4.3 Waves at material interfaces — Practise with help

1. $v = f\lambda = 5.0 \times 0.40 = 2.0\,\text{m/s}$.
2. $40\,\text{cm} = 0.40\,\text{m}$.

If question 1 went wrong, look again at the wave equation in Section 4.1. If question 2 went wrong, remember that the prefix centi means one hundredth.

The law links the angle of incidence and the angle of reflection, and it says which line both angles are measured from.

Say that the two angles are equal, then say that both are measured from the normal.

The angle of reflection equals the angle of incidence. Both angles are measured between the ray and the normal, which is the line at right angles to the surface at the point where the ray hits it. This answer is a model answer for you to compare with your own wording.

The wave has to travel to the wall and then back again, so the total distance is twice the distance to the wall.

Work out the total distance, $2 \times 18 = 36\,\text{m}$. Then use $t = \dfrac{s}{v}$.

The frequency is the same on both sides of the boundary, so use the wave equation twice.

For the first material, work out $\dfrac{5.0}{250}$. For the second, work out $\dfrac{8.0}{250}$.

The higher speed gives the longer wavelength.

The three percentages must add up to 100%.

Add the reflected and absorbed percentages, then subtract the total from 100%.

Half of the energy is transmitted into the second material.

A wave can bounce back, pass through or be taken in by the material. Give the name for each one.

Name reflection, transmission and absorption, and say in one sentence what happens in each case.

Part of the wave can be reflected, so it bounces back from the boundary. Part of the wave can be transmitted, so it passes through into the second material. Part of the wave can be absorbed, so its energy is transferred to the material. Usually all three happen at once. This answer is a model answer for you to compare with your own wording.

Think about the angle the surface makes with the ray. Does that angle change if the surface is tilted?

Say that the normal is at right angles to the surface, so it gives a fixed reference line, then say why the surface does not.

The normal is at right angles to the surface, so it gives a fixed reference line for measuring the angles. The surface itself lies flat along the boundary, so the angle between a ray and the surface depends on the ray and cannot be compared from one diagram to the next. Measuring from the normal gives the angle of incidence and the angle of reflection in the same way at every point. This answer is a model answer for you to compare with your own wording.

The sound travels to the wall and back, so the distance to the wall is half of the distance the sound covers.

Use $d = \dfrac{v t}{2}$, then work out $\dfrac{340 \times 0.50}{2}$.