4.2 - Image height for distant object Flashcards
Why can’t you use M = L/L’ for distant object
Because L = 0 in distant object
What is classed as infinitely distant
5/6m away
What ray is used to calculate image height for distant object
Undeviated ray leaving surface and appearing to come through 2nd nodal point
What are nodal points
Conjugate points along the optical axis of optical system of unit angular magnification
- Angular magnification - light is undeviated
- Conjugate points – one of nodal points is related to the object space and the other to the image space like an object and image pair
What happens to ray directed at 1st nodal point
A ray directed at the first nodal point leaves the system undeviated from the second – as if coming from 2nd nodal point
What is position of N for standard emmetropic reduced eye
+5.56mm from surface of reduced eye
What is position of N’ for standard emmetropic reduced eye
+5.56mm from surface of reduced eye
What is property of nodal points for single surface
Nodal points are coincident for a single surface.
What is an alternative approach for N,N’
Radius of curvature of single surface reduced eye
What is centre of curvature for single surface reduced eye
r = +5.55mm
use F = (n’-n)/r
What happens if ray strikes a spherical surface at right angles to surface
It undeviates…..
cause angle of incidence is 0 so angle of refraction is 0 according to snells law
What is any ray towards centre of curvature classed as
Undeviated…
Nodal points of a single surface have to both lie at centre of curvature – same location as radius of curvature
What is formula for height of distant object for any single surface e.g. spectacle lens or thin lens
h’ = -ntanx / Fe
What is spacing of photoreceptors in middle of fovea
2 to 4 microns from surface of reduced eye
What angle will 6/9 letter subtended and why
7.5’ arc
because 6/6 – SUBTENDS 5’ OF ARC AND 6/9 LETTER IS 1.5X BIGGER SO YOU DO 5 X 1.5 = 7.5
Calculate the retinal image height formed from a 6/9 letter (assumed to be distant) by the emmetropic reduced eye.
- Angle: 6/6 = 5’ arc
6/9 = 7.5’ arc
7.5’ arc to degrees = divide by 60
( Emmetropic reduced eye so n is 1 and Fe is 60 )
- h’ = -(n tan x)/Fe
= -1 tan(7.5/60)/60
= - 0.036mm
What is intermediate image
Image formed by the negative sp lens because that image then acts as the object for the eye and moves onto form final retinal image height h2’
Ray diagram for corrected eye i.e multiple surface with sp lens and eye
- Orignal object is infinitely distant from these parallel rays coming in at fixed angle – these are rays of light that are infinite at spectacle lens
- The spectacle lens then diverges them
- Forming erect virtual image which eye looks at
- To form final retinal image before converging it to
- Form final real image on retina
What does sp lens do
Takes infinitely distant object and produces an image from it so we can calucalte the height of this intermediate image produced by spectacle lens
What is the image produced by sp lens
h1’
What does h1’ i.e. image height from sp lens become
Object for h1’
What equation do we used for object for h1’
M = L2/L2’
This is because this object is finite so need to use transverse mag because its coming from sp lens
How to calculate h2’
- Work out transverse mag:
M = L2/L2’ - h2’ = M x h2 ( intermediate object )
Calculate the retinal image height formed from a 6/6 letter by a spectacle corrected myope if the spectacle lens power is -5DS, the vertex distance 12mm and the power of the reduced eye +60D
Find the image height produced by the spectacle lens - letter and produce an intermediate image of it at h1’ i.e. at Fsp:
- Angle: 6/6 = 5’ arc
5’ arc to degrees = divide by 60 - h1’ = -n tan x /Fsp
= -1 tan(5/60)/-5
= +0.291mm
Trace vergences to find M produced by eye and hence retinal image height:
- Object location:
L1 = 0 (distant object) - Refraction at lens i.e. image vergence:
L’1 = L1 + Fsp
L’1 = 0 + -5 = -5D - Vergence transfer:
L2 = L’1/(1-dL’1)
L2 = -5 / (1 - 0.013 x -5 )= -4.72D - Refraction at eye i.e. image vergence:
L’2 = L2 + Fe
L’2 = -4.72 + 60 = +55.28D - Mag at eye:
M = L1/L1’ X L2/L2’
M = 0/-5 X -4.72 X 55.28
M= -0.085× - Retinal image height:
h’2 = M2 × h2
h’2 =-0.085 × 0.291 = -0.025mm
(h2 = h1’ =0.291mm)
What are principal points P and P’
Conjugate points along the optical axis of unit transverse magnification
What does principal points P and P’ show in terms of height
Incident ray height at first principal plane is same as height of exiting ray on second principal plane
What is a principal plane
Flat surface passing through principal point
For the standard emmetropic reduced eye what is P and p’
Position of P = +0.00mm
Position of P’ = +0.00mm
For a single surface or thin lens, where are the principal points
Always at the surface/lens
Where are ray heights always the same
At P and P’
Where is single reduced eye placed
Mid-way between the principal points of a schematic eye
