lecture 24 - fundamental quantities and basic laws Flashcards
what is the solid angle Ω ?
- the solid (sr) subtended by a cone that marks an area
- size of cone is defined by A/r^2
r= radius of the sphere
what is a cone ?
cone defines a region of space within a sphere
how can solid angle be expressed ?
Ω can be expressed in terms of the semi-angle of the cone
how to calculate semi-angle of the cone ε ?
- dividing the length of cord AB by the radius of the disc
- this gives angle in radians
angle (rad) ε = AB/r
how to calculate cone of solid angle Ω ?
Ω = A/r^2
area A= formed by the intersection with the cone divided by r^2
how to calculate solid angle Ω ?
2π ( 1 - cos ε )
how else can the solid angle ( sr) of a cone of a semi-angle be measured?
Ω = π ε ^2
angle ε must be in radians
what is radiant energy (Q) ?
- the total number of joules that might be produced in a flash of light
- units = Joules (J)
what is radiant flux (Q)?
- rate of flow of energy which is the number of joules produced every second
- units= Joules / s (Watts)
what is radiant emmittance (M) ?
- number of watts emitted( which is joules per second )per unit area of source
- unit= W/m^2
what is irradiance (E) ?
- the amount of radiant flux received per unit area of a given surface
- number of watts received per a unit area of the source
- unit = W/m^2
what is radiant intensity (I) ?
- measured in terms of the radiant flux per unit solid angle in a given direction
- described as intensity of source
- unit=Watts/sr
what is radiance (L) ?
intensity over solid angle which is the amount of radiant flux per unit solid angle in a given direction and size of source that produces that intensity
- unit = Watts/sr/m^2
what are the most important radiometric quantities ?
- Irradiance (E)
- Radiant intensity (I)
- Radiance (L)
what happens if we place a the surface of the receiver which is perpendicular to the line in the direction of illumination which is line from centre of surface to the source?
- when we place the surface of the receiver
- the receiver will be irradiated
- irradiance level will depend on the intensity of the source but also on the actual distance between the source and the receiver
what is the solid angle subtended by the illuminated surface ?
the solid angle subtended by the receiver at the source is the area of the receiver divided by square of the distance
Ω = R/r^2
what is the light flux captured by the surface of the receiver ?
radiant flux captured by receiver is solid angle which receiver subtends at source multiplied by the intensity which is the flux per unit solid angle
radiant flux( Watts) = Ω *I
what is the irradiance of the receiver ?
- radiant flux received per unit area of the receiver
- E=radiant flux capture/ area of the receiver (R)
or E=I/r^2
what are the basic photometric laws ?
. inverse square law
. cosine law
what happens if angle ε is 90 deg ?
- plane of receiver will be along the direction of illumination
- so surface of receiver receives no light at all
what is area of effective area as seen from source ?
A’B’ = AB COS ε
what is solid angle subtended by the illuminated source ?
solid angle = R cos ε/r^2
smaller than it would’ve been if the receiver was perpendicular to direction of illumination
what is the light flux captured by the surface ?
(lm) = Ω *I
what is flux per unit of illuminated surface ?
E=I* cos ε/r^2
what is the inverse square law ?
- the irradiance of a surface element by a point source is proportional to the reciprocal of the square of the distance between the source and the surface element?
what is the equation the summarises the two photometric law ?
E(W/m^2)= I COS ε / r^2
what is the cosine law ?
the irradiance level is proportional to the cosine of the angle between the normal to the surface and the line joining the source and the surface element
what is the Maxwellian viewing system ?
- when we place our eye so that image is in centre of the pupil
- the eye will see the exit pupil of the lens which will have same brightness as LED when viewed directly
- this is known as maxwellien viewing
example
subject sees exit pupil of lens
LED source changed with identical diode that generates the same amount of radiant flux at wavelength of 420 nm?
what will happen to the perceived brightness of source ?
if tunable diode was available which wavelength would generate maximum perceived brightness ?
- brightness will be much darker than yellow light and red light
- answer -
555nm because that is the wavelength for which the eye has maximum sensitivity
