Snell's Law (AQA A Level Physics): Revision Note

A light ray is directed at a vertical face of a glass cube. The angle of incidence at the vertical face is 39°, and the angle of refraction is 25°, as shown in the diagram.

Show that the refractive index of the glass is about 1.5.

[2]

Answer:

Step 1: Write down the known quantities

• Refractive index of air, $n_1=1$

• Refractive index of glass = $n_2$

• Angle of incidence, ${\theta }_1=39°$

• Angle of refraction, ${\theta }_2=25°$

Step 2: Write out Snell's law

$\frac{n_2}{n_1}=\frac{\sin {\theta }_1}{\sin {\theta }_2}$

Step 3: Calculate the refractive index of glass

$n_2=1\times \frac{\sin 39°}{\sin 25°}$ [1 mark]

$n_2=1.489=1.5\left(2\mathrm{s}.\mathrm{f}.\right)$ [1 mark]

• Note: In a ‘show that’ question, the answer should be given to at least one more significant figure than the value given in the question

Always check that the angle of incidence and refraction are the angles between the normal and the light ray. If the angle between the light ray and the boundary is calculated instead, calculate 90 – $\theta$ (since the normal is perpendicular to the boundary) to get the correct angle. 

Also check that your calculator is in degrees and not radians. 

Keeping track of which angles are ${\theta }_1$ and which are ${\theta }_2$ can be tricky. To help, label each angle with the material it is in.

A glass cube is held in contact with a liquid, and a light ray is directed at the vertical face of the cube. When the angle of incidence at the vertical face is 39° and the angle of refraction is 25°, as shown in the diagram, the light ray is totally internally reflected at X for the first time.

(a) Complete the diagram to show the path of the ray beyond X to the air and calculate the critical angle for the glass-liquid boundary. [3]

(b) The refractive index of the glass is 1.489. Calculate the refractive index of the liquid. [2]

Answer:

Part (a):

Step 1: Draw the reflected angle at the glass-liquid boundary

• When a light ray is reflected, the angle of incidence = angle of reflection

• Therefore, the angle of incidence (or reflection) is 90° – 25° = 65° [1 mark]

Step 2: Draw the refracted angle at the glass-air boundary

• At the glass-air boundary, the light ray refracts away from the normal when the ray leaves to air (as air is less dense)

• Due to the reflection, the light rays are symmetrical on either side of the glass (39°) [1 mark]

Step 3: Calculate the critical angle for the glass-liquid boundary

• The question states the ray is “totally internally reflected for the first time” meaning that this is the smallest angle at which TIR occurs 

• Therefore, 65° is the critical angle [1 mark]

Part (b):

Step 1: Write down the known quantities

• Refractive index of glass, n₁ = 1.489 

• Refractive index of liquid = n₂

• Angle of incidence / critical angle, ${\theta }_1={\theta }_c$ = 65°

• Angle of refraction, ${\theta }_2$ = 90°

  • When the angle of incidence = critical angle, the ray travels along the boundary between the media, making it perpendicular to the normal

Step 2: Write out Snell’s law or the equation for critical angle

$\frac{n_2}{n_1}=\frac{\sin {\theta }_1}{\sin {\theta }_2}$ or $\sin {\theta }_c=\frac{n_2}{n_1}$

Step 3: Calculate the refractive index of the liquid

$n_2=n_1\times \frac{\sin {\theta }_1}{\sin {\theta }_2}$ [1 mark]

$n_2=1.489\times \frac{\sin 65°}{\sin 90°}=1.35\left(3\mathrm{s}.\mathrm{f}.\right)$ [1 mark]