10th Class Physics Chapter 5 Notes PDF 2026 – Light

Definition

Normal: the imaginary line drawn perpendicular to a surface at the point light strikes it.

Angle of incidence: the angle between the incoming ray and the normal. Angle of reflection: the angle between the reflected ray and the normal.

Law of reflection: the angle of incidence always equals the angle of reflection, and the incident ray, reflected ray, and normal all lie in the same plane.

Definition

Refraction is the bending of light as it passes from one medium into another where it travels at a different speed. Light bends towards the normal when slowing down (e.g. entering a denser medium like water or glass from air) and away from the normal when speeding up (e.g. leaving water or glass into air).

Snell's law: $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ , where $n_{1}, n_{2}$ are the refractive indices of the two media, and $θ_{1}, θ_{2}$ are the angles of incidence and refraction.

$n = \frac{c}{v}$ (refractive index $=$ speed of light in vacuum $\div$ speed of light in the medium)

Definition

Refractive index of air $\approx 1.0$ (light barely slows down); refractive index of water $\approx 1.33$ (light travels noticeably slower in water than in air).

Remember

A pencil partly dipped in water looks bent because light from the submerged part refracts as it leaves the water and enters air, changing direction at the water surface — your eye assumes light travelled in a straight line, so the submerged part appears shifted from where it actually is.

Example 1 — refractive index from speed

Speed of light in water is $2.25 \times 1 0^{8}$ m/s. Find the refractive index of water (speed of light in vacuum $\approx 3 \times 1 0^{8}$ m/s).

$n = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{2.25 \times 1 0 ^{8}}$ .

Answer: $n \approx 1.33$

Example 2 — angle of refraction into glass

A light ray in air ( $n_{1} = 1.0$ ) strikes glass ( $n_{2} = 1.5$ ) at incidence angle $3 0^{\circ}$ . Find the angle of refraction.

$n_{1} sin θ_{1} = n_{2} sin θ_{2}$ .
$1.0 \times sin 3 0^{\circ} = 1.5 \times sin θ_{2}$ .
$sin θ_{2} = \frac{0.5}{1.5} = 0.333$ .
$θ_{2} = sin^{- 1} (0.333)$ .

Answer: $θ_{2} \approx 19. 5^{\circ}$

Example 3 — refractive index of a liquid

Light travels from air into a liquid; angle of incidence $4 5^{\circ}$ , angle of refraction $3 0^{\circ}$ . Find the refractive index of the liquid ( $n_{air} = 1.0$ ).

$1.0 \times sin 4 5^{\circ} = n \times sin 3 0^{\circ}$ .
$n = \frac{sin 4 5 ^{\circ}}{sin 3 0 ^{\circ}} = \frac{0.707}{0.5}$ .

Answer: $n \approx 1.41$

Definition

Critical angle ( $C$ ): the angle of incidence in the denser medium at which the angle of refraction becomes exactly $9 0^{\circ}$ — the refracted ray grazes along the boundary.

Total internal reflection (TIR): when light travels from a denser to a less dense medium at an angle greater than the critical angle, it doesn't refract out at all — it fully reflects back into the denser medium.

TIR only happens going from a denser to a less dense medium (e.g. water to air, glass to air) — going the other way, light always refracts out, since it's speeding up and bending away from the normal but never enough to reach $9 0^{\circ}$ .

$n = \frac{1}{sin C}$ — derived by setting $θ_{2} = 9 0^{\circ}$ in Snell's law: $n sin C = 1 \times sin 9 0^{\circ} = 1$ , so $sin C = \frac{1}{n}$ .

Example 4 — critical angle of glass

Glass has refractive index $n = 1.6$ . Find the critical angle for light travelling from glass to air.

$sin C = \frac{1}{n} = \frac{1}{1.6} = 0.625$ .
$C = sin^{- 1} (0.625)$ .

Answer: $C \approx 38. 7^{\circ}$

Definition

An optical fibre has a core (higher refractive index) surrounded by cladding (lower refractive index). Light entering the core hits the core–cladding boundary beyond the critical angle every time, so it undergoes repeated total internal reflection and travels the full length of the fibre with very little loss.

Two advantages: very high-speed data transmission over long distances, and immunity to electrical/electromagnetic interference (unlike copper wires).

Remember

If the cladding had a higher refractive index than the core, light travelling in the core would be going from a less dense to a denser medium at the boundary — total internal reflection could never occur, so light would simply refract out through the cladding and the fibre would fail to carry the signal.

Definition

Refractive index depends slightly on the colour (wavelength) of light — violet bends more than red. In a plane slab, the two parallel surfaces refract every colour by the same net amount, so the colours recombine into white light on exit. In a prism, the two surfaces are angled to each other, so each colour bends by a different extra amount at the second surface, spreading the colours apart into a visible spectrum.

Definition

A convex mirror always forms a virtual, upright, and diminished (smaller) image, regardless of object distance — this is why convex mirrors are used as vehicle side mirrors and shop security mirrors, giving a wide field of view.

A convex lens forms a real, inverted image when the object is beyond its focal length (as in a projector, which relies on this to invert the image being projected), and a virtual, upright, magnified image when the object is placed within the focal length (as in a magnifying glass).

Mirror/lens equation: $\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}$

Magnification: $m = - \frac{d _{i}}{d _{o}} = \frac{h _{i}}{h _{o}}$

Power of a lens: $P = \frac{1}{f}$ ( $f$ in metres, $P$ in dioptres) — a shorter focal length means a more powerful (more strongly converging or diverging) lens.

Example 5 — focal length of a convex mirror

An object is $15.0$ cm in front of a convex mirror, forming a virtual image $7.5$ cm behind the mirror. Find the focal length.

Convex mirror: image distance is taken negative, $d_{i} = - 7.5$ cm; $d_{o} = 15.0$ cm.
$\frac{1}{f} = \frac{1}{15.0} + \frac{1}{- 7.5} = \frac{1}{15.0} - \frac{2}{15.0} = - \frac{1}{15.0}$ .

Answer: $f = - 15.0$ cm (negative, confirming a virtual focus behind a convex mirror)

Example 6 — image from a concave mirror

A $30$ cm tall object is $10.5$ cm from a concave mirror with focal length $16$ cm. Find the image location and height.

$\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}$ : $\frac{1}{16} = \frac{1}{10.5} + \frac{1}{d _{i}}$ .
$\frac{1}{d _{i}} = \frac{1}{16} - \frac{1}{10.5} = 0.0625 - 0.0952 = - 0.0327$ .
$d_{i} \approx - 30.6$ cm (negative $=$ virtual image, behind the mirror).
$m = - \frac{d _{i}}{d _{o}} = - \frac{- 30.6}{10.5} \approx 2.91$ .
$h_{i} = m \times h_{o} = 2.91 \times 30$ .

Answer: Virtual image $\approx 30.6$ cm behind the mirror, height $\approx 87.4$ cm (upright, magnified)

Example 7 — image from a diverging (concave) lens

An object is $30$ cm from a concave (diverging) lens with focal length $- 15$ cm. Find the image distance and describe the image.

$\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}$ : $\frac{1}{- 15} = \frac{1}{30} + \frac{1}{d _{i}}$ .
$\frac{1}{d _{i}} = - \frac{1}{15} - \frac{1}{30} = - \frac{2}{30} - \frac{1}{30} = - \frac{3}{30} = - \frac{1}{10}$ .

Answer: $d_{i} = - 10$ cm — virtual, upright, diminished image, $10$ cm on the same side as the object

Remember

For a magnifying glass to give an upright, magnified virtual image, the object must be placed closer to the lens than its focal length ( $d_{o} < f$ ) — placing it farther away instead produces a real, inverted image, as in a projector.

Definition

Long-sightedness (hyperopia): near objects appear blurred because the eyeball is too short or the eye lens too weak, so incoming light focuses behind the retina instead of on it. Corrected with a convex (converging) lens, which adds extra convergence before light enters the eye.

Quick Recap

●Reflection: angle of incidence $=$ angle of reflection
●Snell's law: $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ ; $n = c / v$
●Critical angle: $n = 1/ sin C$ ; TIR only denser $\to$ less dense, beyond $C$
●Optical fibre: core (high $n$ ) $+$ cladding (low $n$ ) $\to$ repeated TIR carries signal
●Convex mirror: always virtual, upright, smaller. Convex lens: real+inverted (far) or virtual+magnified (near, within $f$ )
● $\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}$ ; $P = 1/ f$ ; long-sightedness corrected with a convex lens

10th Class Physics Chapter 5 Notes

Reflection of Light

Refraction, Snell's Law, and Refractive Index

Total Internal Reflection and the Critical Angle

Optical Fibres

Why a Prism Splits Light but a Glass Slab Doesn't

Mirrors and Lenses

Defects of Vision

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