Light — Reflection and Refraction, Lenses & Mirrors (Class 10)

Light Behaves in Predictable Ways — and That's What We Exploit

Light travels in straight lines, but the moment it hits a surface or crosses from one medium to another, it obeys two of the most testable laws in Class 10 physics. The entire chapter — mirrors, lenses, the human eye — is built on just these two laws.

Most students memorise the mirror formula and lens formula separately and treat them as unrelated. They're not. Both come from the same geometry of curved surfaces. Once you see that connection, the chapter stops feeling like a list of formulas and starts making sense.

For CBSE, this chapter consistently carries 5–7 marks in the board exam, usually split between a 2-mark theory question, a 3-mark numerical, and sometimes a 5-mark question on ray diagrams or the human eye. ICSE asks similar questions but tends to go deeper on the human eye and optical instruments. Know your formulas cold, but also know why they work — diagram-based questions reward that understanding.

Key Terms and Definitions

Reflection — when light bounces back from a surface into the same medium. A polished mirror reflects almost all incident light.

Refraction — when light passes from one transparent medium to another and bends at the interface, because its speed changes.

Refractive index (n) — the ratio of the speed of light in vacuum to its speed in a given medium. For glass, n ≈ 1.5, meaning light travels 1.5× slower in glass than in vacuum.

Refractive Index

$n = \frac{c}{v} = \frac{\sin i}{\sin r}$

where $c$ = speed of light in vacuum, $v$ = speed in medium, $i$ = angle of incidence, $r$ = angle of refraction.

Principal axis — the line passing through the centre of curvature and the pole of a spherical mirror or lens.

Focal length (f) — distance from the pole (mirror) or optical centre (lens) to the principal focus. For a concave mirror, f is negative in the sign convention we use.

Optical centre — the central point of a lens through which any ray passes undeviated.

Power of a lens (P) — the reciprocal of focal length in metres: $P = 1/f$ . Unit is dioptre (D). A converging (convex) lens has positive power; a diverging (concave) lens has negative power.

Spherical Mirrors

The Sign Convention

We follow the New Cartesian Sign Convention:

All distances are measured from the pole of the mirror
Distances in the direction of incident light → positive
Distances opposite to incident light → negative
Heights above principal axis → positive; below → negative

Since incident light travels from left to right, the object is always to the left (negative side). For a concave mirror, the centre of curvature and focus are in front of the mirror — also negative. For a convex mirror, they're behind the mirror — positive.

⚠️ Common Mistake

Students often assign the wrong sign to radius of curvature. Remember: for concave mirrors, both R and f are negative in the standard setup. Never mix up which mirror has which sign.

Mirror Formula and Magnification

Mirror Formula

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} = \frac{2}{R}$

Magnification: $m = -\frac{v}{u} = \frac{h'}{h}$

Where $u$ = object distance, $v$ = image distance, $h$ = object height, $h'$ = image height.

If $m$ is negative, the image is inverted (real). If positive, the image is erect (virtual).

Ray Diagrams — the Three Principal Rays

For any spherical mirror, draw two of these three rays to locate the image:

A ray parallel to the principal axis → reflects through (or appears to come from) the focus F
A ray through the centre of curvature C → reflects back along the same path
A ray through (or directed toward) the focus F → reflects parallel to the principal axis

Image Positions for Concave Mirror

Object position	Image position	Nature
At infinity	At F	Real, inverted, highly diminished
Beyond C	Between F and C	Real, inverted, diminished
At C	At C	Real, inverted, same size
Between C and F	Beyond C	Real, inverted, enlarged
At F	At infinity	Real, inverted, highly enlarged
Between F and P	Behind mirror	Virtual, erect, enlarged

Concave mirrors are used in torches, headlights, and shaving/makeup mirrors. Convex mirrors are used as rear-view mirrors because they give a wider field of view and always form erect, diminished images regardless of object position.

Refraction of Light

Snell's Law

$n_1 \sin i = n_2 \sin r$

Or equivalently: $\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$

When light goes from a rarer to a denser medium ( $n_1 < n_2$ ), it bends towards the normal. Going from denser to rarer, it bends away from the normal.

The classic example: a pencil in a glass of water appears bent. The light from the submerged part refracts as it exits water (denser) into air (rarer), bending away from the normal and reaching our eyes at a displaced angle.

Absolute vs Relative Refractive Index

Absolute refractive index = refractive index with respect to vacuum/air (we treat air ≈ vacuum for most problems).

Relative refractive index of medium 2 with respect to medium 1: $_1n_2 = \frac{n_2}{n_1}$

💡 Expert Tip

When light goes from medium 1 to medium 2, the relative refractive index you use is $n_2/n_1$ . Many students accidentally flip this and get an answer that contradicts the direction of bending.

Lenses

Types and Terminology

Convex lens (converging) — thicker at centre, converges parallel rays to a real focus. Focal length is positive.

Concave lens (diverging) — thinner at centre, diverges parallel rays; they appear to come from a virtual focus. Focal length is negative.

Two focal points: $F_1$ (first focal point) and $F_2$ (second focal point) are equidistant from the optical centre on either side.

Lens Formula and Magnification

Lens Formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Magnification: $m = \frac{v}{u}$

Power: $P = \frac{1}{f(\text{in metres})}$ dioptre

Note the lens formula has a minus sign between $1/v$ and $1/u$ — this trips up students who memorise the mirror formula and assume lenses work the same way.

Image Positions for Convex Lens

Object position	Image position	Nature
At infinity	At $F_2$	Real, inverted, point-sized
Beyond $2F_1$	Between $F_2$ and $2F_2$	Real, inverted, diminished
At $2F_1$	At $2F_2$	Real, inverted, same size
Between $F_1$ and $2F_1$	Beyond $2F_2$	Real, inverted, enlarged
At $F_1$	At infinity	—
Between $F_1$ and O	Same side as object	Virtual, erect, enlarged

A concave lens always forms a virtual, erect, diminished image on the same side as the object, regardless of object position.

Combination of Lenses

When two thin lenses are placed in contact: $P = P_1 + P_2 \quad \text{and} \quad \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$

This is why optometrists describe prescriptions in dioptres — powers simply add up.

The Human Eye and Vision Defects

The human eye functions like a convex lens system. The ciliary muscles change the shape of the eye lens, adjusting its focal length — this is called accommodation.

Near point (least distance of distinct vision) = 25 cm for a normal eye. Far point = infinity for a normal eye.

Myopia (Short-sightedness)

The image forms in front of the retina. Distant objects appear blurred. The eyeball is too long or the lens is too converging.

Correction: concave lens (diverging) that shifts the image back to the retina.

Hypermetropia (Long-sightedness)

The image forms behind the retina. Near objects appear blurred. The eyeball is too short or the lens is too flat.

Correction: convex lens (converging) that shifts the image forward to the retina.

Presbyopia

Age-related loss of accommodation — the ciliary muscles weaken and the lens becomes less flexible. Corrected with bifocal lenses.

🎯 Exam Insider

CBSE frequently asks: "A person cannot see objects closer than 1 m. Find the power of lens needed." This is always a hypermetropia problem — the near point has shifted from 25 cm to 100 cm. Use the lens formula with $v = -25$ cm and $u = -100$ cm to get $f$ , then $P = 100/f(\text{cm})$ .

Solved Examples

Example 1 — CBSE Level (2 marks)

An object is placed 15 cm in front of a concave mirror of focal length 10 cm. Find the image distance.

Using sign convention: $u = -15$ cm, $f = -10$ cm.

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15}$ $= \frac{-3 + 2}{30} = \frac{-1}{30}$

So $v = -30$ cm. The image is 30 cm in front of the mirror — real and inverted.

Example 2 — CBSE Level (3 marks)

A convex lens of focal length 20 cm is used as a magnifying glass. An object is placed 15 cm from the lens. Find the image position and magnification.

$u = -15$ cm, $f = +20$ cm.

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-15} = \frac{3 - 4}{60} = \frac{-1}{60}$

So $v = -60$ cm (same side as object → virtual image).

$m = \frac{v}{u} = \frac{-60}{-15} = +4$

The image is virtual, erect, and 4× magnified. This is how a magnifying glass works — object inside focal length.

Example 3 — ICSE/Advanced Level (5 marks)

A person with myopia has a far point of 80 cm. What power of corrective lens does she need? If she later develops presbyopia and her near point shifts to 1.5 m, what power bifocal addition is needed for reading (near point at 25 cm)?

For myopia correction: She can't see beyond 80 cm, so we need a concave lens that forms the image of a distant object ( $u = \infty$ ) at her far point.

$v = -80$ cm $= -0.8$ m, $u = \infty$

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-0.8} - 0 = -1.25 \text{ D}$

For presbyopia correction (reading addition): She needs to see at $v = -1.5$ m (her current near point) when the object is at $u = -0.25$ m.

$\frac{1}{f} = \frac{1}{-1.5} - \frac{1}{-0.25} = -0.667 + 4 = +3.33 \text{ D}$

The bifocal addition needed is approximately +3.33 D.

Exam-Specific Tips

🎯 Exam Insider

CBSE Board (5-7 marks)

At least one 5-mark question typically asks for a ray diagram + formula derivation or two 3-mark numericals. Practice both.
In ray diagrams, always label: object, image, F, C (or $2F$ ), and the principal axis. Missing labels cost marks.
The "power of lens" numerical (1-2 marks) is a guaranteed scoring question — never drop it.

ICSE (similar weightage)

ICSE tends to include more conceptual questions: "Why does a convex mirror always give a diminished image?" Know the geometric explanation, not just the formula result.
Dispersion of light and the spectrum may appear as a short-answer question — know that violet deviates more than red.

Scoring strategy: Mirror/lens formula numericals are 100% formula-based. If you know the sign convention cold, these are full-mark questions. Do not lose even 0.5 marks here.

Common Mistakes to Avoid

⚠️ Common Mistake

Mistake 1 — Forgetting the negative sign in lens formula Mirror formula: $1/v + 1/u = 1/f$ Lens formula: $1/v - 1/u = 1/f$

These are different. Don't substitute $u$ values into the mirror formula when solving lens problems.

⚠️ Common Mistake

Mistake 2 — Wrong sign for focal length Concave mirror: $f$ is negative. Convex lens: $f$ is positive. These are the two most used surfaces in this chapter and students swap them every exam season.

⚠️ Common Mistake

Mistake 3 — Magnification sign confusion $m < 0$ means inverted (real) image. $m > 0$ means erect (virtual). A real image of a real object is always inverted for both concave mirrors and convex lenses.

⚠️ Common Mistake

Mistake 4 — Using cm instead of metres for power $P = 1/f$ requires $f$ in metres. If $f = 50$ cm, then $f = 0.5$ m and $P = +2$ D. Writing $P = 1/50$ gives a wrong answer by a factor of 100.

⚠️ Common Mistake

Mistake 5 — Misidentifying vision defects If a person can't see far → myopia → concave lens correction. If a person can't see near → hypermetropia → convex lens correction. Students often flip the correction lens. Anchor it: concave diverges the rays, pushing the far point back to infinity. Convex converges, pulling the near point closer.

Practice Questions

Q1. An object 5 cm tall is placed 25 cm in front of a concave mirror of focal length 15 cm. Find the image distance, magnification, and image height.

Q2. Light travels from water (n = 4/3) into glass (n = 3/2). If the angle of incidence in water is 30°, find the angle of refraction in glass.

Q3. A convex lens has focal length 10 cm. Where should an object be placed to get a real image twice the size of the object?

Q4. A doctor prescribes –2.5 D lenses for a myopic patient. What is the far point of the patient?

Q5. Two lenses of power +3.5 D and –1.5 D are placed in contact. What is the effective focal length of the combination?

Q6. A concave mirror produces a virtual image 3 times the size of the object. If the object is 10 cm from the mirror, find the focal length.

Q7. The refractive index of glass with respect to water is 1.125. If the refractive index of glass w.r.t. air is 1.5, find the refractive index of water w.r.t. air.

Q8. An object is placed at the centre of curvature of a concave mirror (R = 30 cm). Describe the image completely.

FAQs

Why does a concave mirror form a virtual image only when the object is between F and P?

A concave mirror converges reflected rays. If the object is close enough (inside the focal length), the reflected rays diverge after reflection — they never actually meet in front of the mirror. Your eye traces them back and sees them appearing to come from behind the mirror. That's the virtual image. Move the object beyond F and the reflected rays do converge in front of the mirror — real image.

What is the difference between real and virtual images?

A real image is formed where reflected/refracted rays actually intersect. You can project it on a screen. A virtual image is formed where the rays appear to come from when extended backwards — you can't project it on a screen. A plane mirror always gives a virtual image.

Why is the refractive index of a medium always greater than 1?

Because light travels slower in any medium than in vacuum ( $v < c$ ), so $n = c/v > 1$ always. The refractive index of vacuum itself is exactly 1.

Why do we use a convex mirror as a rear-view mirror and not a concave mirror?

A convex mirror always gives an erect, diminished image for any object position, and it covers a wider field of view than a plane or concave mirror of the same size. A concave mirror would give distorted images for objects at most positions, making it unsafe for driving.

What happens when light travels along the principal axis of a lens?

It passes straight through undeviated. The principal axis is a line of symmetry — there's no bending along it. Only rays at an angle to the principal axis are refracted by the lens surfaces.

Is focal length affected by the colour of light?

Yes — this is called chromatic aberration. Different colours (wavelengths) refract by different amounts (violet more, red less), so they have slightly different focal lengths for the same lens. This causes coloured fringes around images in simple lenses. Achromatic doublets (used in quality optics) correct this.

For a person using reading glasses of +2 D, what is their near point?

Their glasses bring objects at 25 cm (normal near point) into focus for them. The lens with $f = +50$ cm must create an image at their actual near point when object is at 25 cm.

$u = -25$ cm, $f = +50$ cm

$\frac{1}{v} = \frac{1}{50} + \frac{1}{-25} = \frac{1-2}{50} = \frac{-1}{50}$

$v = -50$ cm, so their near point is 50 cm.

Why does a pencil appear bent in a glass of water?

The light from the submerged part of the pencil travels from water (denser, n ≈ 1.33) into air (rarer, n ≈ 1). It bends away from the normal, reaching our eyes at a steeper angle. Our brains interpret light as travelling in straight lines, so they project the submerged pencil to an apparent position that is shifted upwards — making the pencil appear bent at the water surface.