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(I) (a) What is the power of a 23.5 -cm-focal-length lens? (b) What is the focal length of a -6.75 -D lens? Are these lenses converging or diverging?

Short Answer

Expert verified
(a) Power is 4.26 D (converging lens). (b) Focal length is -0.148 m (diverging lens).

Step by step solution

01

Understanding Lens Power

The power of a lens, denoted by \( P \), is given by the equation \( P = \frac{1}{f} \), where \( f \) is the focal length of the lens in meters. Here, the focal length of the given lens is 23.5 cm, which needs to be converted into meters for use in the formula.
02

Convert Focal Length to Meters

Convert 23.5 cm to meters by dividing by 100. Thus, \( f = \frac{23.5}{100} = 0.235 \) meters.
03

Calculate Power of the Lens

Use the formula \( P = \frac{1}{f} \) to find the power of the lens. Substitute \( f = 0.235 \) meters into the equation: \( P = \frac{1}{0.235} \approx 4.26 \) diopters. Since the power is positive, this indicates a converging lens.
04

Understanding Diopter Power and Focal Length

The power of a lens is also given in diopters (D), where \( P = \frac{1}{f} \). For part (b), we need to find the focal length given the power \( P = -6.75 \) D.
05

Calculate Focal Length from Diopters

Rearrange the formula from \( P = \frac{1}{f} \) to \( f = \frac{1}{P} \) to solve for the focal length. Substitute \( P = -6.75 \): \( f = \frac{1}{-6.75} \approx -0.148 \) meters.
06

Determine Lens Type

Since the power for part (b) is negative, the lens is a diverging lens. In general, a positive power indicates a converging lens, while a negative power indicates a diverging lens.

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Key Concepts

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

Focal Length
The focal length of a lens is an important concept in optics. It is defined as the distance between the center of the lens and the point where it converges or diverges light. This point is called the focal point. The focal length is measured in meters and plays a crucial role in determining the "power" of the lens.
For a given lens, focal length affects how much it bends light. A shorter focal length means light converges more quickly, while a longer one allows light to converge more gently. The formula for calculating the power of a lens using its focal length is:
  • Power (P) of a lens is given by the equation: \[ P = \frac{1}{f} \]
where \( f \) is the focal length in meters.
Depending on whether the focal length is positive or negative, the nature of the lens can be determined. Positive focal lengths generally signify a converging lens, while negative focal lengths indicate a diverging lens.
Converging Lens
Converging lenses are often referred to as "positive lenses" due to their positive power. They are designed to bring parallel rays of light to a single focal point. This characteristic is useful in applications like magnifying glasses and cameras. Converging lenses usually have a thicker middle and thinner edges.
  • The power of a converging lens is positive.
  • These lenses are used to correct farsightedness.
  • They cause parallel light rays to meet or "converge" at the lens’s focal point.
When the power of a lens is positive, as in our example with a power of 4.26 diopters, it signifies a converging lens with a focal length calculated as positive. This results in the lens gathering light to create an image at the focal point beyond the lens.
Diverging Lens
Unlike converging lenses, diverging lenses scatter light rays apart, making them appear to diverge from a common focal point on the opposite side of the lens. These lenses are also called "negative lenses" because they have a negative power value. They are often thinner in the middle and thicker at the edges.
  • Diverging lenses are identified by their negative power.
  • They correct nearsightedness by spreading out light before it reaches the eye.
  • Parallel rays of light begin to spread out as they pass through the diverging lens, emerging on the other side as if coming from the focal point.
The lens with a power of -6.75 diopters in our example is a clear case of a diverging lens, as indicated by its negative power. This lens effectively redirects incoming light rays outward, making it appear as if they are originating from a point in front of the lens.

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Most popular questions from this chapter

A woman can see clearly with her right eye only when objects are between \(45 \mathrm{~cm}\) and \(155 \mathrm{~cm}\) away. Prescription bifocals should have what powers so that she can see distant objects clearly (upper part) and be able to read a book \(25 \mathrm{~cm}\) away (lower part) with her right eye? Assume that the glasses will be \(2.0 \mathrm{~cm}\) from the eye.

(III) Given two 12 -cm-focal-length lenses, you attempt to make a crude microscope using them. While holding these lenses a distance \(55 \mathrm{~cm}\) apart, you position your microscope so that its objective lens is distance \(d_{0}\) from a small object. Assume your eye's near point \(N=25 \mathrm{~cm}\). (a) For your microscope to function properly, what should \(d_{\mathrm{o}}\) be? (b) Assuming your eye is relaxed when using it, what magnification \(M\) does your microscope achieve? (c) Since the length of your microscope is not much greater than the focal lengths of its lenses, the approximation \(M \approx N \ell / f_{\mathrm{c}} f_{\mathrm{o}}\) is not valid. If you apply this approximation to your microscope, what \% error do you make in your microscope's true magnification?

A \(200-\mathrm{mm}\) -focal-length lens can be adjusted so that it is \(200.0 \mathrm{~mm}\) to \(206.4 \mathrm{~mm}\) from the film. For what range of object distances can it be adjusted?

(I) A \(680 \times\) microscope uses a \(0.40-\mathrm{cm}\) -focal-length objective lens. If the barrel length is \(17.5 \mathrm{~cm},\) what is the focal length of the eyepiece? Assume a normal eye and that the final image is at infinity.

What is the magnifying power of a \(+4.0-\mathrm{D}\) lens used as a magnifier? Assume a relaxed normal eve.

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