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If the lens of a person’s eye is removed because of cataracts (as has been done since ancient times), why would you expect a spectacle lens of about \({\rm{16 D}}\) to be prescribed?

Short Answer

Expert verified

The lens of the eyes contributes one-third of the eye's refractive power, hence usually a lens of high refractive power \(\left( {{\rm{16 D}}} \right)\) is prescribed to compensate for the loss of the refractive power of the eye due to the removal of the lens.

Step by step solution

01

Concept Introduction

The bending of a wave as it travels from one medium to another is known as refraction. The difference in optical density between the two substances causes bending.

02

Explanation for the lens

Two-thirds of the refractive power of the eye comes from the cornea, approximately the other third comes from the lens. The total refractive power of the eye, assuming an axial eye length (distance between the corneal surface to the retina)\({\rm{2}}{\rm{.35 cm}}\),is\({\rm{63}}\)dioptre. Hence, on average, the lens contributes about\({\rm{23}}\)dioptre.

Also, the person of concern might have not a perfect vision before cataract surgery, maybe he was a near-sighted person, (having an eye with a long axial length), thus the person doing cataracts might not exactly need a lens prescription of\({\rm{23}}\)dioptre but instead, a lens with less refractive power is prescribed for him.

However, in general, upon removing the lens as a procedure for cataract surgery it is quite ordinary for a lens prescription of a high dioptre to be prescribed in order to compensate for the refractive power loss due to the removal of the lens.

03

Calculation for the lens

According to the human anatomy - The distance between the lens and the retina is roughly \({\rm{6}}{\rm{.25 cm}}\). If the focal length of a lens is calculated with the power of \({\rm{16 D}}\), it can be obtained as,

\(\begin{aligned} f &= \frac{{{\rm{100}}}}{{{\rm{16}}\;{\rm{D}}}}\\ &= {\rm{6}}{\rm{.25 cm}}\end{aligned}\).

This proves that the lens with the power \({\rm{16 D}}\) exactly matches our anatomy.

Therefore, due to the distance between the lens and sight receptors the lens \({\rm{16 D}}\) is prescribed.

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Most popular questions from this chapter

An amoeba is 0.305 cm away from the 0.300 cm focal length objective lens of a microscope. (a) Where is the image formed by the objective lens? (b) What is this image’s magnification? (c) An eyepiece with a 2.00 cm focal length is placed 20.0 cm from the objective. Where is the final image? (d) What magnification is produced by the eyepiece? (e) What is the overall magnification? (See Figure 26.16.)

You are using a standard microscope with an \({\rm{0}}{\rm{.10NA 4 \times }}\) objective and switch to a \({\rm{0}}{\rm{.65NA 40 \times }}\) objective. What are the acceptance angles for each? Compare and comment on the values. Which would you use first to locate the target area on of your specimen? (See Figure\({\rm{26}}{\rm{.17}}\).)

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Find the distance between the objective and eyepiece lenses in the telescope in the above problem needed to produce a final image very far from the observer, where vision is most relaxed. Note that a telescope is normally used to view very distant objects.

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