4.4.3 - Superposition Flashcards Preview

A Level - Physics (OCR A) > 4.4.3 - Superposition > Flashcards

Flashcards in 4.4.3 - Superposition Deck (19):
1

What is the principle of superposition?

It states that when two or more waves of the same type meet, the displacement of the resultant wave can be found by adding the displacements of the individual waves.

2

If two waves, A and B, with the same amplitude exist at the same point and are travelling in phase, what will the amplitude of the resultant wave be, and what type of interference is this?

- The resultant wave will be twice that of the individual waves.
- This is known as constructive interference.

3

If two waves, A and B, with the same amplitude exist at the same point and are travelling in anti-phase, what will the amplitude of the resultant wave be, and what type of interference is this?

- The resultant wave will have an amplitude of zero, as the two waves will cancel each other out.
- This is known as destructive interference.

4

In order to calculate a meaningful result using the principle of superposition, what must the two waves have/be?

- A constant phase difference.
- Must be coherent.

5

Define the term 'path difference' and state the units.

(Referred to in relation to the wavelength), is the difference between the distances travelled by two waves arriving at the same point.

Units: metres (m)

6

Define the term 'phase difference' and state the units.

The difference in the phases of two waves of the same frequency.

Units: radians (rad)

7

What does a path difference of any whole number of wavelengths result in?

Waves arriving at the same point in phase.

8

What does a path difference of half a wavelength result in?

A phase difference of π radians.

9

What will determine whether the interference observed between two waves will be constructive or destructive?

The path difference and phase difference between two coherent waves:

Constructive Interference
- Path Difference: whole number of wavelengths
- Phase Difference: 0°, 360°, 720° ... / 0, 2π, 4π ...

Destructive Interference
- Path Difference: odd number of half wavelengths
- Phase Difference: 180°, 540°, 900° ... / 0, 3π, 5π ...

10

How can interference using sound waves be demonstrated?

Using two loudspeakers connected to the same signal generator.

- As you walk along in front of the loudspeakers, you will hear a loud sound where the sound waves reinforce one another and a quiet sound where. the waves partially cancel one another out.
- This variation is clearer if you cover one ear.
- The distance between the loud and quiet frequencies is longer for low frequencies.

11

How can interference using microwaves be demonstrated?

By creating a system in which microwave energy is split along two different paths, A and B, before joining again.

- By adjusting the length of the two paths, it is possible to create either constructive interference or destructive interference when the waves rejoin.

12

Describe Young's double-slit experiment.

1) Young used a monochromatic red light source, which he placed behind a single slit in a piece of black card, X.

2) Light passes through the slit and spreads out by diffraction, until it reaches another obstacle, Y, in which there are two parallel narrow slits. (The light from these two slits is coherent because it comes from a single slit).

3) Progressive waves from both slits that are in phase will interfere constructively. Alternate bright and dark vertical bands ('fringes') are seen on the screen.

13

Why must the double-slit experiment be done in a darkened room?

As the light intensity is low and the fringes are hard to see.

14

Why is monochromatic light used in the double-slit experiment, rather than white light?

Using white light produces blurred fringes as each colour produces its own set of fringes which overlap slightly.

A coloured filter is used to produce fringes with sharp edges so a more accurate value of λ can be found.

(A laser can be used as a source of monochromatic light).

15

How can the percentage error in the fringe separation, x, be reduced?

- By measuring across all fringes and then dividing by the number of fringes.
- Increasing the slit-to-screen distance increases the fringe separation allowing a more precise measurement, but reduces the light intensity at the screen.

16

What is the equation for working out the wavelength of the light source (from the double-slit experiment)?

λ = ax / D

wavelength (m) = ( slit separation (m) * distance between fringes (m) ) / distance from slit to screen (m)

NOTE: This equation only applies if a << D, or the angle away from the central fringe is less than 10°.

17

What is a diffraction grating and what will happen to light that passes through a diffraction grating?

- A piece of optical equipment made from glass, onto which many thousands of very thin, parallel and equally spaced grooves have been accurately engraved using a diamond.
- Light that passes through the grating will be diffracted at different angles based on the wavelength of the incident light and the separation of the grooves.

18

How can the wavelength of light be measured using a diffraction grating?

1) A diffraction grating is put in parallel beam of monochromatic light from a sodium lamp.

2) The angle between the light that passes straight through the grating and the diffracted beam is measured.

3) It is found that there are a number of angles at which light is seen - these are called maxima. (They occur when the path difference between light from adjacent slits is 0, λ, 2λ, 3λ or 4λ... (rather than just λ) and reinforcement (constructive interference) occurs.)

4) The wavelength is found using nλ = d sin θ
- If the grating has X slits per metre, the slit spacing, d= 1/X.

19

What is the diffraction equation?

nλ = d sin θ

order of the maximum * wavelength of the incident monochromatic light (m) = separation of the slits in the grating (m) * sine of the angle that the beam makes with the grating