4.2 - Image height for distant object

Question 1

Why can’t you use M = L/L’ for distant object

Answer 1

Because L = 0 in distant object

Question 2

What is classed as infinitely distant

Answer 2

5/6m away

Question 3

What ray is used to calculate image height for distant object

Answer 3

Undeviated ray leaving surface and appearing to come through 2nd nodal point

Question 4

What are nodal points

Answer 4

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

Question 5

What happens to ray directed at 1st nodal point

Answer 5

A ray directed at the first nodal point leaves the system undeviated from the second – as if coming from 2nd nodal point

Question 6

What is position of N for standard emmetropic reduced eye

Answer 6

+5.56mm from surface of reduced eye

Question 7

What is position of N’ for standard emmetropic reduced eye

Answer 7

+5.56mm from surface of reduced eye

Question 8

What is property of nodal points for single surface

Answer 8

Nodal points are coincident for a single surface.

Question 9

What is an alternative approach for N,N’

Answer 9

Radius of curvature of single surface reduced eye

Question 10

What is centre of curvature for single surface reduced eye

Answer 10

r = +5.55mm

use F = (n’-n)/r

Question 11

What happens if ray strikes a spherical surface at right angles to surface

Answer 11

It undeviates…..

cause angle of incidence is 0 so angle of refraction is 0 according to snells law

Question 12

What is any ray towards centre of curvature classed as

Answer 12

Undeviated…

Nodal points of a single surface have to both lie at centre of curvature – same location as radius of curvature

Question 13

What is formula for height of distant object for any single surface e.g. spectacle lens or thin lens

Answer 13

h’ = -ntanx / Fe

Question 14

What is spacing of photoreceptors in middle of fovea

Answer 14

2 to 4 microns from surface of reduced eye

Question 15

What angle will 6/9 letter subtended and why

Answer 15

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

Question 16

Calculate the retinal image height formed from a 6/9 letter (assumed to be distant) by the emmetropic reduced eye.

Answer 16

1. 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 )

2. h’ = -(n tan x)/Fe
   = -1 tan(7.5/60)/60
   = - 0.036mm

Question 17

What is intermediate image

Answer 17

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’

Question 18

Ray diagram for corrected eye i.e multiple surface with sp lens and eye

Answer 18

• 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

Question 19

What does sp lens do

Answer 19

Takes infinitely distant object and produces an image from it so we can calucalte the height of this intermediate image produced by spectacle lens

Question 20

What is the image produced by sp lens

Answer 20

h1’

Question 21

What does h1’ i.e. image height from sp lens become

Answer 21

Object for h1’

Question 22

What equation do we used for object for h1’

Answer 22

M = L2/L2’

This is because this object is finite so need to use transverse mag because its coming from sp lens

Question 23

How to calculate h2’

Answer 23

1. Work out transverse mag:
   M = L2/L2’
2. h2’ = M x h2 ( intermediate object )

Question 24

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

Answer 24

Find the image height produced by the spectacle lens - letter and produce an intermediate image of it at h1’ i.e. at Fsp:

1. Angle: 6/6 = 5’ arc
   5’ arc to degrees = divide by 60
2. 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:

3. Object location:
   L1 = 0 (distant object)
4. Refraction at lens i.e. image vergence:
   L’1 = L1 + Fsp
   L’1 = 0 + -5 = -5D
5. Vergence transfer:
   L2 = L’1/(1-dL’1)
   L2 = -5 / (1 - 0.013 x -5 )= -4.72D
6. Refraction at eye i.e. image vergence:
   L’2 = L2 + Fe
   L’2 = -4.72 + 60 = +55.28D
7. Mag at eye:
   M = L1/L1’ X L2/L2’
   M = 0/-5 X -4.72 X 55.28
   M= -0.085×
8. Retinal image height:
   h’2 = M2 × h2
   h’2 =-0.085 × 0.291 = -0.025mm
   (h2 = h1’ =0.291mm)

Question 25

What are principal points P and P’

Answer 25

Conjugate points along the optical axis of unit transverse magnification

Question 26

What does principal points P and P’ show in terms of height

Answer 26

Incident ray height at first principal plane is same as height of exiting ray on second principal plane

Question 27

What is a principal plane

Answer 27

Flat surface passing through principal point

Question 28

For the standard emmetropic reduced eye what is P and p’

Answer 28

Position of P = +0.00mm

Position of P’ = +0.00mm

Question 29

For a single surface or thin lens, where are the principal points

Answer 29

Always at the surface/lens

Question 30

Where are ray heights always the same

Answer 30

At P and P’

Question 31

Where is single reduced eye placed

Answer 31

Mid-way between the principal points of a schematic eye