Chapter 4 - Waves Flashcards

Question

The plane of vibration of an electromagnetic wave

Answer 1

The plane that contains the electric field and the direction of the propagation.

Answer 2

Over a period of time, the electric field only oscillates in one direction.

Answer 3

Over a period time, the electric field oscillates in random directions.

Answer 4

1) . A mixture of polarised light and unpolarised light. | 2) . If the plane of polarisation rotates uniformly

Answer 5

Any device that produces plane-polarised light from an unpolarised beam.

Answer 6

It's a polarised used to detect polarised light.

Answer 7

A material which absorbs any light in one particular plane of polarisation allowing transmissions only in the plane of 90 degrees to this.

Answer 8

The reflected ray of a light Ray is always partially plane-polarised. If the reflected ray and the refracted Ray are 90 degrees to each other than the reflected ray is totally plane-polarised. The angle of incidence for this condition = polarising angle

Answer 9

n (the refractive index of the medium) = tan i (i = polarisation angle)

Answer 10

Radio waves etc.. Often emit plane-polarised light whereas light waves do not. Light can be polarised as a result of reflection and selective absorption.

Answer 11

Some crystals exhibit double refraction or birefringence; where in polarised light rays enter the crystal and is split into two plane-polarised beams that have mutually perpendicular planes of polarisation. This is because they refract differently as they travel at different speeds.

Answer 12

The intensity of light is proportional to the (amplitude)^2 ( how intensity varies with the angle) I = I (o) x cos^2 (angle) ``` I = transmitted intensity of light in Wm^-2 I(o) = incident intensity of light in Wm^-2 Angle = angle between the plane of vibration and the analyser's preferred direction. ```

Answer 13

It has a long chain of molecules which selectively absorb light that has electric fields aligned with the molecules.

Answer 14

It allows light with vertically oscillating electric fields to be transmitted and absorb light with horizontally oscillating electric fields.

Answer 15

Overall light intensity and 'glare' from horizontal surfaces will be reduced. Light reflected from horizontal surfaces will be horizontally plane-polarised to some extent.

Answer 16

The stress comes from the molecules being twisted or stretched. They become birefringent. Bright coloured lines are observed in regions of maximum stress when white light is polarised and passed through stressed plastic.

Answer 17

When an objects refractive index depends on the polarization and propagation direction of light. It's the rotated polarised direction and the two components are slowed

Answer 18

It rotates the plane of polarisation of light that passes through it. E.g sugar solution. The more concentrated the solution, the more molecules the oscillations meets and the greater it rotates.

Answer 19

E.g calculator displays and computer monitors The liquid crystal is sandwiched between 2 glass electrodes and is birefringent. The crystal has a twisted structure and causes the plane of polarisation to rotate through 90°. Thus, light can pass through the second polariser, reach the reflecting surface and be transmitted back along the original direction.

Answer 20

With no pd across the electrodes, the LCD appears light. With a pd, it causes molecules to align with the electric field. So less light is transmitted and this section of the LCD will appear darker.

Answer 21

It's used to create a colour image.

Answer 22

The second polariser would absorb all the light passed through the first one, and the screen would appear black.

Answer 23

Transverse waves

Answer 24

It's partially reflected and partially transmitted.

Answer 25

Angle of incidence = angle of reflection | Both the rays lie on the same plane.

Answer 26

When a single ray of light strikes a smooth mirror it produces a single reflected ray.

Answer 27

When a single incident ray hits an uneven surface the reflected rays go in different directions. Objects become visible with a source of light because diffuse reflections have taken place that Scatter light from the source to our eyes.

Answer 28

It's a line showing the direction in which light energy is propagated. It must be at right angles to the wavefront.

Answer 29

Occurs at the boundary between 2 medias. The wave will undergo a change of direction due to a change of wave speed.

Answer 30

If a ray travels into a denser medium, it refracts towards the normal. If a Ray travels into a less dense medium, it refracts away from the normal.

Answer 31

Sin i/ sin r = n | n = constant, for a given frequency.

Answer 32

The ratio is equal to the ratio of the speeds in the different media ``` n(1)/n(2) = sin (o)(2)/ sin (o)(1) = V(2)/V(1) V(2)= speed of wave in medium 2 V(1)= speed of wave in medium 1 ```

Answer 33

Critical ray: the ray which has a refracted angle of 90°. | Critical angle: angle of incidence for the critical ray

Answer 34

Refraction cannot occur so the ray is totally internally reflected at the the boundary and stays in the same medium of which it came from.

Answer 35

``` Sin(0c) = 1/n Sin(0c) = critical angle ```

Answer 36

When a wave passes through apertures they spread out, and they also spread around obstacles

Answer 37

When the wavelength is large in comparison to the size of the aperture. The wavelength needs to be the same order of magnitude as the aperture for the diffraction to be noticeable .

Answer 38

The intensity decreases

Answer 39

It's another application of superposition, for 2 coherent sources having roughly the same amplitude. Within the areas of interference there is constructive and destructive interference, and sections of maximas and minimas.

Answer 40

A monochromatic light source (only gives out one frequency), and the light from the twin slits interferes and patten of light and dark regions (fringes) are seen on the screen.

Answer 41

n x wavelength = d sin(0) ``` n = number of maxima d = distance of the slits ``` We can ignore the sin, but only if it's in radians and the angle is very small

Answer 42

``` s = wavelength x D / d D = distance of light source to screen ```

Answer 43

Maxima: waves are in phase Minima: waves are out of phase and cancel out, thus the dark areas.

Answer 44

When 2 waves are: Of the same amplitude and frequency travelling in opposite direction

Answer 45

When a wave reflects back from a boundary along the route it came. The reflected wave and the incident wave are (nearly) equal amplitude, so they interfere.

Answer 46

Areas of destructive waves, the waves always cancel each other out; they have no amplitude.

Answer 47

Areas of constructive interference. The amplitude is 2A (twice the amplitude of the old wave, as now we add both together). It's the maximum displacement.

Answer 48

All points have the same amplitude - at some point during the wave, the single point will reach the maximum and minimum displacement.

Answer 49

Energy isn't transmitted, but does have energy associated by it.

Answer 50

All points between one node and next one are moving in phase. The wavelength is twice the distance from one node (or antinode) to the next node (or antinode).

Answer 51

What a wave can/can't do. | Specifies the condition that must be met at the edges of the system when standing waves are taking place.

Answer 52

Longitudinal and transverse.

Answer 53

Molecules at an open end must be at a pressure node because the outside of the pipe is at a constant (atmospheric) pressure. However, they are free to move, thus they are at a displacement antinode. The molecules at the closed end of a pipe are at a displacement node because they cannot move past the end of the pipe. However, they still feel the wave from the rest of the pipe which will alternately squash them against the solid end, and pull them away: thus they are at a pressure antinode.