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Refractive Index (Oxford AQA IGCSE Physics)

Revision Note

Leander Oates

Written by: Leander Oates

Reviewed by: Caroline Carroll

Refractive Index

  • The refractive index is a number which is related to the speed of light in the material

refractive space index space equals space fraction numerator speed space of space light space in space straight a space vacuum over denominator speed space of space light space in space the space medium end fraction

  • The refractive index is:

    • always larger than 1

    • always less than the speed of light in a vacuum

    • different for different materials

    • has no units because it is a ratio

  • Objects which are more optically dense have a higher refractive index

    • For example, the refractive index is about 2.4 for diamond

  • Objects which are less optically dense have a lower refractive index,

    • For example, the refractive index is about 1.5 for glass

Calculating Refractive Index

  • The refractive index can also be determined using the angle of incidence and the angle of refraction

    • Also known as Snell's law

n space equals fraction numerator sin space i over denominator sin space r end fraction

  • Where:

    • n = the refractive index of the medium

    • i = the angle of incidence measured in °

    • r = the angle of refraction measured in °

Refraction ray diagram

Refraction of Light, for IGCSE & GCSE Physics revision notes
The angle of incidence and angle of refraction at the air-to-glass boundary

Worked Example

A ray of light approaches a glass block with a refractive index of 1.53. The ray meets the glass at an angle of 15° to the normal.

Calculate the angle between the ray and the normal after it enters the glass block.

Answer:

Step 1: List the known quantities

  • Refractive index of glass, n = 1.53

  • Angle of incidence, i = 15°

Step 2: Write out the appropriate equation

n space equals space fraction numerator sin space i over denominator sin space r end fraction

Step 3: Rearrange the equation to make sin r the subject

sin space r space equals space fraction numerator sin space i over denominator n end fraction

sin space r space equals space fraction numerator sin open parentheses 15 degree close parentheses over denominator 1.53 end fraction space

sin space r space equals space 0.1692

Step 4: Find the angle of refraction by using the inverse sine function

r space equals space sin to the power of negative 1 end exponent open parentheses 0.1692 close parentheses

r space equals space 9.739

r space equals space 10 degree

Examiner Tips and Tricks

There are a lot of common errors that students make when performing these types of calculations. Here are some things to check if you got the wrong answer:

  • fraction numerator sin space i over denominator sin space r end fraction is not the same as i over r

    • You cannot cancel the sine term

  • When multiplying or dividing with sine terms, make sure you close the bracket on your calculator to ensure you get the correct answer

    Your calculator is in degrees and not radians mode

  • The inverse sine function, sin-1, is usually found by pressing 'shift' then 'sin' on your calculator

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    Author
    Reviewer
    Leander Oates

    Author: Leander Oates

    Expertise: Physics

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.