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Corrective Lenses. Determine the power of the corrective contact lenses required by (a) a hyperopic eye whose near point is at 60.0 \(\mathrm{cm}\) and \((b)\) a myopic eye whose far point is at \(60.0 \mathrm{cm} .\)

Short Answer

Expert verified
(a) Hyperopic lens power: -1.67 D; (b) Myopic lens power: -1.67 D.

Step by step solution

01

Identify the Eye Condition (Step 1)

Understand the conditions of the eyes. Hyperopia, also known as farsightedness, means difficulty focusing on nearby objects, so you will need lenses to bring the image closer to the eye. On the other hand, myopia, or nearsightedness, is when distant objects seem blurred, requiring corrective lenses to push the image further away.
02

Set Up the Lens Formula for Hyperopia (Step 2)

For hyperopia, use the lens formula: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] Where \( f \) is the focal length of the lens, \( d_o \) is the object distance, and \( d_i \) is the image distance. Here, \( d_o \) is infinity because we want the person to see at infinity without strain, and \( d_i \) is -60 \( \mathrm{cm} \) (negative because images for hyperopic correction form a virtual image behind the lens).
03

Calculate the Focal Length for Hyperopia (Step 3)

Substitute \( d_o = \infty \) and \( d_i = -60 \mathrm{cm} \) into the lens formula:\[\frac{1}{f} = \frac{1}{\infty} + \frac{1}{-60} = \frac{0}{1} - \frac{1}{60}\]\[\frac{1}{f} = -\frac{1}{60}\]Thus, \( f = -60 \mathrm{cm} = -0.60 \mathrm{m} \).
04

Determine Lens Power for Hyperopia (Step 4)

Calculate the power \( P \) of the lens using the formula:\[ P = \frac{1}{f} \]So, the power needed is:\[ P = \frac{1}{-0.60} \approx -1.67 \mathrm{D} \](Since diopters \((\mathrm{D})\) is the unit for lens power and the value is negative indicating a converging lens.)
05

Set Up the Lens Formula for Myopia (Step 5)

For myopia, use the same lens formula: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]This time, \( d_o = -60 \mathrm{cm} \) (negative because a real object at a finite point is viewed as if it is infinitely far), and \( d_i \) should be infinity to push the image to a further distance.
06

Calculate the Focal Length for Myopia (Step 6)

Substitute \( d_o = -60 \mathrm{cm} \) and \( d_i = \infty \) into the lens formula:\[\frac{1}{f} = \frac{1}{-60} + \frac{1}{\infty} = -\frac{1}{60}\]Thus, \( f = -60 \mathrm{cm} = -0.60 \mathrm{m} \) (negative due to real image formation for myopic eyes).
07

Determine Lens Power for Myopia (Step 7)

Calculate the lens power \( P \) for myopia using:\[ P = \frac{1}{f} = \frac{1}{-0.60} \approx -1.67 \mathrm{D} \]This power is also negative, indicating a diverging lens.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperopia Correction
Hyperopia, commonly referred to as farsightedness, is a condition where individuals struggle to see close objects clearly while distant objects remain in focus. This happens because the light entering the eye focuses behind the retina rather than directly on it. To correct hyperopia, lenses are used to help refract the light so it focuses earlier, directly onto the retina.

The corrective lens needed for hyperopia is a converging lens. These lenses are thicker at the center than at the edges and work by bending the light rays towards each other, thereby reducing the eyes' effort to focus on nearby objects. Typically, the lens power is expressed in diopters (D) and is determined using the lens formula.
  • Light focuses behind the retina
  • Converging lens used for correction
  • Powers expressed in diopters
Myopia Correction
Myopia, or nearsightedness, is when individuals find it difficult to see objects at a distance because the light is focused in front of the retina. Myopic eyes require assistance to disperse the light rays, allowing them to focus further back onto the retina.

Corrective lenses for myopia are diverging lenses. These lenses are thinner at the center and thicker at the edges, helping spread the light rays apart so that the focal point moves backward onto the retina. This adjustment allows clearer vision of distant objects. Power of these lenses is also measured in diopters and is calculated using a specific lens formula.
  • Light focuses before the retina
  • Diverging lens used for correction
  • Negative diopters indicate scattering
Focal Length Calculation
To calculate the focal length of a lens, we apply the lens formula: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance.

In hyperopia correction, the object is considered to be at infinity, meaning the image formed should be at or near the retina for a clear view, which typically requires a negative image distance. For myopia correction, however, the correction must make the image appear infinitely far, effectively representing an image distance approaching infinity.
  • Utilize lens formula for calculations
  • Consider object and image distances
  • Focal length relates to lens type used
Lens Formula
The lens formula is fundamental in lens power calculations and helps determine how much refracting power a lens needs to possess to correct vision issues. It's expressed as \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where \(f\) is focal length, \(d_o\) is object distance, and \(d_i\) is image distance.

For corrective lenses:
  • In hyperopia, \(d_o\) approaches infinity while \(d_i\) is negative (virtual image)
  • In myopia, \(d_o\) is negative, reflecting a corrective achievable distance, with \(d_i\) approaching infinity
  • The formula facilitates the calculation of diopters, indicating lens power
This formula allows for solving practical problems in vision correction, establishing the necessary lens shape and power for specific eye conditions.

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Most popular questions from this chapter

(a) Where is the near point of an eye for which a contact lens with a power of \(+2.75\) diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of \(-1.30\) diopters is prescribed for distant vision?

A spherical, concave, shaving mirror has a radius of curvature of \(32.0 \mathrm{cm} .\) (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virual? (c) Draw a principal-ray diagram showing the formation of the image.

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?

A thick-walled wine goblet sitting on a table can be considered to be a hollow glass sphere with an outer radius of 4.00 cm and an inner radius of 3.40 \(\mathrm{cm}\) . The index of refraction of the goblet glass is \(1.50 .(a)\) A beam of parallel light rays enters the side of the empty goblet along a horizontal radius. Where, if anywhere, will an image be formed? ( \(b\) ) The goblet is filled with white wine \((n=1.37)\) . Where is the image formed?

Recall that the intensity of light reaching film in a camera is proportional to the effective area of the lens. Camera \(A\) has a lens with an aperture diameter of 8.00 \(\mathrm{mm}\) . It photographs an object using the correct exposure time of \(\frac{1}{30} \mathrm{s}\) . What exposure time should be used with camera \(B\) in photographing the same object with the same film if this camera has a lens with an aperture diameter of 23.1 \(\mathrm{mm} ?\)

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