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Chapter 2: Problem 80

A very myopic man has a far point of \(20.0 \mathrm{cm}\). What power contact lens (when on the eye) will correct his distant vision?

Short Answer

Expert verified
The power of the contact lens required to correct the man's distant vision is \(10.0 \mathrm{D}\).

Step by step solution

01

Understand the lens formula and the power of a lens

The lens formula is given by \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), where \(f\) is the focal length of the lens, \(u\) is the object distance, and \(v\) is the image distance. The power of a lens, denoted by \(P\), is defined as the reciprocal of its focal length, i.e., \(P = \frac{1}{f}\). The power is measured in diopters (D).
02

Identify the given values

The far point of the man's vision is given to be \(20.0 \mathrm{cm}\). This means that he can see objects clearly up to a distance of \(20.0 \mathrm{cm}\). Hence, when the contact lens is placed on his eye, the object distance (\(u\)) will be equal to his far point: \(u = 20.0 \mathrm{cm}\).
03

Correcting the distant vision

We want to correct the man's distant vision so that he can see objects at infinity clearly. In this case, the image distance (\(v\)) will be at his far point, which is \(v = 20.0 \mathrm{cm}\).
04

Find the required focal length of the contact lens

Using the lens formula, we can find the focal length (\(f\)) of the contact lens required to correct the man's distant vision: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) Substituting the values of \(u\) and \(v\): \(\frac{1}{f} = \frac{1}{20.0} + \frac{1}{20.0}\) To find \(f\), we solve for \(\frac{1}{f}\): \(\frac{1}{f} = \frac{2}{20.0}\) Now, we find the focal length: \(f = \frac{20.0}{2} = 10.0 \mathrm{cm}\)
05

Calculate the power of the contact lens

Using the power formula \(P = \frac{1}{f}\), we can now find the power of the contact lens required to correct the man's distant vision: \(P = \frac{1}{f}\) Substituting the value of \(f\): \(P = \frac{1}{10.0}\) Converting the focal length from centimeters to meters (since the power is measured in diopters): \(P = \frac{1}{0.10}\) Now, we find the power: \(P = 10.0 \mathrm{D}\) So, the power of the contact lens required to correct the man's distant vision is \(10.0 \mathrm{D}\).

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Most popular questions from this chapter

Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR radiation follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of \(50.0 \mathrm{cm}\) and produces an image of the coils \(3.00 \mathrm{m}\) away from the mirror, where are the coils?

An object of height 3 \(\mathrm{cm}\) is placed at \(25 \mathrm{cm}\) in front of a converging lens of focal length \(20 \mathrm{cm} .\) Behind the lens there is a concave mirror of focal length \(20 \mathrm{cm} .\) The distance between the lens and the mirror is \(5 \mathrm{cm}\). Find the location, orientation and size of the final image.

Can you photograph a virtual image?

An object is located in water \(30 \mathrm{cm}\) from the vertex of a convex surface made of Plexiglas with a radius of curvature of \(80 \mathrm{cm} .\) Where does the image form by refraction and what is its magnification? \(n_{\text {water }}=4 / 3\) and \(n_{\text {Plexiglas }}=1.65\)

The power for normal close vision is 54.0 D. In a vision-correction procedure, the power of a patient's eye is increased by 3.00 D. Assuming that this produces normal close vision, what was the patient's near point before the procedure?
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