Assuming an index of refraction of 1.530, what is the back curve if the total power of the lens is -3.50D and the front curve is +4.00D?

To determine the back curve of a lens, we use the lens maker's formula, which relates the powers of the front and back curves to the total power of the lens and its index of refraction. The formula is given by:

[ P_t = P_f + P_b \times (n - 1) ]

Where:

• ( P_t ) 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.

Rearranging this formula to solve for the back curve power gives us:

[ P_b = \frac{P_t - P_f}{n - 1} ]

In this scenario, we have:

• Index of refraction ( n = 1.530 )

• Total power ( P_t = -3.50D )

• Front curve ( P_f = +4.00D )

Substituting the values into the rearranged formula requires first computing ( n - 1 ):

[ n - 1 = 1.530 - 1 = 0.530 ]

Next, plug the values into the equation