13. Optics Flashcards
Explain why waves refract
Waves refract because the media on either side of the boundary are different.
The speed of the wave depends upon the material hence the wave travels at different speeds on either side of the boundary.
This change in speed causes the direction of travel of the wave to change.
Explain experimental determination of refractive index
— Measure the angle of refraction for various angles of incidence. Calculate the sines of each angle and plot a graph of sin i vs sin r. The gradient will be sin i / sin r, the refractive index.
airμglass = sin i/sin r
Define refractive index
n = speed of light in vacuum/speed of light in the transparent substance
n = wavelength of light in vacuum/wavelength in substance
SUBSTANCE wavelength/speed always in denonminator
SO n >= 1
What is dispersion and why? COMPARE SPEED OF BLUE AND RED THROUGH glass?
— This is the splitting up of light into its different colours (wavelengths).
– For visible light normal dispersion means that the refractive index is higher for blue light than for red.
– The refractive index depends on the wavelength of the incident light! This is why prisms and rainbows work
(WHY?)
What’s partial reflection, where does it occur
NB Partial reflection laso occurs when a light ray in air enters glass or any refractive substance.
- How does a prism split a continuous spectrum of light into its constituent wavelengths?
White light composed of a continuous spectrum of wavelengths from Red 650nm to violet 350nm
The glass prism refracts these different wavelengths by different amounts,
Because the speed of light in glass depends on its wavelength, and the refractive index depends on the wavelength of the incident light.
Hence, each colour in the spectrum is diffracted by a different amount, thus causing the ‘dispersive effect’.
What is total internal reflection ‘TIR’?
If light hits a boundary at the “critical angle” the light refracts along the boundary between the two substances.
If the angle of incidence is increased further, the light undergoes TIR at the boundary, as if the boundary were to be replaced by a plane mirror
(visualised the part of the light wave speeding up by so much that it is able to move “180 degrees” before the other “side” of the wavefront is able to move out of the denser medium.
Consider light travelling inside a more dense medium. At the boundary with a less dense medium it will refract and pass out at a larger angle unless this would be greater than 90o. If the angle it refracts to is 90o then we say the angle of incidence is the critical angle.
What are the conditions for TIR to occur and explain why they hold
The incidence substance has a higher refractice index that the other one.
This is so that the wave refracts away from the normal (by part of the wavefront “speeding up”) upon exiting the first medium.
The angle of incidence exceeds the critical angle (measured from the normal, as always).
Formula for critical angle
sin c = n2/n1
Be able to derive this from “n sin i” rule.
Why do diamonds sparkle when light is directed at them?
White light enters diamond —> refracted into colours of white light spectrum
Diamond has a very high refractive index (n)
—> Colours of the spectrum become highly separated compared to other common substances
—> High n also means a low critical angle
—> So light may be TIR-ed many times before emerging from the diamond
—> Means that there is enormous “dispersion” and spreading out of the colours due to refraction.
Also, by the shape of the diamond, light tends to be scattered rather than going straight through the diamond.
General law of refraction
n1 sin i = n2 sin r
for a ray passing from substance with n1 to substance with n2.
Explain how optical fibres are used for communications
A signal consisting of a light ray(s) is totally internally reflected each time it reaches the fibre boundary, even where the fibre bends (unless the radius of the bend is too small).
At each point where the light reaches the boundary of the fibre, the angle of incidence exceeds the critical angle and thus it tends to be totally internally reflected.
Describe the properties of optical fibres specifically designed for communications
— Allows pulses of light entering at one end to reach a receiver at the other end.
Highly transparent — why? Minimum absorption of light.
Core surrounded by a layer of cladding (lower refractive index, security, jumping signals, scratches).
TIR takes place at boundary between core and cladding.
Core needs to be very narrow to prevent multipath dispersion
Monochromatic light or infrared radiation needs to be used to prevent spectral dispersion
What is multi path dispersion
Multipath dispersion: this causes the signal pulse to become spread out over time as the input signal passes by numerous rays at different angles and hence different path lengths and times.
A pulse of light sent along a wide core would undergo multipath dispersion, because light travelling along the axis of the core travels a shorter distance per metre of fibre than light that repeatedly undergoes total internal reflection.
A pulse of light sent along a wide core would become longer than it ought to be be. It it was too long, it would merge with the next pulse.
What is the purpose of the cladding on a Communications optical fibre? [depends on context - cladding adv. different for medical endoscope]
- Security — To prevent signals from being able to jump from one fibre to another if they come into contact (lack of security). Signals could go from one fibre to another where any unclad fibres come into contact.
- Enabling TIR to be possible — “Most importantly” — to provide a lower refractive index at the other side of the boundary, so that total internal reflection is possible, and its refractive index must be sufficiently low such that the critical angle is large enough to allow a wide enough range of angles of incidence to make sure that the signal can keep on being totally internally reflected throughout its entire ‘journey’.
- Prevent Scratches — Without the cladding, the fibre could get scratched and allow the signal to escape.
[2,3 only for a medical endoscope optical fibre].
What are the advantages of optical fibre cables, compared to electrical signals in wires for COMMUNICATION?
- Glass is very pure, so little signal strength is lost
- Glass is harder to ‘tap’ & is therefore more secure — “tapping” an optical fibre can be detected easily as it destroys the signal for the receiver.
- More channels of information can be transmitted (multiplexing).
COMPARE SPEED OF BLUE AND RED THROUGH glass? (Dispersion)
- For visible light normal dispersion means that the refractive index is higher for blue light than for red.
- Think of a prism’s dispersion effect…
What is spectral dispersion
- Spectral dispersion: this causes the signal pulse to become elongated if the input signal is made up of a number of different wavelengths.
- Each wavelength travels at a different speed so takes a different time to travel through the fibre.
- This effect can be eliminated by using monochromatic radiation.
What problem is caused by spectral dispersion and how can it be solved?
— The difference in speed for the component wavelengths of non-monochromatic light would cause white light pulses in optical fibres to become longer, as the violet component falls behind the faster red component of each pulse.
— So the light (or IR) used must be monochromatic, to prevent pulse merging.
What is meant by an endoscope and its coherent fibre bundle?
Here a large number of fibres is used. The fibre bundle needs to be coherent, the fibre ends at each end are in the same relative positions.
Describe the structure and usage of an endoscope
Medical endoscopes have two budles of fibres, air/water channel, and an instrument channel.
Endoscope inserted into a body cavity, which is illuminated using light sent through one of the fibre bundles.
A lens over the end of the other fibre bundle is used to form an image of the body cavity on the end of the fibre bundle.
The light forming this image travels along the fibres to the other end of the fibre bundle, to the eyepiece, where the image can be observed.
Endoscopes have fibres arranged in coherent bundles — i.e. the fibre ends at each end are in the the same relative positions (very important for correct image generation!).
Explain Coherent (2) light
- Two sources are coherent if they oscillate with a constant phase difference between them.
- This implies that they must have the same wavelength. The phase difference could be zero.
What is monochromatic light
- A source is monochromatic if it is one colour i.e. it contains one frequency.
- Lasers are monochromatic.
- The opposite of this is broadband which contains a wide range of frequencies.
What is the path difference
- This is the difference in path length between two sources and a point expressed in numbers of wavelengths.
- The sources must be coherent.
Describe the experiment carried out by Young
Two coherent light sources are created by diffracting a single source through a single slit, and then passing this light through two further sources.
- If a laser is used the single slit is not required because the laser produces coherent radiation. A pattern of equally bright light and dark bands is observed known as interference fringes.
A pattern of alternating light and dark fringes on a screen placed where the diffracted light from each of the double slits overlap.
The fringes are evenly spaced and parallel to the double slits.
Why won’t the Young’s Double slit experiment work if the single slit is too wide?
If the single slit is too wide, each part of it produces a fringe-pattern which is displaced slightly from the pattern due to the adjacent parts of the single slit….as a result the dark fringes of the double slit pattern become narrower than the bright fringes, and contrast is lost between the dark and the bright fringes.
This is because of the single slit diffraction effect which occurs — if the single slit is too wide, not enough diffraction could occur and so both slits might not be illuminated with maximum intensity.
Explain the required positioning of the double slits for the interference fringes to be formed?
- Each of the double slits must be narrow enough to make light diffract through it by a sufficient amount
- However, if the slits are too wide, or too far apart, then no interference pattern is observed:
- The two slits must be sufficiently close together as well, so that the light from both slits overlap sufficiently on the screen.
Why are the Young’s fringes formed?
Bright fringe — when a maximum brightness is formed, light waves from each slit arrive in phase and interfere constructively.
— The light bands occur where the path difference is a whole number of wavelengths hence constructive interference occurs.
— Phase difference is zero where path difference is a whole number of wavelengths.
Dark fringe (minimum amplitude) — is formed, light waves from each slit arrow 180-degrees out of phase (antiphase) and interfere destructively.
— Whereas the dark bands occur at path differences of (n+1/2) x λ (where n is an integer) so destructive interference occursEquation linking fringe spacing of Young’s interference fringes (w), wavelength of incident light, slit-screen distance (D)
w = Dλ/s
where w is the fringe spacing in metres (m)
D is the distance to the screen in metres (m)
s is the distance between the slits in metres (m)
How would you measure fringe separation of Double slit interference pattern?
Measure across several fringes, from the centre of a dark fringe to the centre of another dark fringe
This is because centres of the dark fringes are easier to locate than the centres of the bright fringes.
Obtain w by dividing your measurement by the number of fringes you measured across.
