# 1. Calculate the power of the eye when viewing an object a distance 0.495 m in front of the eye....

## Question:

1. Calculate the power of the eye when viewing an object a distance {eq}0.495 \, \mathrm{m} {/eq} in front of the eye. Take the lens-to-retina distance to be {eq}2.00 \, \mathrm{cm} {/eq}.

2. A very myopic man has a far point of {eq}34.5 \, \mathrm{cm} {/eq}. What power contact lens (when on the eye) will correct his distant vision?

## Vision and Imaging

The power of a lens is related to its focal length by the following equation:

{eq}\begin{equation*} P = \frac{1}{f} \end{equation*} {/eq}

This power can then be related to the object distance and image distance using the imaging formula:

{eq}\begin{equation*} P = \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \end{equation*} {/eq}

It should be noted that the far point of human eye is defined as the farthest a person can see a clear image. Thus, one uses the imaging equation, allowing for an object at infinity, to find what power is needed to correct this vision:

{eq}\begin{equation*} P = \frac{1}{f} = \frac{1}{\infty} + \frac{1}{d_i} \end{equation*} {/eq}

1. To calculate the power of the lens, we use:

{eq}\begin{align*} P &= \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_I} \\ \\ &= \frac{1}{0.495} +...

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