3.3.2.3
Total internal reflection requires two conditions
Refraction & Total Internal Reflection — AQA A-Level Physics
Key Definition
Total internal reflection (TIR) — A special case of refraction where all light is reflected back into the denser medium at the boundary.
- Two conditions must BOTH be met for TIR to occur:
- 1. The angle of incidence must be greater than the critical angle ($\theta > \theta_c$).
- 2. Light must be travelling from a denser medium to a less dense medium ($n_1 > n_2$).
- When TIR occurs, the law of reflection applies: angle of $incidence = angle$ of reflection.
- If either condition is not met, refractionThe change in direction of a wave as it passes from one medium to another, caused by a change in wave speed. (not TIR) occurs.
Worked Example
A glass cube (n = 1.489) is in contact with a liquid. Light enters from air at 39 degrees, refracts to 25 degrees in the glass, and totally internally reflects for the first time at the glass-liquid boundary. Calculate the refractive indexThe ratio of the speed of light in a vacuum to the speed of light in a medium. Determines how much light bends on entering the medium. of the liquid.
Show Solution
1
Find the critical angleThe angle of incidence at which the refracted ray travels along the boundary (angle of refractionThe change in direction of a wave as it passes from one medium to another, caused by a change in wave speed. = 90 degrees). For angles greater than this, total internal reflection occurs.
The refracted ray in glass hits the glass-liquid boundary. The angle with the normal at that boundary is:
$$\theta_c = 90^{\circ} - 25^{\circ} = 65^{\circ}$$Since TIR occurs for the first time, this is the critical angleThe angle of incidence at which the refracted ray travels along the boundary (angle of refractionThe change in direction of a wave as it passes from one medium to another, caused by a change in wave speed. = 90 degrees). For angles greater than this, total internal reflection occurs..
2
Apply the critical angleThe angle of incidence at which the refracted ray travels along the boundary (angle of refraction = 90 degrees). For angles greater than this, total internal reflection occurs. equation
$$\sin \theta_c = \frac{n_2}{n_1}$$
$$n_2 = n_1 \sin \theta_c = 1.489 \times \sin 65^{\circ}$$3
Calculate
$$n_2 = 1.489 \times 0.9063 = 1.35 \text{ (3 s.f.)}$$
Answer
$n_{\text{liquid}} = 1.35$
Related:Diffraction