Optical Ed Articles » Compensated and Effective Power

When we compensate an Rx, we do so because the lens changes power as moves closer or farther away from the eye. We are making an Rx stronger or weaker depending on whether the lens is moved closer farther away.

*Remember that a lens that is moved closer to the eye will gain in minus power and a lens that moves further from the eye will gain plus power.

A glasses wearer tells you that her Rx is unchanged, but she wants newer, smaller glasses. Her old glasses fit at 13 mm, and the new ones that you fit her with are at 9 mm. If her right lens Rx is -12.00 DS, what will the compensated power for her new right lens be? Round to the nearest 1/8 diopter step.

We need to solve for the compensated power which is going to compensate the lens when moved away from the face or towards the face.

$D_{new} = \frac{D_{rx}}{(1 - dD_{rx})}$

$D_{new}$ is the new power

$D_{rx}$ is the sphere power

$d$ is the change in vertex in meters. It is negative if moved away from the face and positive if moved toward the face.

$D_{new} = \frac{-12.00}{[1 - (0.004)(-12.00)]}$

$D_{new} = \frac{-12.00}{[1 - (-0.048)]}$

$D_{new} = \frac{-12.00}{1.048]}$

$D_{new} = -11.45\ D$

Since this is compensated we need to round this to a dioptric value. -11.50 D.

What if the original Rx had cylinder? We need to put the Rx on the optical cross to find the major meridians of power.

If the Rx is +13.00 – 3.00 X 180, they were refracted at 10 mm but the glasses fit at 13, what is the compensated Rx?

The two major meridians of power are +13.00 X 180 and +10.00 X 90. Solve for each of these independently.

$D_{new} = \frac{13.00}{[1 - (-0.003)(13.00)]}$

$D_{new} = \frac{13.00}{[1 - (-0.039)]}$

$D_{new} = \frac{13.00}{1.039}$

$D_{new} = +12.51$

$D_{new} = \frac{10.00}{[1 - (-0.003)(10.00)]}$

$D_{new} = \frac{10.00}{[1 - (-0.03)]}$

$D_{new} = \frac{10.00}{1.03}$

$D_{new} = +9.71$

So we have +12.51 X 180 and +9.71 X 90. If you pull this off of the optical cross you will have +12.50 -2.75 X 180

When we talk about effective power we are referring to the power the wearer will feel if the lens is moved closer or farther away.
The formula is the same except that the numerator is now $[(1+dD_{rx})]$

Using the first example of the -12.00 moved 4 mm closer to the eye. What would the effective power be.
$D_{new} = \frac{-12.00}{[1 + (0.004)(-12.00)]}$

$D_{new} = \frac{-12.00}{[1 + (-0.048]}$

$D_{new} = \frac{-12.00}{0.952}$

$D_{new} = -12.61\ D$

So the power that the wearer will experience if a -12.00 D lens is moved closer to the eye is -12.61 D.