Optical- Scattering Flashcards

1
Q

What happens when light traverses a boundary with a change in refractive index?

A

Some light is reflected. Remainder is refracted because of velocity change

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2
Q

What is specular reflection?

A

Reflection from a polished flat surface

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3
Q

What is diffuse reflection?

A

Reflection from a rough surface (so in many directions)

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4
Q

Why is reflection needed?

A

The electric and magnetic vectors for incident and refracted rays do not match across the boundary. A third ray needed to make up difference (reflected ray). This may have change in phase as well as intensity.

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5
Q

Reflection in polycrystalline materials

A

Some light reflected at every interface crossed. Interfaces are randomly oriented so net effect is to scatter light everywhere. The smaller the crystals (down to λ/2n for maximum scattering) the more boundaries and the more light is scattered. The bigger the refractive index difference the more the light is scattered.

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6
Q

Titania paints

A

Opacifier in white paints typically TiO2 with highnrefractive index of 2.2 compared with resin in which it is embedded. More finely ground titania means more effective

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7
Q

Why are polycrystalline ceramics generally opaque/translucent?

A

Because of light scattering at boundaries

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8
Q

How come polycrystalline IR lenses can be made?

A

Use cubic symmetry materials because their r.i is independent of orientation (e.g ZnS, CaF2, NaCl not Al2O3). No r.i change means no refraction or reflection (scattering). Must eliminate porosity which causes scattering.

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9
Q

Reflection at normal incidence formula

A

Ireflected/Iincident=(n2-n1)^2/(n2+n1)^2

Reflectivity independent of direction of light travel (high to low or low to high r.i)

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10
Q

Problem with high refractive index glass

A

Can have reflectivities of 1/9 per surface or more. For several components, losses through reflection become large. Multiple reflections also add to background and spoils image.

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11
Q

Antireflection coatings

A

2 interfaces at which reflection can occur. Make coating λ/4 thick so extra path of light reflected from coating/substrate interface is half a wavelength (strictly λ/4nc as nc affects wavelength). Makes two rays equal in intensity is appropriate choice of coating r.i. Total destructive interference should occur so no reflection.

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12
Q

Ideal refractive index for coating

A

rt(nairxns)
nair is r.i of air
ns is r.i of substrate
Sometimes not many coating materials with low enough r.i exist

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13
Q

Limitation of coating thickness

A

Coating can only have correct thickness at one wavelength and angle of incidence but fortunately reflectivity varies slowly with wavelength and angle

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14
Q

Why are some coating slightly purple?

A

Designed for optimum performance in middle of visible spectrum and reflect more at the blue and red end giving net colour of purple.

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15
Q

Solutions to limitations of coatings

A

Using 2 layers can generate 2 minima in reflectivity within visible range so giving a lower overall (greenish) reflectivity. Can use multiple coatings (low and high r.i material). Can use graded r.i mimicking a fly’s eye.

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16
Q

Methods of application of multilayers

A

From vapour phase (off line in vacuum systems or on line).
By solution routes and evaporation.
Such deposition routes involve extreme conditions and costing material’s properties can be different to bulk properties.
Typically slow and costly processes
Can affect chemical and mechanical durability

17
Q

Reflecting layers

A

Make multilayer sandwiches with repeat distance of λ/2 thick so constructive interference occurs. Reflection can be enhanced if refractive changes are large. Can build up several layers to produce very high reflectivities. Can build in errors in sequence giving band pass filters transmitting some wavelengths and reflecting others.