Light passes from air (n = 1.00) into glass (n = 1.50) at an angle of incidence of 40^. Find the angle of refraction. [2 marks]

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How does light bend between materials, and when is it totally internally reflected?

Refraction and the electromagnetic spectrum: refractive index and Snell's law, total internal reflection and the critical angle, optical fibres, and the regions and properties of the electromagnetic spectrum.

A focused answer to the OCR H556 content on refraction and the electromagnetic spectrum, covering refractive index and Snell's law, total internal reflection and the critical angle, optical fibres and their use, and the regions, ordering and shared properties of the electromagnetic spectrum.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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Quick answer

The refractive index of a material is $n = \frac{c}{v}$ , the ratio of the speed of light in a vacuum to the speed in the material; it is always greater than 1. At a boundary, Snell's law gives $n_1\sin\theta_1 = n_2\sin\theta_2$ . When light travels from a denser to a less dense medium, beyond the critical angle $\theta_c$ (where $\sin\theta_c = \frac{n_2}{n_1}$ ) it is totally internally reflected; this traps light in optical fibres. The electromagnetic spectrum runs from radio waves (longest wavelength, lowest frequency) through microwaves, infrared, visible light, ultraviolet, X-rays to gamma rays (shortest wavelength); all travel at the speed of light in a vacuum and are transverse.

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What this dot point is asking
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Examples in context
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What this dot point is asking

OCR wants you to define and use refractive index, apply Snell's law at a boundary, explain total internal reflection and calculate the critical angle, describe optical fibres and their use, and recall the regions, ordering and shared properties of the electromagnetic spectrum.

The answer

Refractive index and Snell's law

Total internal reflection and the critical angle

Total internal reflection is loss-free reflection, which is why it is exploited in optical fibres, prisms in binoculars and periscopes, and the cat's-eye reflectors on roads.

Optical fibres

The electromagnetic spectrum

Speed of light in a material

A type of glass has a refractive index of $1.52$ . Find the speed of light in this glass and its wavelength there if the wavelength in air is $5.9 \times 10^{-7}\ \text{m}$ . Take $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$ .

step 1: Speed in the glass

$v = \frac{c}{n} = \frac{3.0 \times 10^{8}}{1.52} = 1.97 \times 10^{8}\ \text{m s}^{-1}$ .

step 2: Wavelength in the glass

Frequency is unchanged, so the wavelength scales with speed: $\lambda_{\text{glass}} = \frac{\lambda_{\text{air}}}{n} = \frac{5.9 \times 10^{-7}}{1.52} = 3.9 \times 10^{-7}\ \text{m}$ .

step 3: Interpret

Light slows to about two-thirds of its vacuum speed and its wavelength shortens by the same factor, while the colour (frequency) stays the same.

Examples in context

Optical fibres carry the world's internet and telephone traffic, using total internal reflection to send light pulses over thousands of kilometres with low loss. Endoscopes use bundles of fibres to image inside the body. Microwave ovens and mobile phones use the microwave band; thermal cameras detect infrared; sterilisation and tanning use ultraviolet; medical imaging and security scanners use X-rays. The dispersion of white light into a spectrum by a prism arises because refractive index varies slightly with wavelength.

Try this

Q1. Define the refractive index of a material. [1 mark]

Cue. The ratio of the speed of light in a vacuum to the speed of light in the material, $n = \frac{c}{v}$ .

Q2. Light passes from air ( $n = 1.00$ ) into glass ( $n = 1.50$ ) at an angle of incidence of $40^{\circ}$ . Find the angle of refraction. [2 marks]

Cue. $\sin\theta_2 = \frac{1.00 \times \sin 40^{\circ}}{1.50} = 0.429$ , so $\theta_2 = 25^{\circ}$ .

Q3. List the regions of the electromagnetic spectrum in order of increasing frequency. [2 marks]

Cue. Radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.

Exam-style practice questions

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

OCR 20193 marksA ray of light travels from glass of refractive index

1.52

into water of refractive index

1.33

. The angle of incidence in the glass is

35^{\circ}

. Calculate the angle of refraction in the water.

Show worked answer →

Apply Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ , rearranged for the refraction angle.

$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{1.52 \times \sin 35^{\circ}}{1.33} = \frac{1.52 \times 0.574}{1.33} = 0.656$ .

So $\theta_2 = \sin^{-1}(0.656) = 41^{\circ}$ .

Markers reward the correct form of Snell's law, the substitution, and the angle about $41^{\circ}$ (bent away from the normal, since the ray enters a less dense medium).

OCR 20213 marksThe core of an optical fibre has a refractive index of

1.50

and the cladding has a refractive index of

1.45

. Calculate the critical angle at the core-cladding boundary and explain why a cladding of slightly lower index is used.

Show worked answer →

At the critical angle the refracted ray travels along the boundary, so $\sin\theta_c = \frac{n_2}{n_1} = \frac{1.45}{1.50} = 0.967$ .

$\theta_c = \sin^{-1}(0.967) = 75^{\circ}$ .

The cladding has a slightly lower index so that total internal reflection occurs at the boundary, trapping light in the core, while keeping the critical angle high enough that only rays close to the axis travel on, reducing multipath dispersion and protecting the core surface.

Markers reward $\sin\theta_c = \frac{n_2}{n_1}$ , the angle about $75^{\circ}$ , and an explanation linking the lower-index cladding to total internal reflection and signal quality.

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Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)