Lecture 29- Light flow through the eye Flashcards

1
Q

What is a troland ?

A

measure of retinal illuminance- takes into account the seize of the pupil and luminance of the object

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2
Q

What is the most expensive part of the data projector ?

looking at screen luminance of projected image

A

projection lens - which captures a lot of light from this slide and is projected to form an image on a screen- the screen is usually a lambertian screen - important seen equally bright to people in audience

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3
Q

What is the intensity of light towards projection lens ?

A

area x luminance
then multiply by the solid angle (which lens subtends at the slide ) ( pi u2) - we have total light flux captured by this lens from this slide

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4
Q

What happens to some of the total light flux ?

A
  • will be lost by scatter and absorption within the lens
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5
Q

So , what is the total light flux that is transmitted to image space ?

A

t (transmittance ) x light flux captured by the lens

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6
Q

Where is the light flux distributed ?

A

over the area of the image , area of image a’ = area of object a x square of transverse magnification
***

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7
Q

Why do you need a lot of light in this slide ?

A

to ensure the image seen in audience is of sufficient luminance

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8
Q

What will the illuminance level in the plane of the screen ?

A

total luminous flux captured by lens and transferred to image space divided by area of image

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9
Q

What does the the amount off light per unit are of this screen depends on ?

A
  • brightness of this image which is L
  • depends on size of lens - solid angle - lens subtends at the slide piu²
  • how much light absorbed by this lens - t
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10
Q

What happens if light is distributed over a small image ?

A

-will correspond to a high illuminance

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11
Q

What happens if light is distributed over a large image (so large Magnification )?

A
  • will correspond to a small illuminance

* ***

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12
Q

What is the screen colour/light ?

A

usually white

-has a certain reflectance

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13
Q

What are majority of the screen ?

A

lambertian surfaces

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14
Q

Why are they lambertian surfaces ?

A

because they should appear to be equally bright wherever the subject is in the audience
-for e.g could be both perpendicular to the screen as well as at large angles subtended from lens

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15
Q

What will the total light flux that is reflected back by the screen will be?

A

the total flux it receives x reflectance of screen

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16
Q

What is the total flux emitted by a lambertian source of L’ - which is the luminance of a screen?

A

luminance x area of image x pi

17
Q

What is the formula for image size ?

A

A ‘ = A x M²

18
Q

Overall what will the luminance of the screen equal to (equation ) ?

A

L’ = TRsLu²(1/M2)

brighter slide- brighter luminance of screen
proportional u² - because its poroptional to the solid angle of which the light is captured by the lens then divided by the square of transverse mag - which means larger image - smaller luminance

19
Q

What is the luminance of the screen ?

A

L’

20
Q

How does image size affect screen luminance?

A

Let the reflectance of the screen be, Rs, and the luminance of projected image on the screen be, L’ . Since most screens are Lambertian, the total luminous flux reflected by the screen from an image of area, A’, equals pi x L’ x A’

21
Q

What is the troland ?

A

a measure of retinal illuminance

22
Q

Why was troland introduced ?

A

introduced to avoid the need to know light losses within each eye and parameters such as refractive index and image size

23
Q

What is the retinal illuminance equation in trolands ?

A

E (trolands) = PA(mm2)*L (cd/m2)

L- luminance of object
PA- area of pupil in mm²

24
Q

What is one troland ?

A
  • the retinal illuminance which results when the eye uses a screen of luminance 1cd/m² through a pupil having an area of 1 mm²

for e.g
If the pupil area is 10 mm2, for example, then the same object luminance (i.e., 1 cd/m2) would generate a retinal illuminance of 10 trolands.

25
Q

Explain the pupil size and retinal illuminance via simple lens system :

A
  • lens approximates pupil
  • relationship of image and object in terms of light levels is virtually the same
    -light flux captured by the lens - luminous flux = ALpiu²
    -Light flux transmitted to image space: luminous flux’ = TALpiu² - the flux that is transmitted towards the retina
  • Image size (mag) - nu x n’u’ = transverse magnification
    Mag - A’ = AM² WHERE M = nu/n’u’
  • total flux transferred towards retina divide by area of image- E = TLPIu²/ M²
26
Q

What is the image illuminance for retina ?

A

E = TLPIu²/ M²

27
Q

What is the derivation of the retinal illuminance equation \?

A
  • E = TLPIu²/ M²
  • E = TLpi (n’/n)2(u’)2
  • Let PA = pi x a (area of pupil)
  • E = T(n’/f’)² (LPA)
  • Retinal Illuminance (trolands) = LPA
28
Q

What is the actual change in image plane position ?

A
  • difference between true image distance and focal lens very small - reason why can express u’ as the radius of the pupil of the eye divided by fovea length f’
29
Q

What is the summary of formulas ?

A

Illuminance (lm/m2)- amount of light per unit area of a given service recieved from a source of intensity I theta . - this formula captures both inverse square law and cosine law
E(lm/m2)= I cosθ/r²

-Luminance (lm/sr/m2) or (cd/m2)
L (cd/m2)= I /Acosθ
A - area of source projected onto a plane perpendicular to the direction of interest - Acosθ

-Lambertian sources / surfaces ?
, L= Io/A (constant)
- the luminance is constant and independent of the direction of measurement which means the surface appears to be equally bright irrespective of the direction is viewed at
-for illuminated lambertian surface
L= E x R/pi
A- area
L -luminance 
total luminous flux- which source emits = pi x LA

-Retinal Illuminance (td)- Retinal Illuminance, E(trolands) = LPA