How do you calculate the total power of a lens with a front curve of +8.00D and a back curve of -5.00D, assuming an index of refraction of 1.530?

How do you calculate the total power of a lens with a front curve of +8.00D and a back curve of -5.00D, assuming an index of refraction of 1.530?

• A. +2.00D
• B. +1.50D
• C. +3.00D
• D. +4.00D

Answer: (C)

To calculate the total power of a lens, you can use the formula: \[ P = P_f + P_b + (n - 1) \left( \frac{P_f \cdot P_b}{n} \right) \] Where: - \( P \) is the total power of the lens, - \( P_f \) is the front curve power, - \( P_b \) is the back curve power, - \( n \) is the index of refraction. In this case: - The front curve, \( P_f \), is +8.00D. - The back curve, \( P_b \), is -5.00D. - The index of refraction, \( n \), is 1.530. First, we can add the front and back contributions: \[ P_f + P_b = +8.00D + (-5.00D) = +3.00D. \] Next, we calculate the interaction effect: \[ (n - 1) = 1.530 - 1 = 0.530. \] Now we compute \( \frac{P_f \cdot P_b}{n} \): \[ P_f \

To calculate the total power of a lens, you can use the formula:

[ P = P_f + P_b + (n - 1) \left( \frac{P_f \cdot P_b}{n} \right) ]

Where:

• ( P ) is the total power of the lens,

• ( P_f ) is the front curve power,

• ( P_b ) is the back curve power,

• ( n ) is the index of refraction.

In this case:

• The front curve, ( P_f ), is +8.00D.

• The back curve, ( P_b ), is -5.00D.

• The index of refraction, ( n ), is 1.530.

First, we can add the front and back contributions:

[ P_f + P_b = +8.00D + (-5.00D) = +3.00D. ]

Next, we calculate the interaction effect:

[ (n - 1) = 1.530 - 1 = 0.530. ]

Now we compute ( \frac{P_f \cdot P_b}{n} ):

[ P_f \