Skip to main content
SingaporePhysicsSyllabus dot point

How does light bounce off mirrors and bend when it passes between materials?

State the laws of reflection and refraction, define refractive index, and describe total internal reflection

A focused answer to the O-Level Physics outcome on reflection and refraction. The law of reflection, refraction and Snell's law, refractive index, the critical angle, and total internal reflection with its uses.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to state the law of reflection, to describe and calculate refraction using the refractive index (Snell's law), and to explain total internal reflection and the critical angle with their uses. The big idea is that light travels in straight lines, reflects predictably off surfaces, and bends when it changes speed crossing a boundary between materials.

The answer

The law of reflection

When light reflects off a surface, the angle of incidence equals the angle of reflection, both measured from the normal (the line perpendicular to the surface):

angle of incidence=angle of reflection\text{angle of incidence} = \text{angle of reflection}

The incident ray, the reflected ray, and the normal all lie in the same plane. A plane mirror forms an image that is upright, the same size, laterally inverted, and as far behind the mirror as the object is in front.

Refraction

Refraction is the bending of light as it passes from one material to another, caused by a change in its speed. Going into a denser material (such as air to glass), light slows down and bends toward the normal. Going into a less dense material, it speeds up and bends away from the normal.

Refractive index and Snell's law

The refractive index nn of a material measures how much it bends light:

n=sin⁡isin⁡rn = \frac{\sin i}{\sin r}

where ii is the angle of incidence and rr the angle of refraction, from the normal. A larger refractive index means more bending. Glass has n≈1.5n \approx 1.5.

Total internal reflection

When light travels from a denser to a less dense material, increasing the angle of incidence eventually makes the refracted ray bend to 90∘90^\circ. The angle of incidence for which this happens is the critical angle. Beyond the critical angle, no light escapes and it is all reflected back inside, a process called total internal reflection. This needs two conditions: the light is in the denser medium, and the angle of incidence exceeds the critical angle.

Examples in context

Example 1. Optical fibres. A thin glass fibre carries light by total internal reflection: each time the light hits the fibre wall it strikes above the critical angle, so it reflects rather than escapes, bouncing along the fibre even around gentle bends. This carries internet and phone signals as pulses of light over long distances.

Example 2. A straw looking bent. A straw in a glass of water looks bent at the surface because light from the underwater part refracts as it leaves the water into the air, bending away from the normal. Your eyes trace the rays back in straight lines, so the submerged part appears displaced, an everyday sign of refraction.

Try this

Q1. State the law of reflection. [2 marks]

  • Cue. The angle of incidence equals the angle of reflection, both measured from the normal.

Q2. Light passes from air into a material at 40∘40^\circ and refracts to 25∘25^\circ. Calculate the refractive index. [2 marks]

  • Cue. n=sin⁡40∘sin⁡25∘=0.6430.423=1.52n = \dfrac{\sin 40^\circ}{\sin 25^\circ} = \dfrac{0.643}{0.423} = 1.52.

Q3. State the two conditions needed for total internal reflection. [2 marks]

  • Cue. The light must be in the denser medium, and the angle of incidence must be greater than the critical angle.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA ray of light passes from air into glass of refractive index 1.51.5. The angle of incidence in air is 30∘30^\circ. (a) State the law that relates the angles. (b) Calculate the angle of refraction in the glass.
Show worked answer →

(a) Snell's law: n=sin⁡isin⁡rn = \dfrac{\sin i}{\sin r}, where ii is the angle of incidence and rr the angle of refraction, both measured from the normal.

(b) Rearrange: sin⁡r=sin⁡in=sin⁡30∘1.5=0.501.5=0.333\sin r = \dfrac{\sin i}{n} = \dfrac{\sin 30^\circ}{1.5} = \dfrac{0.50}{1.5} = 0.333, so r=sin⁡−1(0.333)=19.5∘r = \sin^{-1}(0.333) = 19.5^\circ.

Markers reward Snell's law stated with angles from the normal, the correct rearrangement, and the angle of refraction (about 19.5∘19.5^\circ), smaller than the incident angle because light bends toward the normal entering glass.

Original5 marks(a) Define the critical angle. (b) The critical angle for a glass-air boundary is 42∘42^\circ. State what happens to a ray inside the glass that hits the boundary at (i) 30∘30^\circ and (ii) 50∘50^\circ. (c) State one use of total internal reflection.
Show worked answer →

(a) The critical angle is the angle of incidence (inside the denser medium) for which the angle of refraction is exactly 90∘90^\circ.

(b) (i) At 30∘30^\circ, which is less than the critical angle, the ray mostly refracts out of the glass (with some weak reflection). (ii) At 50∘50^\circ, which is greater than the critical angle, the ray undergoes total internal reflection and stays inside the glass.

(c) Total internal reflection is used in optical fibres (or in prismatic periscopes and binoculars). (Any one.)

Markers reward the critical angle defined via a 90∘90^\circ refraction, refraction below the critical angle and total internal reflection above it, and a valid use such as optical fibres.

Related dot points