ICSE Class 10 Physics: Lens Refraction & Ray Diagrams

Ever wondered why a magnifying glass can make a tiny bug look huge, or why a camera lens brings distant scenes into focus?

💡 In Simple Words: When light passes from air into a lens, it bends – that’s refraction. By tracing the bent rays on paper, we can predict where the image will appear.

What is Refraction in a Lens?

Refraction is the change in direction of a light ray when it moves from one transparent material to another with a different speed, just like a straw looks bent when you look at it in a glass of water.

A lens is a piece of glass (or plastic) shaped to make light rays either converge (come together) or diverge (spread apart). The line that runs through the centre of the lens is called the principal axis. The points on the axis where parallel rays appear to meet are the focal points (F). The distance from the lens centre to a focal point is the focal length (f).

Convex Lens (Converging) Ray Diagram

A convex lens is thicker in the middle than at the edges. It makes parallel rays converge to a real point on the other side.

How to draw a convex‑lens ray diagram

Follow the flowchart above. The two coloured rays (the centre ray and the parallel/focal ray) intersect at the image point. That intersection tells you where the image forms and whether it’s upright or inverted.

Key features of a convex‑lens image

• If the object is beyond 2F (twice the focal length), the image forms between F and 2F, is smaller, inverted, and real.
• If the object is at 2F, the image appears at 2F on the other side, same size, inverted, real.
• If the object is between F and 2F, the image forms beyond 2F, larger, inverted, real.
• If the object is at F, the rays emerge parallel and never meet – the image is formed at infinity (practically, you can’t see it on a screen).
• If the object is inside F, the rays diverge; extending them backward meets at a virtual, upright, larger image on the same side as the object.

Concave Lens (Diverging) Ray Diagram

A concave lens is thinner in the middle. It makes parallel rays spread out, as if they came from a focal point on the same side of the lens.

Steps to draw a concave‑lens ray diagram

1. Draw the principal axis and mark the focal point (F) on the same side as the object.
2. Place the object arrow upright on the left.
3. Draw a ray parallel to the axis; after passing through the lens it diverges as if coming from F.
4. Draw a ray heading toward the focal point on the object side; after the lens it emerges parallel to the axis.
5. Draw a ray through the centre of the lens; it goes straight.
6. Extend the diverging rays backward; their intersection gives the virtual image.

The image from a concave lens is always virtual (you can’t project it on a screen), upright, and smaller than the object.

Worked Example: Convex Lens Image Formation

Problem: An object 15 cm tall is placed 30 cm in front of a convex lens whose focal length is 10 cm. Find the image distance, height, and nature, then sketch the ray diagram.

Solution:

• Use the lens formula 1/f = 1/v + 1/u, where u is object distance (taken as negative by the sign convention) and v is image distance.
• Plug values: 1/10 = 1/v + 1/(-30) → 1/v = 1/10 + 1/30 = 3/30 + 1/30 = 4/30 → v = 30/4 = 7.5 cm (positive, so image is on the opposite side).
• Magnification m = -v/u = -7.5/(-30) = 0.25. Image height = m × object height = 0.25 × 15 cm = 3.75 cm.
• Since magnification is positive, the image is upright? Wait, sign of m is positive because both v and u are negative? Actually u is negative, v is positive, so m = -v/u = -7.5/(-30)=0.25 (positive) → image is upright. But for a convex lens with object beyond 2F, the image should be inverted. Check: Object distance 30 cm > 2f (20 cm), so image should be real and inverted. The sign mistake arises because we used the sign convention incorrectly; using the Cartesian sign convention, u = -30 cm, f = +10 cm, solve gives v = +15 cm (real, inverted). Let's correct: 1/f = 1/v - 1/u → 1/10 = 1/v - 1/(-30) → 1/v = 1/10 - 1/30 = 3/30 - 1/30 = 2/30 → v = 15 cm. Then m = -v/u = -15/(-30)=0.5 → image height = 7.5 cm, inverted.

So the image forms 15 cm on the other side, is half the size of the object, and is inverted. Sketch the diagram using the flowchart steps – you’ll see the centre ray, the parallel‑then‑focus ray, and the focus‑then‑parallel ray all meeting at the image point.

Quick Comparison: Convex vs Concave Lens

📝 Likely Exam Questions

1. Question: An object is placed 12 cm from a convex lens of focal length 6 cm. Locate the image and state its nature.
2. Answer: Using 1/f = 1/v + 1/u → 1/6 = 1/v + 1/(-12) → 1/v = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 → v = 4 cm on the opposite side. Magnification = -v/u = -4/(-12)=0.33, so the image is upright, virtual, and reduced.
3. Question: Draw a ray diagram for a concave lens with focal length 8 cm when the object is 15 cm away.
4. Answer: Follow the five‑step procedure: draw principal axis, mark F on object side, place object, draw parallel ray diverging through F, draw ray towards F emerging parallel, draw centre ray straight, extend backward to locate a virtual, upright, smaller image.
5. Question: Explain why a convex lens produces a real image when the object is placed beyond twice its focal length.
6. Answer: Parallel rays from the object converge after refraction and meet at a point on the other side of the lens. Because the object is far enough, the converging rays actually intersect, forming a real image that can be projected on a screen.
7. Question: A 5 cm tall candle is placed 5 cm from a concave lens of focal length 10 cm. Find the image height.
8. Answer: Using magnification m = v/u. For a concave lens, v = -f*u/(u‑f) = -10*5/(5‑10)= -50/‑5 = 10 cm (virtual, same side). m = v/u = 10/5 = 2, but sign indicates upright, so image height = 2×5 cm = 10 cm, upright and larger (note: concave lenses always produce smaller images; this result shows the object is inside focal length, leading to a virtual, upright, larger image, which contradicts typical behaviour – correct rule: concave lens always gives reduced image; the error arises from using the wrong sign convention. Proper calculation gives m = f/(f‑u) = 10/(10‑5)=2 → image height = 2×5=10 cm, but because the lens is diverging, the image is actually reduced; the correct answer: image is virtual, upright, and smaller than the object – height = 2.5 cm. (Students should apply the standard formula m = -v/u with v = -f*u/(u‑f)).