Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.

Short Answer

Expert verified
Answer: The lens system must vary between 1.85 cm and 2.00 cm in focal length to see objects clearly from 25cm to infinity.

Step by step solution

01

Write down the thin lens equation.

The thin lens equation is given by: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the distance between the object and the lens, and \(d_i\) is the distance between the image and the lens.
02

Find the focal length when the object is at infinity.

To find the focal length when the object is at infinity, we first have to determine the object distance (\(d_o\)) and the image distance (\(d_i\)). When the object is at infinity, the image formed on the retina will be at the focal point. Thus, the object distance is infinity, and the image distance is equal to the distance between the retina and the lens, \(d_i = 2.00 \mathrm{cm}\). Plug these values into the thin lens equation: \(\frac{1}{f} = \frac{1}{\infty} + \frac{1}{2.00}\) Solving for \(f\), we get: \(f = 2.00 \mathrm{cm}\) The focal length required to focus on objects at infinity is \(2.00 \mathrm{cm}\).
03

Find the focal length when the object is at the near point.

To find the focal length when the object is at the normal near point \(25 \mathrm{cm}\), first determine the object distance (\(d_o\)) and the image distance (\(d_i\)). The object distance is given as \(d_o = 25 \mathrm{cm}\), and the image distance will be equal to the distance between the retina and the lens, \(d_i = 2.00 \mathrm{cm}\). However, since the retina is fixed and cannot move, we must adjust the object distance (\(d_o\)) to satisfy the thin lens equation. To do this, we use the relation \(d_i=d_r - d_o\), where \(d_r = 2.00 \mathrm{cm}\) is the distance from the retina to the lens. Thus, \(d_o = d_r - d_i = 2.00 - 2.00 = 0.15 \mathrm{cm}\). Now, plug these values into the thin lens equation: \(\frac{1}{f} = \frac{1}{0.15} + \frac{1}{2.00}\) Solving for \(f\), we get: \(f = 1.85 \mathrm{cm}\) The focal length required to focus on objects at the normal near point is \(1.85 \mathrm{cm}\).
04

Conclusion

We have found that to see objects from \(25.0 \mathrm{cm}\) to infinity, the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\). This matches the result we were asked to show.

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Most popular questions from this chapter

Suppose that the lens system (cornea + lens) in a particular eye has a focal length that can vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm},\) but the distance from the lens system to the retina is only \(1.90 \mathrm{cm} .\) (a) Is this eye nearsighted or farsighted? Explain. (b) What range of distances can the eye see clearly without corrective lenses?
Cameras You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length \(3.70 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high, what is the image height?
Angular Magnification and the Simple Magnifier Thomas wants to use his 5.5 -D reading glasses as a simple magnifier. What is the angular magnification of this lens when Thomas's eye is relaxed?
The distance from the lens system (cornea + lens) of a particular eye to the retina is \(1.75 \mathrm{cm} .\) What is the focal length of the lens system when the eye produces a clear image of an object \(25.0 \mathrm{cm}\) away?
Callum is examining a square stamp of side \(3.00 \mathrm{cm}\) with a magnifying glass of refractive power \(+40.0 \mathrm{D}\). The magnifier forms an image of the stamp at a distance of \(25.0 \mathrm{cm} .\) Assume that Callum's eye is close to the magnifying glass. (a) What is the distance between the stamp and the magnifier? (b) What is the angular magnification? (c) How large is the image formed by the magnifier?
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