CBSE Class 10 Science · 2026–27
Why does a coin look higher in water, and how can one lens make either a screen image or a magnified view? Follow a ray, predict its path, then check your prediction with a calculation. Work through the short topics in order or return to the sign and image maps when you need them.
- Trace refraction through a boundary and a parallel-sided slab.
- Use Snell’s law and absolute/relative refractive index with clear medium order.
- Construct and interpret convex/concave lens images.
- Apply Cartesian signs, lens formula, magnification and power with units.
- Explain and evaluate simple ray-tracing/focal-length observations.
01Why does a straw look bent?
A straw in a glass can look kinked at the water surface even though it is straight. The light reaching your eye changes direction as it leaves the water. Refraction is the change in the path of light at a boundary between media with different refractive indices.
Draw a normal, a line perpendicular to the surface at the point where the ray arrives. Measure both the angle of incidence i and angle of refraction r from this normal, never from the surface.
- Into a medium with higher refractive index: the ray slows and bends towards the normal.
- Into a medium with lower refractive index: a transmitted ray bends away from the normal.
- At normal incidence, the ray continues straight even though its speed changes.
“Optically denser” means higher refractive index, not necessarily greater mass per unit volume. For these diagrams, use transparent ordinary media and a single colour of light.
Pause. Think. Try.
02A rule that predicts the bend
For a fixed pair of media and a fixed colour, the incident ray, refracted ray and normal lie in one plane. Their angles obey Snell’s law:
n₁ sin i = n₂ sin r, or n₂/n₁ = sin i/sin r.
The ratio is not i/r. Sines are numerical values associated with angles. When sine values are supplied, substitute those values directly; you do not need to guess the angles.
Try the reasoning: air is approximated as n₁=1.00. For one ray, sin i=0.64 and sin r=0.40. Then n₂=0.64/0.40=1.60. Refractive index has no unit because it is a ratio. Snell’s law also explains the straight path at normal incidence: i=0° gives r=0° for transmission.
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03Refractive index is a speed comparison
The absolute refractive index of a medium is n=c/v, where c≈3.00×108 m/s is the speed of light in vacuum and v is its speed in the medium. For classroom calculations, the speed in air is usually approximated by c.
A material with n=1.60 has v=(3.00×108)/1.60=1.875×108 m/s. A larger n means a smaller speed, not faster light.
For light passing from medium 1 to medium 2, the relative refractive index is n₂₁=n₂/n₁=v₁/v₂. Always state which medium is compared with which. If n₁=1.25 and n₂=1.60, n₂₁=1.28. Reversing the direction gives the reciprocal, 1/1.28≈0.781.
Frequency stays the same across a stationary boundary. Speed and wavelength change. Index values depend on the material and the colour used; a table value describes stated conditions.
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04Through a glass slab: parallel, but shifted
A rectangular glass slab has two parallel refracting faces. A ray entering obliquely from air bends towards the first normal. At the second face it leaves glass for air and bends away from the second normal.
Because the faces are parallel and the surrounding medium is the same, the emergent ray is parallel to the original incident direction: the angle of emergence equals the angle of incidence. It is usually displaced sideways. The shortest perpendicular distance between the emergent ray and the original ray’s extended path is the lateral displacement.
The displacement depends on slab thickness, incidence angle and refractive index. Do not assume that doubling an angle or index exactly doubles the shift. At normal incidence there is no sideways displacement.
Reading a diagram: extend the incident path as a dashed reference, draw the emergent physical ray with an arrow, and measure the shift between these two parallel lines.
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05Why a coin can look higher
Light from a submerged coin travels from water into air and bends away from the normal. Your eye traces the arriving rays backwards along straight lines. Those backward extensions appear to start above the actual coin: a virtual image is seen nearer the water surface.
This explains a pool looking shallower and the underwater part of a pencil appearing displaced. The object has not moved, and the eye does not send rays out to it.
A simple observation: fix your viewing position while a partner adds clean water to a bowl containing a coin. A coin hidden by the rim may become visible because a new refracted path reaches your eye. Use a stable bowl and wipe spills. The model is an observation of light, not a reliable way to judge swimming depth.
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06Meet the two lenses and their landmarks
A lens is a transparent optical element bounded by two surfaces, at least one curved. We use the thin spherical-lens model in air: rays close to the principal axis, and lens thickness small compared with object and image distances.
- A convex lens, thicker in the middle, converges rays parallel to its principal axis.
- A concave lens, thinner in the middle, makes those rays diverge.
The optical centre O is the reference point for distance measurements. A ray through O is drawn undeviated in the thin-lens approximation. The principal axis passes through O and the centres of curvature of the spherical surfaces.
Each lens has two principal foci, one on either side. Their distance from O is the magnitude of the focal length. For our left-to-right light diagrams, label the left focus F₁ and the right focus F₂. A convex lens has a real focus for an incoming parallel beam; the concave lens’s diverging rays have a virtual focus found by backward extensions.
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07Three useful rays for a convex lens
A large number of rays leave every object point. To locate its image, draw two rays whose paths you know:
- A ray parallel to the principal axis refracts through F₂.
- A ray through O continues straight in the thin-lens model.
- A ray through F₁ emerges parallel to the axis.
Start both construction rays at the same object point, usually the top of an upright arrow. If the actual refracted rays meet, they locate a real image point. If only their dashed backward extensions meet, the image point is virtual.
Do not join a ray from the top of the object to a ray from the bottom and call their crossing the image. Draw physical rays with arrows; dashed extensions are a tracing aid, not light travelling backwards.
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08The convex-lens image map
Move an illuminated arrow towards a convex lens. The image changes when the object crosses 2 F₁ and F₁. Use two rays to understand the table rather than memorising six disconnected statements.
| Object position | Image position | Size and nature |
|---|---|---|
| Very far away | Near F₂ (on the focal plane) | Highly diminished; real, inverted |
| Beyond 2 F₁ | Between F₂ and 2 F₂ | Diminished; real, inverted |
| At 2 F₁ | At 2 F₂ | Same size; real, inverted |
| Between F₁ and 2 F₁ | Beyond 2 F₂ | Enlarged; real, inverted |
| At F₁ | At infinity in the ideal model | Emerging rays parallel; no finite screen position |
| Between F₁ and O | Same side as object | Enlarged; virtual, upright |
A camera uses a real image, a projector can make an enlarged real image, and a magnifying glass uses a virtual upright image. At the focus, parallel rays do not meet at a finite distance. “Image at infinity” names the limiting case.
Covering half a lens does not remove half the image: rays from every object point can still pass through the uncovered portion. The whole image remains, usually dimmer.
Pause. Think. Try.
09A concave lens makes a smaller upright view
For a real object in front of a concave lens, use these rays: a parallel ray emerges as if from F₁; a ray through O goes straight; a ray directed towards F₂ emerges parallel to the axis.
The outgoing rays diverge. Extend them backwards with dashed lines. Their apparent meeting point lies between O and F₁, on the object’s side. The image is virtual, upright and diminished for every finite real-object position in this model.
As a very distant object moves farther away, its image approaches F₁ and becomes very small. A concave lens therefore does not act as an ordinary magnifying glass. It is useful as the diverging corrective element in the school model of myopia.
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10Put signs on the drawing before calculating
Use the New Cartesian sign convention. Place the real object to the left and let incident light travel left to right. Measure every horizontal distance from O.
- Right of O: positive; left of O: negative.
- Above the axis: positive height; below it: negative height.
- A real object on the left has u<0.
- Convex lens: f>0. Concave lens: f<0.
- Real image on the right: v>0. Virtual image on the left: v<0.
Write a signed “given” line before substituting. A 30 cm distance in front of the lens becomes u=−30 cm. The minus sign specifies a direction; it does not mean a negative physical length.
Mirror comparison: lens distances start at O; mirror distances start at the pole P. A real lens image is on the outgoing side, unlike a real mirror image on the incident side. Reusing a mirror sketch can give a plausible-looking wrong sign.
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11The lens formula and what magnification tells you
For a thin lens, 1/f = 1/v − 1/u. An often useful rearrangement is 1/v = 1/f + 1/u. Keep u, v, f in the same unit. This formula is used for both convex and concave lenses; the signs carry the difference.
Linear magnification: m=hᵢ/hₒ=v/u. A positive m means an upright image; for the real objects here it is virtual. A negative m means inverted, and here real. The magnitude |m| compares sizes: greater than 1 means enlarged; less than 1 means diminished.
For example, m=−2 means an inverted image twice the object’s height. It does not mean a smaller image because −2 is less than 1. Use the magnitude for size and the sign for orientation.
The mirror relations differ: 1/f=1/v+1/u and m=−v/u. Choose the optical device before choosing the formula. Deriving either formula is not needed for this course’s core exercises.
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12Worked example: put a projector image on a screen
An upright 2 cm arrow is 18 cm in front of a convex lens of focal length 12 cm. Where is the image, and how tall is it?
- Signed data: u=−18 cm, f=+12 cm, hₒ=+2 cm.
- Lens relation: 1/v=1/12+1/(−18)=3/36−2/36=1/36.
- Image distance: v=+36 cm, on the other side of the lens.
- Magnification: m=36/(−18)=−2.
- Image height: hᵢ=m hₒ=−2×2=−4 cm.
The image is real, inverted and 4 cm tall. A screen 36 cm to the right receives it. The object is between f and 2 f, so an enlarged real image beyond 2 f is exactly what the ray map predicts.
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13Worked example: a magnifying-glass view
Place an upright 1 cm symbol 10 cm in front of a convex lens with focal length 15 cm.
- u=−10 cm, f=+15 cm.
- 1/v=1/15−1/10=(2−3)/30=−1/30.
- v=−30 cm: the image is on the same side as the symbol.
- m=(−30)/(−10)=+3; hᵢ=+3 cm.
The image is virtual, upright and three times as tall. You view it through the lens; it is not projected onto a screen at that virtual position.
Why does a converging lens give a virtual image? The object is inside the focal distance. The rays still bend towards one another relative to their incoming directions, but they leave diverging. Their backward extensions meet. “Convex” describes the lens, not a guarantee of a real image for every object position.
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14Worked example: a diminishing lens
An upright 3 cm arrow stands 36 cm in front of a concave lens with focal-length magnitude 18 cm.
- u=−36 cm, f=−18 cm, hₒ=+3 cm.
- 1/v=−1/18−1/36=−3/36=−1/12.
- v=−12 cm.
- m=(−12)/(−36)=+1/3.
- hᵢ=(1/3)×3=+1 cm.
The image is virtual, upright, 1 cm tall and 12 cm to the left of O, between the lens and its left focus. All three observations match the concave-lens ray map.
Answer check: if you obtain a real enlarged image for an isolated concave lens with a real object in air, revisit the signs and formula before accepting the arithmetic.
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15Power: convert centimetres first
The power of a lens measures its convergence or divergence: P=1/f, with f in metres. The unit is the dioptre (D), equivalent to m−1.
A convex lens with f=+40 cm has f=+0.40 m, so P=1/0.40=+2.5 D. A concave lens with f=−25 cm has f=−0.25 m and P=−4.0 D.
Positive power denotes a converging lens; negative power denotes a diverging lens. A greater magnitude means a stronger effect on a parallel beam. Thus −4 D is a stronger diverging power than −2 D, although −4 is the smaller signed number.
To go backwards, use f=1/P in metres, then convert if needed. For example, P=+1.25 D gives f=+0.80 m=+80 cm. These are optical calculations, not a method for choosing someone’s spectacles.
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16Make an observation that tests the model
Find an approximate focal length. With a teacher’s setup, point a convex lens towards a distant, well-lit tree or building. Move a white screen on the other side until a sharp inverted image appears. Measure from O to the screen. A sufficiently distant object gives an image near the focal plane, so this distance approximates f. Repeat and compare readings.
Use an ordinary distant object, never the Sun, and do not look through the lens towards it. A distant object is an approximation to infinity, so report a sensible measurement uncertainty rather than an exact mathematical focal length.
Trace a slab ray. Outline a rectangular glass block on paper. Mark an oblique incident path and use a ray box, or teacher-guided upright pins and careful sighting, to find the emergent path. Remove the block, join the paths, draw normals and measure i, r, e. Keep pins upright and handle them carefully. The testable prediction is i≈e with a sideways shift; experimental readings need not match perfectly.
When describing an experiment, name what you observed, what you measured and the relation you tested.
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17Explore further: two thin lenses in contact
For two thin lenses in close contact in air, the equivalent power is the algebraic sum: P=P₁+P₂. Add powers, not focal lengths.
For example, +3.0 D and −1.0 D combine to give +2.0 D, equivalent to a converging lens of f=+0.50 m. With separated lenses, the simple sum is generally insufficient: spacing matters.
This is a useful extension of the power idea. The core path through this chapter is complete before this extra application.
Pause. Think. Try.
Your formula shelf
| Idea | Relation | Remember |
|---|---|---|
| Absolute refractive index | n=c/v | Same colour; n dimensionless, speeds in the same units. |
| Relative index / Snell law | n₂₁=n₂/n₁=v₁/v₂=sin i/sin r | Light travelling from medium 1 to 2; angles measured from the normal. |
| Thin-lens formula | 1/f=1/v −1/u | Thin spherical lens, paraxial rays; use consistent distance units and Cartesian signs. |
| Linear magnification | m=hᵢ/hₒ=v/u | Signed heights/distances; |m| compares sizes. |
| Lens power | P=1/f | f in metres; P in dioptres (D). Convex positive, concave negative. |
| Thin lenses in contact: extension | P=P₁+P₂ | Thin lenses in contact in air; not for arbitrary separated lenses. |
Work through the numbers
Light travels in a transparent block of index 1.60. Find its speed, taking c=3.00×10^8 m/s.
n=c/v, so v=c/n=(3.00×108)/1.60=1.875×108 m/s. The answer is smaller than c, as expected.
A 2 cm object is 18 cm in front of a convex lens of focal length 12 cm. Find image position, magnification and height.
u=−18 cm, f=+12 cm. 1/v=1/12−1/18=1/36; v=+36 cm. m=v/u=−2; hᵢ=−4 cm. The image is real, inverted and 4 cm tall, 36 cm on the other side.
A 1 cm symbol is 10 cm in front of a convex lens of focal length 15 cm. Find its image.
u=−10 cm, f=+15 cm. 1/v=1/15−1/10=−1/30. Thus v=−30 cm, m=+3 and hᵢ=+3 cm. The virtual upright image lies 30 cm from the lens on the object’s side.
A 3 cm arrow is 36 cm in front of a concave lens of focal-length magnitude 18 cm. Find its image.
u=−36 cm, f=−18 cm. 1/v=−1/18−1/36=−1/12. Thus v=−12 cm, m=+1/3 and hᵢ=+1 cm: virtual, upright and diminished.
A lens has power −2.5 D. Find its focal length in metres and centimetres.
f=1/P=1/(−2.5)=−0.40 m=−40 cm. It is a diverging lens.
Bring it all together
Mix the ideas. Try each question before opening its explanation.
Write it. Explain it.
Use paper for a calculation, a labelled diagram or a short explanation. Then compare the reasoning. Suggested marks help you check how complete your answer is.
A ray enters a rectangular glass slab from air at an oblique angle and leaves into air. Explain the two bends and the emergent direction. Describe the lateral displacement so a reader could draw it. · 3 marks
It bends towards the normal on entering higher-index glass and away from the normal on leaving for air. With parallel faces and the same outside medium, the emergent ray is parallel to the original incident direction. Lateral displacement is the perpendicular distance between the emergent ray and the straight extension of the original incident ray.
- Normals perpendicular to both faces
- Arrowed real ray and dashed original-path extension
A projector lens has focal length +10 cm. An upright 2 cm slide is 15 cm in front of it. Find image distance, signed magnification, image height and screen position. · 4 marks
u=−15 cm, f=+10 cm. 1/v=1/10−1/15=1/30, so v=+30 cm. m=30/(−15)=−2. hᵢ=−2×2=−4 cm. Put the screen 30 cm beyond the lens; the image is real, inverted and 4 cm tall.
- 1: signed givens and correct lens relation
- 1: v=+30 cm
- 1: m=−2 and hᵢ=−4 cm
- 1: interpretation/screen position
A team uses a convex lens with f=12 cm. It tries object distances 30 cm, 24 cm, 18 cm and 8 cm. Which gives a same-size real image? Which gives an enlarged real image? Which gives an upright magnified view? Explain whether covering half the lens removes half of any image. · 4 marks
24 cm=2 f gives the same-size real image. 18 cm is between f and 2 f and gives an enlarged real image. 8 cm is inside f and gives the upright magnified virtual view. Covering half the lens reduces brightness but does not remove half the image, since rays from each object point can pass through the uncovered region.
- 1: 24 cm with 2 f reasoning
- 1: 18 cm with between f/2 f reasoning
- 1: 8 cm with inside f/virtual reasoning
- 1: complete-but-dimmer image with ray-bundle reason
An upright 3 cm object is 30 cm before a concave lens of focal length −15 cm. Find the image distance, height and nature. · 3 marks
u=−30 cm. 1/v=−1/15−1/30=−1/10, so v=−10 cm. m=(−10)/(−30)=+1/3 and hᵢ=+1 cm. The image is virtual, upright and diminished on the object’s side, between O and F₁.
- 1: correct formula/signs and v=−10 cm
- 1: m=+1/3 and hᵢ=+1 cm
- 1: correct nature and position