transverse magnification
ratio of image height to object height
M = h’/h - L/L’
image height / object height = object vergence / image vergence
-4D myope looks at a 10p coin of 24mm diameter -25cm from the eye. what is the retinal image sieze (assume +60D power and axial length = 23.81mm)
l = -25cm so L = -4 (1/-0.25)
L’ = L + Fe = -4 + 60 = 56D
M = L/L’ = -4/56 = -0.071 x mag
image height:
M = h’ / h
h’ = M x h = -1.71mm
calculating magnification with multiple lenses eg: spectacle wearers rule
image formed by one surface acts as an object for the next
h1’ = h2 , h2’ = h3
calculate the mag of 2 spectacle lenses - A corrective lens with a power -5D. Reduced eye power = +64D$.SThe distance between the lens and the eye = 0.012m
“finite object” placed at -0.25m from the spectacle lens
l1 = -25cm so L1 = -4D
L1’ = L1 + Fsp = -4 + -5D = -9D
L2 = L1’ / (1-d x L1’) = -8.12D
(L2 = -9 / 1- (0.012 x -9)
L2’ = L2 + Fe = -8.12D + 64D = +55.88D
M = L1/L1’ x L2/L2’ = 0.069 x
nodal points
conjugate points along the optical axis of unit angular magnification
a ray directed at the first nodal point leaves the system from the second point undeviated
in emmetropic eyes N and N’ = 5.56mm away from surface
alternative approaches to finding nodal points N and N’ - equation
F = n’ - n / r
eg: r = 5.55mm so the Nodal Point is located -5.56mm behind the corneal surface
a ray travelling through the centre of curvature travelling along a surface normal remains undeviated
calculating the height of an image of a distant object equation
h’ = -k’ tan w’ (angle of refraction)
h’ = -n tan w’ / fe
calculate the retinal image height formed from a 6/9 letter assumed to be distant by the emmetropic reduced eye of power 60
6/6 substends 5 minute sof arc
6/9 = 1.5 x > so 5 x 1.5 = 7.5
w = 7.5/60 degrees
h’ = -1 x tan(7.5/60) / 60
h’ = -0.036mm
calculate retinal image height of a corrected eye (considering distant and finite objects)
effect of spectacle lens = distant object
h’1 = -ntanw / Fsp = h2
effect of eye - finite object
M = h’2/h2 = L2/L’2 = K/K’
h’2 = (K/K’) -n tan w / F sp
h’2 = magnification of eye x height of image formed by spectacle lens?
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
image height due to spectacle lens:
w = 5’ = 5/60 degrees (6/6)
h’1 = -ntanw/ Fsp =
-1 x tan (5/60) / -5.00 = 0.291mm
Mag produced by eye and retinal image height:
L1 = 0
L’1 = L1 + Fsp = -5.00 D
L2 = L’1 / (1-dL’1) = -4.72D
L’2 = L2 + Fe = 55.28D
M2 = L2/L’2 = -0.085 x
h’2 = M2 x h2 = -0.085 x 0.291 = -0.025mm
principle points
conjugate points along the optical axis of unit transverse magnification
incident ray height at first principle plane is same as height of exiting ray on second principle plane
P/ P’ = 0mm from surface
always at the surface/lens for thin lenses
ray heights at P always equal P’
