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Measurements on the cornea of a person's eye reveal that the magnitude of the front surface radius of curvature is \(7.80 \mathrm{~mm},\) while the magnitude of the rear surface radius of curvature is \(7.30 \mathrm{~mm}\) (see Figure 25.18 ), and that the index of refraction of the cornea is 1.38 . If the cornea were simply a thin lens in air, what would be its focal length and its power in diopters? What type of lens would it be?

Short Answer

Expert verified
Focal length is 0.9917 cm, power is 100.83 diopters, and it's a converging lens.

Step by step solution

01

Identify the Formula

To find the focal length of a thin lens, we use the lens maker's formula: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]where \(f\) is the focal length, \(n\) is the index of refraction, \(R_1\) is the radius of curvature of the front surface, and \(R_2\) is the radius of curvature of the back surface.
02

Plug in the Values

Substitute the given values into the lens maker's formula. Here, \(n = 1.38\), \(R_1 = 7.80\, \text{mm} = 0.780\, \text{cm}\), and \(R_2 = -7.30\, \text{mm} = -0.730\, \text{cm}\). The negative sign for \(R_2\) indicates the direction of curvature opposite to R1.\[ \frac{1}{f} = (1.38 - 1) \left( \frac{1}{0.780} - \frac{1}{-0.730} \right) \]
03

Simplify the Equation

Simplify the equation step by step. Calculate the values in the parenthesis:1. \( \frac{1}{0.780} \approx 1.282 \)2. \( \frac{1}{-0.730} \approx -1.370 \)Then plug these into the equation:\[ \frac{1}{f} = 0.38 \times (1.282 + 1.370) \]\[ \frac{1}{f} = 0.38 \times 2.652 = 1.00836 \]
04

Calculate Focal Length

Calculate the focal length \(f\) from the reciprocal of \(1/f\).\[ f = \frac{1}{1.00836} \approx 0.9917\, \text{cm} \]
05

Find the Power of the Lens in Diopters

The power \(P\) of a lens in diopters is the reciprocal of its focal length in meters.First, convert the focal length from centimeters to meters:\[ 0.9917\, \text{cm} = 0.009917\, \text{m} \]Then compute the power:\[ P = \frac{1}{0.009917} \approx 100.83\, \text{diopters} \]
06

Determine the Type of Lens

Since the power of the lens is positive, it indicates that the lens is converging.
07

Final Answer

The focal length of the lens is approximately \(0.9917\, \text{cm}\) (or \(9.917\, \text{mm}\)), its power is \(100.83\, \text{diopters}\), and it is a converging lens.

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Key Concepts

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

Lens Maker's Formula
The lens maker's formula is an essential tool in optics used to determine the focal length of a lens. This formula considers the physical properties of the lens — its curvature and material — to calculate how it focuses light. Mathematically, it's expressed as: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]where:- \(f\) is the focal length,- \(n\) represents the index of refraction,- \(R_1\) is the radius of curvature of the lens's front surface,- \(R_2\) is the radius of curvature of the back surface.In practice, this formula allows us to predict how a lens will function based on its shape and the material from which it is made. For example, in the human eye, understanding these properties helps optometrists design corrective lenses.
Focal Length
Focal length is the distance over which parallel light rays are brought to focus. In the context of lenses, it plays a critical role in determining how strongly the lens converges (focuses) or diverges (spreads out) light. - **Convergence and Divergence:** - A positive focal length means the lens is converging, great for forming images. - A negative focal length implies a diverging lens, which spreads light outwards. - **Unit Conversion:** - Typically measured in centimeters (cm) or meters (m), but often interpreted in millimeters (mm) for precision in small optical devices. Understanding focal length is key to designing and understanding imaging systems, whether in photography, microscopy, or vision correction.
Lens Power
The power of a lens in optics is a measure of its ability to converge or diverge light, expressed in diopters. Diopter is a unit of lens power, defined as the reciprocal of the focal length in meters:\[ P = \frac{1}{f} \]- **Positive Power:** - Indicates a converging lens, which means it can aid in correcting hyperopia (farsightedness).- **Negative Power:** - Suggests a diverging lens, often used to address myopia (nearsightedness).This measurement is essential for prescriptions in eyewear, providing the necessary corrective power to compensate for vision impairments.
Index of Refraction
The index of refraction (or refractive index) quantifies how much light slows down when passing through a material compared to vacuum. It fundamentally affects how a lens bends light, determined by:\[ n = \frac{c}{v} \]where:- \(c\) is the speed of light in a vacuum,- \(v\) is the speed of light in the material.- **Higher Index:** - Means light slows down more, leading to greater bending of the light rays.- **Importance in Lenses:** - Crucial for lens design, affecting both focal length and power. - Commonly tested when designing eyewear and optical instruments, ensuring lenses meet specific light-bending requirements.A deeper understanding of the index helps in selecting the right materials for lenses to achieve desired optical properties.

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Most popular questions from this chapter

A frog can see an insect clearly at a distance of \(10 \mathrm{~cm}\). At that point the effective distance from the lens to the retina is \(8 \mathrm{~mm} .\) If the insect moves \(5 \mathrm{~cm}\) farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? A. \(0.02 \mathrm{~cm},\) toward the retina B. \(0.02 \mathrm{~cm}\), away from the retina C. \(0.06 \mathrm{~cm}\), toward the retina D. \(0.06 \mathrm{~cm}\), away from the retina

Given that frogs are nearsighted in air, what is most likely to be true about their vision in water? A. They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. B. They are less nearsighted because the cornea is less effective at refracting light in water than in air. C. Their vision is no different because only structures that are internal to the eye can affect the eye's ability to focus. D. The images projected on the retina are no longer inverted because the eye in water functions as a diverging lens rather than a converging lens.

A telescope is constructed from two lenses with focal lengths of \(95.0 \mathrm{~cm}\) and \(15.0 \mathrm{~cm},\) the \(95.0-\mathrm{cm}\) lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification of the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall and \(3.00 \mathrm{~km}\) away. \((\mathrm{c}) \mathrm{What}\) is the angular size of the final image as viewed by an eye very close to the eyepiece?

A thin lens with a focal length of \(6.00 \mathrm{~cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at infinity and that the lens is very close to the eye.

A camera is focused on an object that is \(1.2 \mathrm{~m}\) from the lens. The distance between the CCD image sensor and the lens is \(5 \mathrm{~cm}\). What is the focal length of the lens?

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