FOR OPTOMETRIC OPTICS

1. PRISM
(DIVIDING,RESOLVING &COMPOUNDING)
OPTOM ASKAR PK

3. Prentice's rule
Prentice's rule, named so after the optician Charles F.
Prentice, is a formula used to determine the amount of induced
prism in a lens:
• P=cf
• where:
• P is the amount of prism correction (in prism dioptres)
• c is decentration (the distance between the pupil centre and the
lens's optical centre, in millimetres)
• f is lens power (in dioptres)

6. Problem
• Eg: 1.
Calculate the thickness difference of a 4∆ prism
made in spectacle crown glass, 44mm in
diameter.
P=4 ∆, d= 44mm, n= 1.523
g= pd/100(n-1) ?

7. Prism
Base
Direction

8. Eg:R 2 PRISM 160
Or 2 prism base up & out along
160

9. L 2 PRISM 340
Or L 2 PRISM BASE DOWN & OUT ALONG 340

10. DIVIDING (SPLITTING) PRISM
When prism is prescribed it is usually divided equally b/w the two
eyes
• Weight more evenly distributed
• Reduces thickness of the intended lens
• Minimizes chromatic aberrations
Eg:r 3 prism base in causes right eye to deviate outwards by 3 prism
If we place 1.50 prism in before each eye,the required total deviation will still be
achieved.
Therefore R 3 Prism Base In=R 1.5 B In Combined With L 1.5 B In

11. DIVIDING PRISM-HORIZONTAL MERIDIAN
• A single prism in front of one eye may be represented by 2 prism infront of each eye with
their bases either both in or both out.
• >>>>>>>>>>>

12. DIVIDING PRISM-VERTICAL MERIDIAN
• Eg:L 2 Prism Base Down Prescribed
The L eye has to rotate upwards to view the image
This separates the visual axes by required angle
If we place 1 prism base down before each eye,the required separation would not be
achieved.
Ie,the eefect would be neutralised as b oth would rotate up by the same amount
therefore it is necessary to place 1 prism base up before the R eye & 1 prism base
down before the L to achieve the correction if we wish to divide the prism equally.

13. EXAMPLES OF DIVISION OF PRISM
POWER
• ?? R 3 P Base In L Plano
• ANSWER:R 1.50 P BASE IN L 1.50 P BASE IN
• ?? R 6 P Base Down L Plano
• ANSWER:R 3 P BASE DOWN L 3 P BASE UP

14. • Eg: R 3 PD BASE UP L 4 PD BASE DOWN
• ??WHAT SINGLE PRISM REPLACES THESE 2 PRISM FOR R.E
• ANSWER:DIVIDE THE AMOUNT EQUALLY B/W 2 EYES
• L 4 BASE DOWN IS EQUIVALENT TO R 4 BASE UP
• THEREFORE TOTAL FOR R EYE= R 7 BASE UP
• R 3.50 PD BASE UP L 3.50 PD BASE DOWN

15. ??R 5 PD BASE UP L 4 PD BASE IN
• R 5 PD BASE UP=R 2.50 PD BASE UP L 2.50 PD BASE DOWN
• L 4 PD BASE IN=R 2 PD BASE IN L 2 PD BASE IN
• ANSWER IS :R 2.50 PD BASE UP L 2.50 PD BASE DOWN
R 2 PD BASE IN L 2 PD BASE IN

16. R 4 P BASE UP L 2 P BASE OUT
• R 2 P BASE UP 1 P BASE OUT
L 2 P BASE DOWN 1 P BASE OUT

17. COMPOUNDING & RESOLVING PRISM
• COMPOUND R 3 P BASE UP L 4 P BASE IN INTO A SINGLE PRISM
INFRONT OF THE RIGHT EYE???
L 4 P BASE IN EQUIVALENT TO R 4 BASE IN

18. • OR =ROOT OF 3 ^ 2 + 4 ^ 2 = 5 P
• TAN ROH = RH/OH ( Pythagoras Theorem )
• TAN ROH = 3 / 4 =0.75
• ROH = 36 DEGREE 52 ‘ANSWER:SINGLE PRISM R 5 P BASE UP &
IN ALONG 36 DEGREE 52 ‘

19. COMPOUND 1P BD & 2 P BI INTO A SINGLE PRISM
INFRONT OF THE RIGHT EYE???
• OR=ROOT OF 1 ^ 2 + 2 ^ 2=2.236 PRISM
• TAN ROH=1 / 2 = 0.50 ,ROH=26 DEGREE 34’ (THE RESULTANT IN
OLD ENGLISH NOTATION LIES ALONG180 – ROH=153 DEGREE
26’)
• ANSWER IS:2.236 P BD & IN @153 DEGREE 26’

22. RESOLVING POWER
• RESOLVE 4 P 30 DEGREE INTO HORIZONTAL & VERTICAL COMPONENETS

23. • OV=OR SIN 30
• OH=OR COS 30
• OV=4 SIN 30 OH = 4 COS 30
• OV= 2 P BU OH = 3.46 P BI