Lecture 23 - Means of Light

Question 1

What is light?

Answer 1

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)

Question 2

What is the the wavelike representation of light ?

Answer 2

accounts for a number of diffraction and interference phenomena

Question 3

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

Answer 3

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

Question 4

When does light exist ?

Answer 4

only when we can detect it

Question 5

What is one wavelength indicated by in a graph?

Answer 5

a sinusoidal variation in electric field associated with a photon

Question 6

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

Answer 6

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

Question 7

What is c in a vacuum?

Answer 7

2. 998 x 10 ^8 ms-1

- high due to v being high

Question 8

What does each photon /and each particle have?

Answer 8

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

Question 9

What is SE equation?

Answer 9

SE = hv

Question 10

What does the higher frequency mean?

Answer 10

the higher the energy

Question 11

What is h?

Answer 11

planks constant

6.626 x 10_34 Js

Question 12

What are the 3 different ways of describing light?

Answer 12

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

Question 13

What are the 3 typical components of a photometric system ?

Answer 13

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.

Question 14

What is a very important element?

Answer 14

detector of radiation

Question 15

What are the decetors of the human eye ?

Answer 15

photoreceptor pigments

Question 16

How is the most important detector of the eye formed?

Answer 16

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

Question 17

What does the average human retina contain ?

Answer 17

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

Question 18

What is at low light levels?

Answer 18

rod receptors indicated by small diameter photoreceptors

Question 19

What is the density of photoreceptor in centre fovea ?

Answer 19

• lots of cones a

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

Question 20

What do the cone receptors do?

Answer 20

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

Question 21

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

Answer 21

more rods

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

Question 22

What is the ideal detector of radiation ?

Answer 22

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

Question 23

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

Answer 23

• 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

Question 24

How can we measure the spectrum responsively of the detector?

Answer 24

• 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

Question 25

What is the equation for detector spectral responsivityy ?

Answer 25

- 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

Question 26

how to measure Response linearity ?

Answer 26

- 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.

Question 27

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

Answer 27

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

Question 28

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

Answer 28

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.

Question 29

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

Answer 29

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)

Question 30

What does 1 lumen equal to (luminous flux) ?

Answer 30

1 / 683 watt ( radiant flux)

Question 31

What does 1 watt generate ?

Answer 31

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

Question 32

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

Answer 32

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

Question 33

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

Answer 33

1 lumen = 1/683 1 watt= 683 lumens efficacy of this source A = 683 / 1

Question 34

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?

Answer 34

683 x 0.06 = 40.98 lumens produced by source B efficacy of this source is = 40.98 / 1 (radiant flux) =40.98

Question 35

3. 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?

Answer 35

- use a table ( on slide) - add source a and b together you get 683 + 40.98 = 724 4. -the efficacy of this source will be 724/2 = 362

Question 36

How is the efficacy of light worked out ?

Answer 36

luminous flux output / power consumed ( radiant flux )

Question 37

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

Answer 37

555nm

Question 38

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

Answer 38

true

Question 39

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

Answer 39

false

Question 40

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

Answer 40

false

Question 41

what is the unit for luminous flux?

Answer 41

lumen

Question 42

what is the unit for radiant flux?

Answer 42

watts