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Q10PE

Expert-verifiedFound in: Page 956

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Unless otherwise stated, the lens-to-retina distance is 2.00 cm****.**

**What was the previous far point of a patient who had laser vision correction that reduced the power of her eye by 7.00 D****, producing normal distant vision for her?**

The previous far point of a patient who had laser vision correction was ${d}_{\circ}=0.143m$.

**The thin lens equations can be used to examine the image formation by the eye quantitatively. The power of the lens is equal to the reciprocal of the focal length,**

**${\mathit{P}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{f}}{\mathbf{=}}{\left(\frac{1}{{d}_{\circ}}+\frac{1}{{d}_{i}}\right)}$……………(1)**

**Where d _{0} and d_{i} is the distance between the object and eye as well as the distance from the lens to the retina.**

The power of the patient’s eyes is P = 7.00 D.

The lens-to-retina distance is, ${d}_{i}=2.00cm\left(\frac{1m}{100cm}\right)=0.02m$.

It is known that for the normal person, the power for the distant vision from which the object is at infinity is P = 50 D.

The power of the lens is equal to the reciprocal of the focal length, which is represented as,

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

$\frac{1}{{d}_{0}}=P-\frac{1}{{d}_{i}}\phantom{\rule{0ex}{0ex}}=\frac{1}{P-\frac{1}{{d}_{i}}}\phantom{\rule{0ex}{0ex}}=\left(50D+7D\right)-\frac{1}{0.02m}\phantom{\rule{0ex}{0ex}}=7D$

This will give the value of ${d}_{\circ}=0.143m$

Therefore, the value for the distance is obtained as ${d}_{\circ}=0.143m$.

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