CBSE Class 10 Science — Light: Reflection and Refraction

Chapter Overview & Weightage

Light: Reflection and Refraction is Chapter 10 in CBSE Class 10 Science. It is the most important physics chapter in Class 10, forming the direct foundation for Class 12 optics and ranking as one of the highest-scoring chapters in board exams.

This chapter carries 15–18 marks in CBSE Class 10 Science board exams. Mirror and lens formula numericals, image formation for different object positions, and the new Cartesian sign convention are the highest-weightage topics. At least one 5-mark numerical and one diagram question appear in every board exam.

What this chapter covers:

Reflection of light: laws, plane mirrors, spherical mirrors
Mirror formula, magnification
Refraction: laws, Snell’s law, refractive index
Lens formula, lens power
Image formation in mirrors and lenses for different object positions
Sign convention (New Cartesian)

Key Concepts You Must Know

Sign Convention (New Cartesian)

Everything is measured from the optical centre (lens) or pole (mirror):

Distances measured in the direction of incident light: positive
Distances measured against the direction of incident light: negative
Object is always placed to the left: $u$ is always negative
Heights above the principal axis: positive; below: negative

Laws of Reflection

Angle of incidence = Angle of reflection (both measured from the normal)
Incident ray, reflected ray, and normal are all in the same plane

Spherical Mirrors

Concave mirror: Reflecting surface curves inward (like a bowl facing you). Focal length $f$ is negative.

Convex mirror: Reflecting surface curves outward. Always gives virtual, erect, diminished images. Focal length $f$ is positive.

Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

Relationship: $f = \dfrac{R}{2}$ (focal length = half of radius of curvature)

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

$m > 0$ : virtual, erect image
$m < 0$ : real, inverted image
$|m| > 1$ : enlarged; $|m| < 1$ : diminished

Laws of Refraction (Snell’s Law)

Incident ray, refracted ray, and normal at the point of incidence are in the same plane
$n_1 \sin i = n_2 \sin r$ (Snell’s law)

Absolute refractive index ( $n$ ) of a medium = $\dfrac{\text{speed of light in vacuum}}{\text{speed of light in medium}} = \dfrac{c}{v}$

Optically denser medium: Higher $n$ → light bends toward the normal when entering.

Optically rarer medium: Lower $n$ → light bends away from the normal when entering.

Important Formulas

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

m = \frac{h'}{h} = -\frac{v}{u}

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

m = \frac{h'}{h} = \frac{v}{u}

P = \frac{1}{f(\text{in metres})} \quad \text{(power in dioptres, D)}

n_1 \sin i = n_2 \sin r

n = \frac{c}{v} = \frac{\lambda_0}{\lambda_m}

n_{21} = \frac{n_2}{n_1} = \frac{\sin i}{\sin r}

Solved Previous Year Questions

PYQ 1 — Mirror Formula Numerical

Q: An object is placed at 10 cm in front of a concave mirror of focal length 20 cm. Find the image distance and nature of the image. (CBSE 2023)

Solution:

Given: $u = -10$ cm (object to the left), $f = -20$ cm (concave mirror: $f$ negative).

Using mirror formula:

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-10} = -\frac{1}{20} + \frac{1}{10} = \frac{-1+2}{20} = \frac{1}{20}

$v = +20$ cm.

Nature: $v$ is positive → image is behind the mirror → virtual and erect.

Magnification: $m = -\frac{v}{u} = -\frac{20}{-10} = +2$ → image is twice as large (enlarged).

PYQ 2 — Lens Formula

Q: A convex lens of focal length 30 cm forms an image on a screen 60 cm away from the lens. Find the object distance. (CBSE pattern)

Solution:

$f = +30$ cm (convex lens), $v = +60$ cm (image on same side as real image → positive).

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

\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{60} - \frac{1}{30} = \frac{1-2}{60} = -\frac{1}{60}

$u = -60$ cm (object 60 cm to the left of lens).

PYQ 3 — Refractive Index

Q: The speed of light in glass is $2 \times 10^8$ m/s. Find the refractive index of glass. (Speed of light in vacuum = $3 \times 10^8$ m/s.) (CBSE — 2 marks)

Solution:

n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5

The refractive index of glass is 1.5 — light travels at $\frac{2}{3}$ its vacuum speed in this glass.

Image Formation Table — Concave Mirror

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

Difficulty Distribution

Difficulty	Topic	Marks
Easy (20%)	Laws of reflection/refraction; refractive index formula	1 mark
Medium (45%)	Mirror/lens formula numericals; magnification; image position	2–3 marks
Hard (35%)	Image formation table; ray diagrams; power calculations; multi-step	4–5 marks

Expert Strategy

The most common error in Class 10 optics: forgetting the sign of $u$ . Object distance $u$ is ALWAYS negative (object to the left). Write $u = -30$ cm, not $u = 30$ cm. Use this after every substitution: check that your substituted values match the sign convention.

The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ uses a minus sign (not plus like the mirror formula). This trips many students. To remember: Lens = $\frac{1}{v} \mathbf{-} \frac{1}{u}$ ; Mirror = $\frac{1}{v} \mathbf{+} \frac{1}{u}$ .

Common Traps

Trap 1 — Using the wrong formula for mirrors vs lenses: Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ . Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . The sign difference is crucial. Using the lens formula for a mirror problem gives a wrong answer.

Trap 2 — Forgetting that convex mirror always gives virtual image: No matter where the object is, a convex mirror always forms a virtual, erect, diminished image behind the mirror. If your calculation gives a real image for a convex mirror, your signs are wrong.

Trap 3 — Focal length sign: Concave mirror: $f$ is negative. Convex mirror: $f$ is positive. Convex lens: $f$ is positive. Concave lens: $f$ is negative. Students mix these up under exam pressure.

Trap 4 — Power of lens in wrong units: Power $P = 1/f$ requires $f$ in metres, not cm. If $f = 30$ cm, convert: $f = 0.30$ m, then $P = 1/0.30 = 3.33$ D. Using $f = 30$ (in cm) gives $P = 0.033$ D — wrong by a factor of 100.