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Optics > lec 28 - light flow through optical systems > Flashcards

lec 28 - light flow through optical systems Flashcards

1
Q

what is total luminous flux when dealing with a lambertian surface and if the surface is L?

A
  • when dealing with a lambertian surface and if the luminance of the surface is L
  • you instantly know the total luminous flux which the surface emits
    which is equal to πLA
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2
Q

what is the intensity in a lambertian surface?

A
  • the intensity in the direction theta is equal to the intensity in a direction normal to the surface multiplied by cosine of angle theta
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3
Q

what is the unit for total luminous flux?

A

lumens

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4
Q

what is the unit of luminance?

A

lm/sr/m2 or cd/m2

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5
Q

what is the relationship between surface illuminance (E) and luminance (L) of a lambertian surface?

A
  • suppose we have an area of size A- so its a surface which is lambertian
  • it also has reflectance R
  • the surface receives its light because it is illuminated by light - an as result we have luminance L
  • therefore the total luminous flux received by the surface + E A (lm)
  • if reflectance of surface is less than 1 then only a certain fraction of light will be scattered back in every direction
  • let the reflectance of the surface be R
  • then the total luminance flux radiated over a hemisphere
    will be = total amount of radiant flux x reflectance or RXEXA
  • if the surface is lambertian then the total amount of flux must be equal to πLA
  • this gives us the expression between the luminance of the surface and the illuminance of the surface
    L= R* (E/π)
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6
Q

what happens in the majority of surfaces which have a high reflectance ?

A
  • the luminance of the surface is about 1/3 of illuminance

- e.g. white paper which has a reflectance of about 0.9

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7
Q

what is the expression between the luminance surface and illuminance ?

A

L= R* (E/π)

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8
Q

example 1 - you are given a photometer that measures luminance ( cd/m2) and piece of matt paper that reflect most of the light it receives. how would you go about arranging to measure the maximum luminous intensity of a small tungsten halogen lamp? how would you go about using this set up to calibrate an illuminance meter?

A
  • place the white diffuser at a certain distance away which we can measure accurately
  • we then eliminate the the white diffuser and measure using the photometer the luminance at an angle
  • remember that the luminance of a lambertian surface is independent of angle of measurement
  • we know that the illuminance level in the plane of this white diffuser is equal to the intensity of the source divided by the square of the distance
    -E=I/d2
  • this allows us to express the intensity of the lamp as being the illuminance level multiplied by d squared
    -I=E* d2
  • since this is a lambertian surface then the relationship between the illuminance on the surface and its luminance is given by this expression
    L = R * ( E/π)
  • this allows us to derive the illuminance as E=πL/R
  • once we know the illuminance of the surface we can work out the intensity of the source
    I= E*d2
  • also if we know the illuminance of the surface, we can take any illuminance instrument and place the sensor head perpendicular to the direction of illumination at this point where the luminance of the surface was measured
  • the instrument would read the estimated illuminance using a measure of luminance
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9
Q

what happens when you form the image of an object?

what is the luminance of the image ?

A
  • object is illuminated and its got a certain luminance
  • area of image also has luminance
    you have to
    I. calculate light flux, captured by the lens
    II. calculate light flux, transmitted by the lens
    III. calculate area of the image
    IV. calculate luminance L’, of the areal image L’=T(n’/n)2 L
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10
Q

what is the luminance of an image?

A
  • the luminance of image in terms of this constant is given by this expression
  • L’= T (n’/n)2 L
  • the luminance of an image is given by a constant T which is always less then one
  • the refractive indices are usually one
  • this means that the luminance of the image is not greater than luminance of object
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11
Q

how does the magnification of the image affect the relationship between object and image luminance ?

A
  • magnification doesn’t affect the luminance of the image
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12
Q

what do we want to know when forming an image?

A
  • when we from an image, we want to know not only the luminance of the image L’ but the illuminance
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13
Q

why is knowing the illuminance important?

A
  • determines the exposure you need in the camera

- the time that shutter should stay open to capture the appropriate amount of light

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14
Q

how do we calculate image illuminance?

A

the expression is
E=L(n’/n)2 x ( T x π/4 ) x ( D/f’)2

  • we have the illuminance level
    in the plane of the image which proportional to the luminance of the object
  • this is because the brighter the object the higher the illuminance level in the plane of the image
  • transmittance of lens and refractive indices are constants
  • D2 is proportional to the area of the pupil
  • the larger the pupil the more light is captured by the lens
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15
Q

what is f’ / D replace by ?

A
  • replaced by the f-number of the photographic lens which is the N value
  • replace by N
  • this changes illuminance expression to
    E=L(n’/n)2 x ( T x π/4 ) x (1/N)2
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16
Q

what happens when f-number of lens doesn’t change?

A
  • we can see that provided the f-number of the lens which the ration of the focal length to the diameter of the lens doesn’t change the illuminance level is the same
17
Q

what does image illuminance vary with ?

A
  • image illuminance varies only with object luminance and the square of the reciprocal of the f-number of the lens
18
Q
  • how much can you change f-number ( N-value ) in a lens?
A
  • a typical lens allows you to change the f-number setting which is the value of N from 1.4 to 16
19
Q

what is the typical human eye N-value?

A
  • the typical human eye has large value of N around 5.6 to 8
20
Q

what is the relationship between the f-number and pupil size?

A
  • if the pupil size is large, the f -number is small

- if pupil size is small, the f-number is large

21
Q

what is the illuminance level in the plane of the image proportional to ?

A
  • the illuminance level in the plane of the image is proportional to (1/N)2
  • it is also the total exposure which is the illuminance multiplied by the exposure time t
22
Q

what is the total exposure?

A
  • it is also the total exposure which is the illuminance multiplied by the exposure time t
23
Q

how is total exposure controlled?

A
  • which can be controlled in a good camera from 2 sec to 250ms shutter time
  • as you go from one setting to next you double exposure time
24
Q

what happens as you go from on f-number setting to the next?

A
  • as you go from one f-number setting to the next you half the exposure or the illuminance level
  • this means that we can change the aperture and exposure time we end up with the same exposure
25
Q

why is it useful that we end up with the same exposure

when we change aperture or exposure time?

A
  • this is useful because
  • as you have a large f-number , then diameter of lens is small and you need large exposure time, but also means that the depth of field which is those objects in the field view which remain in focus is very large because we are using a small aperture
  • things will be in focus over a large distance
26
Q

what settings do we need when taking a portrait?

A

when taking portrait and only to wish subject face in focus we require a small depth of field
- for a small depth of field we use a small f-number which implies large aperture , then depth of filed will be limited to object of interest

27
Q

what to do if we want to rapaidly capture changing scenery?

A
  • if we wish to capture rapidly changing scenery - we need short exposure number and small f-number setting
28
Q

why are lenses with small f-number expensive?

A
  • difficult to correct the aberrations when the size of the lens becomes comparable to its focal length
29
Q

how does the illuminance formula apply to the human eye?

A
  • E=L(n’/n)2 x ( T x π/4 ) x (1/N)2
  • since N=f’/D
    then
    L(n’/f’)2 x T ( πD2/4)
  • ( πD2/4) this is the area of the pupil of the eye- where D is diameter of pupil - we can replace this with are of pupil
    -E= T X (n’/f’)2 x L x P
  • T= transmittance
  • (n’/f’)2 = constant
    -L= luminace can be kept same= tells you bright object is
  • P= area of pupil
30
Q

what is the true retinal illuminance (lm/m2) directly proportional to ?

A

the true retinal illuminance (lm/m2) directly proportional to pupil size and object luminance

31
Q

what does larger pupil size mean?

A

more light captured from object

Optics (34 decks)

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