First, you must notice that the power in this lens is +14.00D on the 045th meridian and +12.00D on the 135th meridian. You have to determine the effective power in each meridian, then create an Rx again.

EFFECTIVE POWER. This is a plus lens moved toward the face. It will be effectively weakened.

• On the 45 meridian:

The change in vertex distance is 3 mm, which equals 0.003 m. The lens moved toward the face, so d is positive. Dl is +14.00. What is De?

     Dl
De = ---------------
     1 + dDl

     +14.00
De = --------------------------
     1 + (+0.003)(+14.00)

     +14.00
De = --------------
     1.042

De = +13.44

The effective power on the 45th meridian is +13.44.
 
• On the 135 meridian:

The change in vertex distance is 3 mm, which equals 0.003 m. The lens moved toward the face, so d is positive. Dl is +12.00. What is De?

     Dl
De = ---------------
     1 + dDl

     +12.00
De = --------------------------
     1 + (+0.003)(+12.00)

     +12.00
De = --------------
     1.036

De = -11.58

The effective power on the 90 meridian is +11.58D, not +12.00.

The 45 meridian is +13.44, the 135 meridian is +11.58D, so the effective power that the wearer is experiencing is +13.44 -1.86 x 045.
 

COMPENSATED POWER: We need more plus get what the Doctor intended.

• On the 45th meridian:

The change in vertex distance is 3 mm, which equals 0.003 m. The lens moved toward the face, so d is positive. Dl is +14.00. What is Dc?

     Dl
Dc = ---------------
     1 - dDl

     +14.00
Dc = --------------------------
     1 - (+0.003)(+14.00)

     +14.00
Dc = --------------
     0.958

Dc = +14.61

The power needed on the 045 meridian rounds to +14.62D.
 
• On the 135 meridian:

The change in vertex distance is 3 mm, which equals 0.003 m. The lens moved toward the face, so d is positive. Dl is +12.00. What is Dc?

     Dl
Dc = ---------------
     1 - dDl

     +12.00
Dc = --------------------------
     1 - (+0.003)(+12.00)

     +12.00
Dc = --------------
     0.964

Dc = +12.45

The power needed on the 135 meridian rounds to +12.50D.

EFFECTIVE POWER. This is a plus lens moved toward the face. It will be effectively weakened.

The change in vertex distance is 10 mm, which equals 0.010 m. The lens moved toward the face, so d is positive. Dl is +13.00. What is De?

COMPENSATED POWER. This is a plus lens moved toward the face. The trial contact lens needs more plus power.

The change in vertex distance is 10 mm, which equals 0.010 m. The lens moved toward the face, so d is positive. Dl is +13.00. What is Dc?

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