Question

The contact lens prescription for a nearsighted person is -4.00 D and the person has a far point of 22.5 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?

Short answer

The power of the tear layer is -0.444 D.

Step-by-step solution

How do contact lenses work?

Contact lenses operate by bending light to bring it into focus on the retina, where vision receptors collect data and send it to the brain.

Given information and Formula to be used

$P_{Lens}$= -4.00 D

The power of the lens is equal to the reciprocal of the focal length is :

$$\begin{gathered} P=\frac{1}{f} \\ P=\frac{1}{d_0}+\frac{1}{d_i} \end{gathered}$$………….(1)

What is the power of the tear layer?

Rearrange the distance between the far point and the person's eyes and solve for it:

Equation (1) is obtained by substituting data values.

$$\begin{gathered} P=\frac{1}{d_0}+\frac{1}{d_i} \\ =\frac{1}{0.02m}+\frac{1}{0.225m} \\ =54.444D \end{gathered}$$

The regular eye's power is 50 D, as we all know. We use the following relationship to determine the normal eye's power:

$$\begin{gathered} P_{Normal}=P_{Tear}+P_{Calc}+P_{Lens} \\ {P}_{Tear}=P_{Normal}-P_{Calc}-P_{Lens} \\ {P}_{Tear}=50D-54.444D-\left(-4D\right) \\ {P}_{Tear}=-0.444D \end{gathered}$$

Therefore, the power of the tear layer is -0.444 D.