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Optics > Lecture 23 - Means of Light > Flashcards

Lecture 23 - Means of Light Flashcards

1
Q

What is light?

A

Transverse electromagnetic waves that extend from gamma rays (high energy) to radio waves (low energy).
-If undisturbed, these propagate in straight lines called “rays” (geometrical optics)

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2
Q

What is the the wavelike representation of light ?

A

accounts for a number of diffraction and interference phenomena

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3
Q

What is the particle or quantum nature of light is particularly useful to?

A

describe the exchange of energy when light is absorbed and scattered.

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4
Q

When does light exist ?

A

only when we can detect it

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5
Q

What is one wavelength indicated by in a graph?

A

a sinusoidal variation in electric field associated with a photon

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6
Q

What is the total distance traveled every second of one wavelength (SPEED OF LIGHT) ?

A

c= lamda (WAVELENGTH) x v (frequency of radiation)

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7
Q

What is c in a vacuum?

A
  1. 998 x 10 ^8 ms-1

- high due to v being high

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8
Q

What does each photon /and each particle have?

A
  • has small amount of energy associated by it and it is only determined by its frequency nu
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9
Q

What is SE equation?

A

SE = hv

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10
Q

What does the higher frequency mean?

A

the higher the energy

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11
Q

What is h?

A

planks constant

6.626 x 10_34 Js

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12
Q

What are the 3 different ways of describing light?

A
  • wavelike nature
  • rays
  • particle aspect of light which allows us to understand the energy in each photon.
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13
Q

What are the 3 typical components of a photometric system ?

A
  1. Source- of electromagnetic radiation which may be the sun (most important), a light emitting diode, a candle or an electric lamp- it then travels the light until it hits a detector which is the time when light is detected.- detecting light means absorbing energy- which means destroying photons.
  2. A detector - which might be the human eye, or a photocell, or a tomato plant
  3. A filter or modulator -which is anything in between these two, such as the atmosphere, the macular pigment in the eye, or a slit aperture or filter in an optical system- (it absorbs mostly short wavelength of lights) - after absorption of these filters including lens of human eye- light is converted into electrical signals by absorption in photoreceptor pigments.
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14
Q

What is a very important element?

A

detector of radiation

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15
Q

What are the decetors of the human eye ?

A

photoreceptor pigments

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16
Q

How is the most important detector of the eye formed?

A

by combining. signals from long wavelengths ( L cones ) and medium wavelengths (M )
-They are usually in ratio of 2L : 1M = Photopic

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17
Q

What does the average human retina contain ?

A

twice as many long wavelength cones (L CONES) than M cones

18
Q

What is at low light levels?

A

rod receptors indicated by small diameter photoreceptors

19
Q

What is the density of photoreceptor in centre fovea ?

A
  • lots of cones a

- Note the size of cone receptors in the fovea and the absence of rod receptors.

20
Q

What do the cone receptors do?

A

increase rapidly in diameter and end up having a larger diameter than surrounding rod photoreceptors

21
Q

What is the density of photoreceptors 1.2 mm in the periphery (RHS section)?

A

more rods

-Note the large number of rods in the near periphery and the fewer, but much larger cones.

22
Q

What is the ideal detector of radiation ?

A

-Spectral responsivity, V(lamda), linear response, large dynamic range and high sensitivity.

23
Q

What are the good properties of the ideal detector of radiation ?

A
  • spectral repsonsiovity- which is the spectral range over which you get a measurable response
  • the response linearity of the detector - its dynamic range which is arranged by levels over which the detector produces a linear response
  • The sensitivity of detector- which is the smallest amount of radiant flux for which the detector produces a measurable response
24
Q

How can we measure the spectrum responsively of the detector?

A
  • a graph
  • use silicion light detector
  • we allow light of a particular wavelength to fall on its surface and measure the signal generated
  • usually when you produce a spectrum the amount fo radiant flux over a narrow range of wavelength varies as you move the slit over the visual spectrum
  • so you may get more light in the long wavelength region os spectrum opposed to short
  • so need to divide the signal the detector produces in response to this wavelength of light by the amount of radiant flux at that wavelength which falls onto the surface of the detector
25
Q

What is the equation for detector spectral responsivityy ?

A

-
V(lamda)= S(lamda)/lamda(lamda)s(lamda)

divide signal by the radiant flux of that particular wavelength

  • then repeat for each wavelength of interest
  • then end up with spectral responsitivity of detector
26
Q

how to measure Response linearity ?

A
  • suppose we allow light of a particular wavelength i.e lamda 1 - to fall on the surface of detector and measure the signal S1
    -Then move the slit to another wavelength lamda 2 and measure signal S2
    -If you were to add the 2 fluxes together we would have a linear detector if the signal generated to the sum of the 2 wavelengths. - which is
    S1 + S2
    -this also applies when wavelength of light is the same - doesn’t have to change- if we put same light on surface of detector - we would double the signal to measure.
27
Q

What is the LINK between radiometric and photometric units (i.e., the link between the unit of luminous flux and the watt)?

A

the link between the luminous flux (what eye perceives) and radiant flux measured in watts

28
Q

What is the LINK between radiometric and photometric units (i.e., the link between the unit of luminous flux and the watt)?

A

the link between the luminous flux (what eye perceives) and radiant flux
-Need a detector of radiation
-In the case of a human eye - the V lamda response (shown in graph on slide) - it is suavely represented normalised with respect to maximum
-the maximum is measure at 555nm - this particular standardised spectral luminous efficiency function corresponds to the eye which relies on 2L cones than M cones
-Now this is the key point of conversion of radiant flux to luminous flux
-If we start with 1/683 of one watt (radiant flux) at the wavelength of
555 nm, which is the peak of the human eye’s response curve, represents a luminous flux of 1 lumen.
-All other wavelengths are weighted according to the V() response.

29
Q

What is the LINK between radiometric and photometric units (i.e., the link between the unit of luminous flux and the watt)?

A

the link between the luminous flux (what eye perceives) and radiant flux
-Need a detector of radiation
-In the case of a human eye - the V lamda response (shown in graph on slide) - it is suavely represented normalised with respect to maximum
-the maximum is measure at 555nm - this particular standardised spectral luminous efficiency function corresponds to the eye which relies on 2L cones than M cones
-Now this is the key point of conversion of rasdain flux to luminous flux
-If we start with 1/683 of one watt (radiant flux) at the wavelength of
555 nm, which is the peak of the human eye’s response curve, represents a luminous flux of 1 lumen.
-All other wavelengths are weighted according to the V(lamda) response.( sensitivity of human eye)

30
Q

What does 1 lumen equal to (luminous flux) ?

A

1 / 683 watt ( radiant flux)

31
Q

What does 1 watt generate ?

A

683 lumens ( but multiplied by the corresponding sensnity of the eye for that wavelength )

32
Q

What happens if sensitivity of the eye is 20 % of the maximum ?

A

multiply 683 by 0.2lumens for the same amount t fo radiant flux

33
Q

Source A produces monochromatic light at 555 nm. How many watts of radiant flux should the source emit to generate 683 lumens?

A

1 lumen = 1/683

1 watt= 683 lumens

efficacy of this source A = 683 / 1

34
Q

Source B produces monochromatic light at 460nm. The sensitivity of the eye at 460nm is only 6% of the maximum. How much luminous flux do we now have from source B if emits the same amount of radiant flux as source A?

A

683 x 0.06 = 40.98 lumens produced by source B

efficacy of this source is = 40.98 / 1 (radiant flux) =40.98

35
Q
  1. Which of the two sources is most energy efficient?

4. How many lumens of light do we have if we combine the two sources of light?

A
  • use a table ( on slide)
  • add source a and b together you get 683 + 40.98 = 724
  1. -the efficacy of this source will be 724/2 = 362
36
Q

How is the efficacy of light worked out ?

A

luminous flux output / power consumed ( radiant flux )

37
Q

What wavelength of the light will have the highest possible efficacy ?

A

555nm

38
Q

The eye makes use of four classes of photoreceptor in order to measure the amount of light over four distinct wavelength regions

A

true

39
Q

A normal eye is most sensitive to light of wavelength 555 nm when the ambient light level is high

A

false

40
Q

1 watt of radiant flux at 505 nm (where rods respond best to light) produces 683 lumens of light

A

false

41
Q

what is the unit for luminous flux?

A

lumen

42
Q

what is the unit for radiant flux?

A

watts

Optics (34 decks)

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