spread the rays slightly before they enter the eye.
• Long-sightedness (Hyperopia): The eye cannot focus on nearby objects because the light would converge behind the retina.
  • Correction: Use a Converging lens to begin bending the rays inward before they enter the eye.

How to remember: Short-sighted = can see short distances (nearby) but not far → needs diverging lens to spread rays. Long-sighted = can see long distances (far) but not near → needs converging lens to focus rays.

In the exam, you may be asked to draw a ray diagram showing the correcting lens in front of the eye with rays converging on the retina (worth 3–4 marks). Always show: the lens type, rays entering the lens, and rays meeting on the back of the eye (retina).

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Key Equations

While ray diagrams are the primary tool, the following concept is used for magnification:

$$Linear\ Magnification = \frac{\text{Image Height}}{\text{Object Height}}$$ OR $$Linear\ Magnification = \frac{\text{Image Distance }(v)}{\text{Object Distance }(u)}$$

• Units: Magnification has no units (it is a ratio).
• $v$: Distance from center of lens to image (m or cm).
• $u$: Distance from center of lens to object (m or cm).

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Common Mistakes to Avoid

• ❌ Wrong: Measuring focal length from the edge of the lens or the surface of the glass.
• ✓ Right: Always measure distances ($f$, $u$, and $v$) from the vertical center line of the lens.
• ❌ Wrong: Forgetting to draw arrowheads on rays.
• ✓ Right: Every ray must have at least one arrowhead to show the direction of light travel.
• ❌ Wrong: Thinking an image is only "same size" regardless of position.
• ✓ Right: An image is only the same size as the object when the object is placed at exactly $2f$ (two focal lengths away).
• ❌ Wrong: Using the surface of the lens as the reference for angles.
• ✓ Right: Like all optics, if measuring angles of refraction, measure from the normal (though ray diagrams usually focus on the focal point).

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Exam Tips

1. Use a Sharp Pencil and Ruler: Inaccurate lines in ray diagrams can lead to the wrong image position, which loses marks.
2. Dashed vs. Solid Lines: Use solid lines for real light rays and dashed (dotted) lines for virtual rays (extrapolations) and virtual images.
3. The "Center" Rule: If you are unsure where to draw a ray, the ray passing through the exact center of any thin lens always goes straight through without bending. This is often the easiest ray to draw correctly.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A converging lens is used to form an image of a small toy car.

(a) Define the term focal length of a converging lens. [2]

(b) State three characteristics of the image formed by a converging lens when the object (toy car) is placed at a distance greater than twice the focal length from the lens. [3]

Worked Solution:

(a)

1. The focal length is the distance between the lens and the principal focus. The distance from the centre of the lens to the principal focus $\boxed{}$ This is the basic definition.

How to earn full marks:

• State "distance" or equivalent (e.g., "length").
• Correctly relate this distance to the "lens" and the "principal focus" (or "focal point").

(b)

1. The image will be inverted because the object is beyond 2f. Inverted $\boxed{}$ This indicates the orientation of the image.

2. The image will be real because the object is beyond f. Real $\boxed{}$ This indicates that the image can be projected.

3. The image will be diminished because the object is beyond 2f. Diminished (smaller than the object) $\boxed{}$ This describes the size of the image relative to the object.

How to earn full marks:

• State inverted.
• State real.
• State diminished or smaller.

Common Pitfall: Many students confuse the focal length with the object or image distance. Remember the focal length is a property of the lens itself, while object and image distances depend on the setup. Also, be sure to use the terms "real" and "inverted" correctly when describing images.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

A student is using a converging lens as a magnifying glass to observe an ant. The ant is placed closer to the lens than the focal length.

(a) State two differences between a real image and a virtual image. [2]

(b)

On the diagram, draw two rays from the top of the ant to locate the position of the image formed by the lens. [4]

Worked Solution:

(a)

1. A real image can be projected onto a screen, a virtual image cannot. Real image can be projected onto a screen; virtual image cannot $\boxed{}$ This is the key distinction between the two types of images.

2. A real image is formed by the actual intersection of light rays, a virtual image is formed by the apparent intersection of extrapolated rays. Real image is formed by actual intersection of rays; virtual image by extrapolated rays $\boxed{}$ This explains how each type of image is formed.

How to earn full marks:

• Clearly state that a real image can be projected, but a virtual image cannot.
• State that a real image is formed by actual rays, while a virtual image is formed by extrapolated rays.

(b)

1. Draw a ray from the top of the object, parallel to the principal axis, refracted through the focal point on the other side of the lens. Ray parallel to the principal axis, refracted through the focal point $\boxed{}$ This is a standard ray diagram construction.

2. Draw a ray from the top of the object, passing straight through the centre of the lens, without being refracted. Ray through the centre of the lens, unrefracted $\boxed{}$ This is another standard ray diagram construction.

3. Extend both rays backwards (on the same side of the lens as the object) using dashed lines. Rays extended backwards with dashed lines $\boxed{}$ This is necessary to locate the virtual image.

4. The intersection of the backward extensions marks the top of the virtual image. The image is upright and magnified. Intersection of rays correctly indicates the top of the image, image is upright and magnified $\boxed{}$ This correctly identifies the image location and characteristics.

How to earn full marks:

• Ray 1: Correctly drawn and refracted ray parallel to the principal axis passing through the focal point.
• Ray 2: Correctly drawn and unrefracted ray through the centre of the lens.
• Rays Extended: Both rays must be extended backwards using dashed lines.
• Image Location: The intersection of the extended rays must correctly locate the top of the virtual image. The image must be upright and magnified.

Common Pitfall: When drawing ray diagrams for virtual images, remember to extend the rays backwards from the lens using dashed lines. The virtual image is formed where these extended lines intersect. Also, make sure your rays are accurately drawn, passing through the focal point or the center of the lens as required.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A student investigates the focal length of a converging lens. They set up the lens, a light source (object), and a screen, and adjust the positions of the lens and screen until a sharp image of the light source is formed on the screen. The distance $u$ between the light source and the lens, and the distance $v$ between the lens and the screen, are measured. The student repeats the procedure for different values of $u$.

(a) State the relationship between $u$, $v$, and the focal length $f$ of the lens. [1]

(b) The student obtains the following data:

$u$ / cm	$v$ / cm
20.0	20.0
25.0	16.7
30.0	15.0

Use the data to calculate the focal length $f$ of the lens. Show your working. [4]

(c) Describe how the student could improve the accuracy of their experiment to determine the focal length. [3]

Worked Solution:

(a)

1. The lens equation relates object distance, image distance, and focal length. $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ $\boxed{}$ This is the standard lens equation.

How to earn full marks:

• State the correct lens equation relating f, u, and v.

(b)

1. Consider the first data point. When u = 20.0 cm and v = 20.0 cm, the object is at 2f, so the image is also at 2f. Therefore, f = u/2 = v/2. When $u = 20.0 \text{ cm}$ and $v = 20.0 \text{ cm}$, $f = \frac{u}{2} = \frac{20.0 \text{ cm}}{2} = 10.0 \text{ cm}$ $\boxed{}$ This correctly uses the special case where u = v = 2f.

2. Alternatively, apply the lens formula using the first set of data. $\frac{1}{f} = \frac{1}{20.0 \text{ cm}} + \frac{1}{20.0 \text{ cm}}$ $\boxed{}$ Substitute the given values.

3. Simplify the right-hand side. $\frac{1}{f} = \frac{2}{20.0 \text{ cm}} = \frac{1}{10.0 \text{ cm}}$ $\boxed{}$ Simplify the fraction.

4. Invert to find f. $f = 10.0 \text{ cm}$ $\boxed{10.0 \text{ cm}}$ Invert both sides to solve for f.

How to earn full marks:

• Correctly use the data point where u = v = 20.0 cm, and deduce that f = 10.0 cm.
• OR correctly substitute u and v into the lens equation.
• Correctly simplify the equation and isolate f.
• State the final answer with the correct unit (cm).

(c)

1. Repeat the experiment for more values of u and v, and then plot a graph of 1/u against 1/v. The y-intercept of this graph will be equal to 1/f. Take more readings and plot a graph of $\frac{1}{u}$ against $\frac{1}{v}$. The y-intercept is $\frac{1}{f}$. $\boxed{}$ This uses a graphical method to improve accuracy.

2. Use a travelling microscope to measure the distances u and v more accurately, reducing parallax error. Use a travelling microscope to measure $u$ and $v$ more accurately. $\boxed{}$ This reduces parallax errors.

3. Ensure the lens, light source, and screen are all aligned on the principal axis of the lens. Ensure lens, light source, and screen are aligned on the principal axis. $\boxed{}$ This minimizes distortion and improves image quality.

How to earn full marks:

• Take more readings and plot a graph (1/u vs 1/v) and use the y-intercept to find 1/f.
• Use a travelling microscope to measure u and v more accurately.
• Ensure the lens, light source, and screen are aligned on the principal axis.

Common Pitfall: Make sure you understand the lens equation and how to apply it correctly. A common mistake is forgetting to invert after finding 1/f to get the actual value of f. Also, remember to include the correct units (cm in this case) in your final answer.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A person is long-sighted. This means they have difficulty focusing on objects that are close to their eyes.

(a) Explain what is meant by the term "long-sighted". [3]

(b) State the type of lens that is used to correct long-sightedness. [1]

(c)

On the diagram of the eye, draw two rays from a near object to show how the image is formed behind the retina in a long-sighted person. Label the cornea, lens, and retina. [3]

(d) Explain how the lens stated in part (b) corrects long-sightedness. [2]

Worked Solution:

(a)

1. Long-sightedness (hyperopia) is a defect of vision where a person can see distant objects clearly, but has difficulty focusing on near objects. A person can see distant objects clearly but has difficulty focusing on near objects. $\boxed{}$ This explains the basic symptoms.

2. The eye lens cannot refract light strongly enough to focus light from near objects onto the retina. The eye lens cannot refract light strongly enough to focus light from near objects onto the retina. $\boxed{}$ This describes the problem with the eye's focusing ability.

3. The image of near objects is formed behind the retina. The image of near objects is formed behind the retina. $\boxed{}$ This specifies where the image is formed.

How to earn full marks:

• State that distant objects are seen clearly, but near objects are blurry.
• Explain that the eye lens is unable to refract light strongly enough.
• State that the image of near objects is formed behind the retina.

(b)

1. A converging lens is used to correct long-sightedness. Converging lens (convex lens) $\boxed{}$ This identifies the correct type of lens.

How to earn full marks:

• State "converging lens" or "convex lens".

(c)

1. Draw two rays from the object entering the eye. These rays should be refracted by the cornea and the lens. Two rays drawn from the object, refracted by the cornea and lens. $\boxed{}$ This sets up the ray diagram.

2. The rays should converge behind the retina. The point where they converge should be clearly behind the retina. Rays converge behind the retina. $\boxed{}$ This shows the effect of long-sightedness.

3. Cornea, lens, and retina are correctly labelled. Cornea, lens, and retina are correctly labelled. $\boxed{}$ This ensures all parts are identified.

How to earn full marks:

• Two rays are drawn correctly, entering the eye and being refracted.
• The rays converge behind the retina.
• The cornea, lens, and retina are correctly labelled.

(d)

1. A converging lens placed in front of the eye helps to refract the light rays more before they enter the eye. The converging lens refracts the light rays more before they enter the eye. $\boxed{}$ This explains the initial effect of the lens.

2. This allows the eye lens to focus the light rays onto the retina, correcting the long-sightedness. This allows the eye lens to focus the light rays onto the retina. $\boxed{}$ This explains how the converging lens helps the eye to focus properly.

How to earn full marks:

• Explain that the converging lens pre-refracts the light rays.
• Explain that this enables the eye lens to focus the rays correctly onto the retina.

Common Pitfall: When explaining long-sightedness, remember that the problem is the eye cannot focus near objects, and the image is formed behind the retina. Don't confuse this with short-sightedness. Also, be clear that the converging lens helps the eye to pre-focus the light, so the eye's lens can then focus it correctly on the retina.