Refractive Index (Oxford AQA IGCSE Physics): Revision Note

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

$n=\frac{\sin i}{\sin r}$

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

$\sin r=\frac{\sin i}{n}$

$\sin r=\frac{\sin \left(15°\right)}{1.53}$

$\sin r=0.1692$

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

$r={\sin }^{-1}\left(0.1692\right)$

$r=9.739$

$r=10°$

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:

• $\frac{\sin i}{\sin r}$ is not the same as $\frac{i}{r}$

  • 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