Waves

Question 1

Define refractive index

Answer 1

The ratio of the speed of light in one medium to another.
n = c_i / c_r

Question 2

Give another equation for refractive index

Answer 2

n = sin(i) / sin(r)

Question 3

Define Frequency

Answer 3

Number of oscillations in a given time period

Question 4

What is frequency equal to, and hence what is the equation for the velocity of the wave?

Answer 4

f = 1 / T
Hence, V = λf or V = λ/T

Question 5

Define the terms in phase and antiphase

Answer 5

2 waves at once, if they’re in sync, they’re in phase. If they are 1/2 λ out of phase, they’re in antiphase. Otherwise, they’re out of phase.

Question 6

How do you calculate the displacement of a wave at any given point?

Answer 6

S = Acos(ωt)
Where A is amplitude, t is time, and ω is angular frequency, i.e. How many times a revolution happens per second in radians (2πf)

Question 7

Describe how a wave reflects at a fixed end?

Answer 7

It flips and undergoes a phaseshift of π, so it cancels out the unreflected wave. There is no oscillation, and a node is formed.

Question 8

Describe the structure of a standing waves.

Answer 8

Node - point of no displacement.
Antinode - point of maximum displacement.
There is always 1 more node than the numbered overtone, and antinodes always match the overtone number.
e.g. the 2^nd overtone has 2 antinodes and 3 nodes.

Question 9

In a standing wave, what is the distance between 2 adjacent nodes?

Answer 9

1/2λ, so 1λ is 3 nodes and 2 antinodes.

Question 10

What is special about waves in rods?

Answer 10

The standing wave structure is reversed. Where there would be a node, there is now an antinode.

Question 11

Explain a way to calculate the speed of sound in air

Answer 11

1. Place a resonance tube into a bigger tube filled with water
2. Hold a tuning fork above the resonance tube
3. Move the tube until the amplitude of the wave is at its maximum, then measure the distance from the surface of the water to the top of the tube. This is L₁.
4. As there is a fixed end (the surface of the water), there is a node there, so there is an antinode at the tuning fork. That’s 1/2λ. As the tuning fork is not exactly at the top of tube, we add a little correction, so therefore 1/2λ = L₁ + k
5. Repeat the process until you find another point of maximum amplitude. This is your L₂, it will be about 3/4λ. So, 3/4λ = L₂ + k
6. Using algebra to cancel k, you’ll find that L₂ - L₁ = 1/2λ. From here, solve for λ.
7. Use c = fλ for the speed of sound in air.

Question 12

Describe a problem with using a resonance tube to find the speed

Answer 12

The uncertainty is very large due to not knowing the exact position of the standing wave.

Question 13

When do waves meet in phase? Then when do they meet in antiphase?

Answer 13

In phase - Path difference = nλ
In antiphase - Path difference = (n + 1/2)λ

Question 14

How can you calculate path difference?

Answer 14

Calculate the difference in distance the wave has traveled from both source, then use the difference as a ratio to original λ.

Question 15

Define what makes 2 wave sources coherent

Answer 15

There is a constant phase difference

Question 16

Describe the terms constructive and destructive interference?

Answer 16

Constructive - Waves are in phase, which leads to a larger superposition amplitude
Destructive - Waves are in antiphase, which leads to a lower superposition amplitude

Question 17

What is an interference pattern?

Answer 17

The superposition of waves when there is more than one source

Question 18

Describe how the size of a gap affects waves that undergo diffraction

Answer 18

The smaller the gap, the larger the spreading of the wave. It does not affect its speed, wavelength or frequency.

Question 19

Describe how Young’s double slit experiment is performed

Answer 19

1. A coherent light source is fired at two very small slits
2. The light diffracts through the slits, and produces a pattern of light and dark fringes on the opposite screen. This is an interference pattern.

Question 20

Describe how an interference pattern is produced when doing Young’s double slit experiment

Answer 20

For each slit, if the light reaches the screen in phase, then a light fringe is produced. If the light reaches the screen in antiphase, then a dark fringe is produced instead.

Question 21

Explain how to derive the formula nλ = dsinθ

Answer 21

1. Construct a right angled triangle between two waves that are in phase, where d (the distance between slits) is the hypotoneuse, λ is the opposite and θ is the angle made at the slit between one bright fringe and the next.
2. sinθ = o/h so subbing in and rearranging, you get λ = dsinθ. For a path difference between waves of nλ, the formula becomes nλ = dsinθ_n.

Question 22

Explain how to derive the formula λ = Sd/L

Answer 22

1. Construct a right angled triangle where L (the distance from the slits to the screen) is the adjacent and S (the distance between fringes on the screen) is the opposite.
2. tanθ = S/L. As Sinθ ≈ tanθ at small angles, λ/d = S/L
3. Rearrange for λ = Sd/L

Question 23

How do you calculate d given the number of lines per metre?

Answer 23

1 / number of lines per metre = seperation between lines, i.e. d

Question 24

Explain thin film interference.

Answer 24

1. When light enters thin translucent film, some of it will reflect off the front surface and some off the back of the film.
2. The light that reflects off of the back of the film has refracted, so there will be a phase difference between the two waves
3. As the second wave now leaves the film, the two waves are now possibly out of phase. This can lead to contructive, destructive or no interference depending on the phase shift.

Question 25

Explain how dark spots can be produced in thin films

Answer 25

1. When the film becomes very thin, the front of the surfaces causes a hard reflection of the light. This causes the phase of the front wave to be reversed. 2. As the film is very thin now, there is very little phase difference between the two waves as the second wave leaves the film. 3. As the first wave is now in reverse to the second wave, they are in antiphase, and destructive interference occurs.