Optics Flashcards

Review key terms, formulae, experiments and typical exam questions for the optics module.

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1
Q

What is the refractive index of a material?

A

The ratio of sini/sinr for light travelling from a vacuum into that material.

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2
Q

Why does refraction occur?

A

Refraction occurs because the speed of light is different in each substance, the amount of refraction depends on the speed of the wave in each substance.

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3
Q

The refractive index is given by ns=c/cs what does this show?

A

The smaller the speed of light in a substance the greater the refractive index.

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4
Q

Define the terms in n1sin1=n2sin2.

A

refractive index of material 1, sine of incident angle, refractive index of material 2, sine of refracted angle

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5
Q

How does light refract when entering a more optically dense material?

A

It refracts towards the normal.

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6
Q

How does light refract when entering a less optically dense material?

A

It refracts away from the normal.

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7
Q

What is total internal reflection?

A

When light reflects inside a material.

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8
Q

What are the requirements for total internal reflection?

A

1.The incident substance has a larger refractive index than the other substance. 2.The angle of incidence exceeds the critical angle.

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9
Q

How do we calculate the critical angle of a substance?

A

The refracted angle is 90 degrees using Snell’s Law: n1sin1=n2sin2 sin1 = n2/n1 (as sin2=sin90 = 1)

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10
Q

How do optical fibres work?

A

The light ray is totally internally reflected each time it reaches the fibre boundary even when the fibre bends. At each point where the light ray reaches the boundary the angle of incidence exceeds the critical angle.

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11
Q

Why do optical fibres need to be narrow and have core cladding?

A

Total internal reflection occurs at the core cladding boundary, if there was no cladding the light would cross from one fibre to another leading to a signal not being secure. The core has to be narrow to prevent multipath dispersion - this is when the light ray can reflect a different amount of times before reaching the end of the fibre.

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12
Q

What is monochromatic light?

A

Light of a single wavelength

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13
Q

What is coherent light?

A

Light waves with constant phase difference and the same frequency.

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14
Q

Describe Young’s Double Slit experiment.

A

Light was passed through two slits and then an interference pattern (areas of light and dark) was viewed on a screen opposite.

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15
Q

Why are bright fringes formed in Young’s Double Slit Experiment?

A

The light from each slit reinforces and the light arrives at the screen in phase.

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16
Q

Why are dark fringes formed in Young’s Double Slit Experiment?

A

The light from each slit cancels out and the light arrives at the screen 180 degrees out of phase.

17
Q

What do the symbols in the interference equation stand for (w=λD/s)?

A

w = fringe spacing λ = wavelength D = distance to screen s = slit separation

18
Q

What is 1nm in metres?

A

1 x 10-9

19
Q

We can see an interference pattern from two wave sources when the two waves are coherent. When are waves coherent?

A
  1. When they have a constant phase difference. 2. When the waves have the same wavelength.
20
Q

Using the equation w=λD/s, does red light produce a greater or a smaller fringe separation than blue light?

A

It will produce a greater fringe separation as the wavelength is larger (providing D and s are kept constant).

21
Q

Why is it important never to look along a laser beam, even after reflection?

A

A laser light source produces coherent, monochromatic EM waves that superpose and focus to a very fine point. Intense concentration of such radiation on a tiny spot of the retina would destroy it.

22
Q

What interference pattern is produced when white light is shone through double-slits instead of monochromatic light?

A

1) A central white fringe where every colour contributes. 2) Further fringes are a spectrum with blue on the inner side and red on the outer side where wavelengths diffract by different amounts and wavelengths do not overlap.

23
Q

Describe single-slit diffraction

A

A central fringe where intensity is greatest at the centre with subsidiary minima as a result of secondary wavelets. Each of the outer fringes are the same width but peak intensity of each fringe decreases with distance from the centre. (Width of the central diffraction fringe, W=2λD/a where ‘D’ is the slit difference and ‘a’ is the slit width.)

24
Q

What does each term of the equation dsinθ=nλ stand for?

A

d = distance between gaps in the diffraction grating θ = angle between normal at the grating and the specific maxima. n = order of maxima (central is 0) λ = wavelength

25
Q

What pattern does a diffraction grating give?

A

Bands of light opposite the grating starting with a central band and then moving outwards. It can also split visible light into its constituent colours (CDs do this to light as they act like a diffraction grating).

26
Q

How is a line emission spectra formed?

A

A glowing gas emits light at specific wavelengths. The wavelengths of the lines are characteristic of the chemical element that produced the light. The emitted light is shone through a prism and displayed as a spectrum with fine lines corresponding to the wavelengths.

27
Q

How is a line absorption spectra formed?

A

A continuous spectrum of light is shone through a glowing gas. The gas absorbs light of specific wavelengths so the transmitted light is missing these wavelengths. The light is shone through a prism and a spectrum shows dark lines where the elements have been absorbed.

28
Q

For a diffraction grating how do you find the largest order of maxima that is visible using the following formula; dsinθ=nλ?

A

You assume sinθ= 1 as you will not see any maxima beyond 90 degrees as this would be behind the grating. This simplifies the formula to: d=nλ The is solved for n: n= d/λ Remember to always round the n down as if it was rounded up then the angle would be greater than 90 degrees which is impossible!!!