Page 112 - The Indian Optician Digital Edition May-June 2023

surfaces), and adding to this the minimum
thickness which has been specified for the lens .
1
If the lens is positive the minimum thickness
lies at the edge of the lens. The edge thickness,
or edge substance as it is frequently known, is
denoted by t .
E
If the lens is negative, the minimum
substance lies at the optical centre of the
lens. The axial or centre thickness (or centre
substance), is denoted by t .
C
Figures 1 and 2 illustrate how the thickness of
different forms of positive and negative lenses
is made up. The accompanying rules should be
self evident from the geometry of the figures.

bi-convex lens: t = z + z + t
C
1
E
2
t = t - (z + z )
2
C
E
1
plano convex lens: t = z + t
C
1
E
t = t - z 1 FIGURE 1 - THICKNESS OF PLUS LENSES
C
E
curved form: t = z - z + t The semi aperture of the lens, y = 20mm, so
C
2
E
1
t = t - (z - z ) on substituting into the sag formula:
E
C
2
1
bi concave lens: t = z + z + t z = 52.3 - √(52.3 - 20 ).
2
2
1
E
C
2
t = t (z + z ) = 52.3 - √(2735.29 - 400)
2
E
1
C
= 52.3 - 48.32
plano concave lens: t = z + t so, z = 3.98mm.
C
2
E
t = t - z 2
E
C
The centre thickness, t = z + t = 4.98mm.
minus curved lens: t = z - z + t C E
C
1
2
E
t = t - (z - z ) When a lens has two curved surfaces,
2
C
E
1
the sum or difference of the sags must be
Note that for both plus and minus curved calculated, i.e., the formula must be applied
lenses the geometry of these two forms twice.
demonstrates that the statement:
(ii) Calculate the edge substance of a -10.00 DS
z + t = z + t C lens, made in curved form in spectacle crown
E
2
1
is valid for all curved lenses. glass, (n = 1.523), the surface powers of which
are +4.00 DS and 14.00 DS. The lens is 44mm in
EXAMPLES: diameter and has a central thickness of 1.0mm.
(i) Calculate the centre thickness of a plano We have, t = 1.0mm F = +4.00 D
C
1
convex lens made in spectacle crown glass (n = y = 22mm F = -14.00 D
1.523). The power of the convex surface is +10.00 n = 1.523 2
DS, the diameter of the lens 40mm and the
edge substance 1mm. Proceeding as before, r = 523 / 4 = 130.75mm
1
The radius of curvature of the curved surface r = -523 / -14 = 37.36mm
2
2
2
must first be found from: z = 130.75 - √(130.75 - 22 ) = 1.86mm
1
z = 37.36 - √(37.36 - 22 ) = 7.17mm
2
2
2
r (mm) = 1000(n - 1) / F
Now, t = z - z + t = 7.17 - 1.86 + 1.0, so the
E
2
1
C
r = 523 / 10 = 52.3mm. edge thickness, t = 6.31mm.
E
108 | THE INDIAN OPTICIAN | MAY-JUNE 2023 LENS TALK