field of view?

[lee staniforth]

Hi,Question for you all...

1. +10.00DS plano-convex lenticular CR39 lens is to be supplied. The aperture is 34mm. If the vertex distance is 12mm and the eyes centre of rotation is 13mm from the cornea, then find the angular field of view.

2. If the above lens has a centre thickness of 6.5mm then find the front surface power.

Any help would be great

Thanks:drop:

[HarryChiling]

1) +10.00DS plano-convex lenticular CR39 lens is to be supplied. The aperture is 34mm. If the vertex distance is 12mm and the eyes centre of rotation is 13mm from the cornea, then find the angular field of view.

There is a real and apparent field of view. The apparent field of view depends of the size of lens (aperture in this case) and the distance from the center of rotation, the real field of view also takes into consideration the power. Lets start with the apparent and then the real:

y = 1/2 of aperature in mm
s = distance from back vertex to center of rotation in mm
f = focal length of the lens in mm
D = power in diopters


Apparent FOV:

tan(A) = y/s
tan(A) = 17mm/25mm
tan(A) = 0.68
A = 34

The apparent field of view is 2*A or 68 degrees.

Real FOV:

f = 1000/D
f = 1000/10
f = 100mm

tan(R) = y/s - y/f
tan(R) = 17mm/25mm - 17mm/100mm
tan(R) = 0.68 - 0.17
tan(R) = 0.51
R = 27

The real field of view is 2*R which is 54 degrees.

2) If the above lens has a centre thickness of 6.5mm then find the front surface power.

First jot down all your known variables

Back Vertex Power (Ft) = +10
Center thickness (t) = 0.0065m
Back Surface Power (F2) = +0.00

You need to find the front surface power of the lens, well plug your known values into the back vertex power equation:

Ft = [F1 + F2 - (t/n) * F1 * F2] / [1 - (t/n) * F1]

seems daunting but our back surface power is +0.00 so we can cut down a lot of this equation

Ft = F1 / [1 - (t/n) * F1]
Ft * [1 - (t/n) * F1] = F1
Ft - (t/n) * F1 * Ft = F1
Ft = F1 + (t/n) * F1 * Ft
Ft = F1 * (1 + (t/n) * Ft)
F1 = Ft / (1 + (t/n) * Ft)

F1 = 10 / (1 + (0.0065/1.5) * 10)
F1 = 10 / (1 + 0.04)
F1 = 10 / 1.04
F1 = 9.58D

So the front surface power would be 9.58D, I made an assumption that the lens is a 1.5 index since no index was given but feel free to sub it out if the index is some other material.

[lee staniforth]

Harry. you are a star.. thank you.