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Chapter 26: Problem 77

A concave lens has a focal length of \(-32 \mathrm{cm}\). Find the image distance and magnification that result when an object is placed \(29 \mathrm{cm}\) in front of the lens.

Short Answer

Expert verified
The image distance is approximately \(310\, \text{cm}\) and the magnification is \(-10.69\).

Step by step solution

01

Identify Lens Formula

The lens formula relates the object distance \(u\), the image distance \(v\), and the focal length \(f\) of a lens. It is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]Here, \(f = -32\, \text{cm}\) and \(u = -29\, \text{cm}\) (object distance is usually taken as negative in lens formula).
02

Substitute Values into Lens Formula

Substitute the given values into the lens formula:\[\frac{1}{-32} = \frac{1}{v} + \frac{1}{-29} \]Now, solve for \(\frac{1}{v}\).
03

Solve for Image Distance \(v\)

First, rearrange the formula to solve for \(\frac{1}{v}\):\[\frac{1}{v} = \frac{1}{-32} - \frac{1}{-29} \]Calculate each fraction:- \(\frac{1}{-32} = -0.03125\)- \(\frac{1}{-29} = -0.03448\) (approx.)Subtract:\[ -0.03125 + 0.03448 = 0.00323 \]Therefore, \( \frac{1}{v} = 0.00323 \), so \( v \approx \frac{1}{0.00323} \approx 310\, \text{cm} \).
04

Calculate Magnification

Magnification \( m \) is calculated using the formula:\[ m = \frac{v}{u} \]Substitute the values:\[ m = \frac{310}{-29} \approx -10.69 \]Thus, the magnification is approximately \(-10.69\).

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Key Concepts

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

Understanding Concave Lens
A concave lens, often called a diverging lens, is a type of lens that has a thinner center compared to its edges. This unique shape causes light rays that pass through it to spread out or diverge. Unlike converging lenses, which bring light rays to a focal point, concave lenses do the opposite.
When parallel light rays hit a concave lens, they are refracted in such a manner that they appear to be diverging from a focal point on the lens's front side. Because of this property, concave lenses are used in situations where light needs to be spread over a broader area, such as in eyeglasses for people who are near-sighted. It鈥檚 crucial to remember that concave lenses always form virtual images, which means the images cannot be projected onto a screen as they appear to be behind the lens.
Exploring the Lens Formula
The lens formula is a simple mathematical equation that connects the focal length of a lens with the distances of the object and the image from the lens. It is written as: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where:
  • \( f \) is the focal length, and for concave lenses, it is negative.
  • \( v \) is the image distance.
  • \( u \) is the object distance, usually taken as negative according to convention.
This formula is pivotal in optics as it helps determine where an image will form based on a given object's position and the lens鈥檚 focal length. Solving problems using the lens formula involves substituting known values and solving for the unknown variable. Remember to account for sign conventions since they dictate the direction of the measurements involved.
Decoding Focal Length
Focal length plays a critical role in understanding lenses. It is the distance between the lens and its focal point, where light rays appear to meet (or appear to diverge from, in the case of concave lenses). The focal length is an essential parameter that influences how lenses bend light. For concave lenses, the focal length is always negative.
In practical terms, the focal length informs us about how powerful the lens is in terms of converging or diverging light. A longer focal length implies less bending, while a shorter one means a stronger optical power. For this exercise, knowing the focal length allows us to predict where an image will be formed. When the object is positioned closer to the lens than its focal length, the concave lens forms a smaller, upright, and virtual image on the same side as the object.

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Most popular questions from this chapter

The focal length of a lens is inversely proportional to the quantity \((n-1),\) where \(n\) is the index of refraction of the lens material. The value of \(n,\) however, depends on the wavelength of the light that passes through the lens. For example, one type of flint glass has an index of refraction of \(n_{\mathrm{r}}=1.572\) for red light and \(n_{\mathrm{v}}=1.605\) in violet light. Now, suppose a white object is placed \(24.00 \mathrm{cm}\) in front of a lens made from this type of glass. If the red light reflected from this object produces a sharp image \(55.00 \mathrm{cm}\) from the lens, where will the violet image be found?

A submerged scuba diver looks up toward the calm surface of a freshwater lake and notes that the Sun appears to be \(35^{\circ}\) from the vertical. The diver's friend is standing on the shore of the lake. At what angle above the horizon does the friend see the sun?

(a) Imagine for a moment that the Earth has no atmosphere. Over the period of a year, is the number of daylight hours at your home greater than, less than, or equal to the number of nighttime hours? (b) Repeat part (a). only this time take into account the Earth's atmosphere.

While studying physics at the library late one night, you notice the image of the desk lamp reflected from the varnished tabletop. When you turn your Polaroid sunglasses sideways, the reflected image disappears. If this occurs when the angle between the incident and reflecsted rays is \(110^{\circ},\) what is the index of refraction of the varnish?

A light beam undergoes total internal reflection at the interface between medium \(A,\) in which it propagates, and medium \(\mathrm{B}\), on the other side of the interface. Which medium has the greater index of refraction? Explain.
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