/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 When an object is located \(46 \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When an object is located \(46 \mathrm{cm}\) to the left of a lens, the image is formed \(17 \mathrm{cm}\) to the right of the lens. What is the focal length of the lens?

Short Answer

Expert verified
The focal length of the lens is approximately 12.42 cm.

Step by step solution

01

Understand the Lens Formula

The lens formula relates the object distance \(u\), the image distance \(v\), and the focal length \(f\) of the lens. The formula is \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).
02

Identify Given Values

The problem states that the object distance \(u = -46\, \mathrm{cm}\) (negative because it is on the left) and the image distance \(v = 17\, \mathrm{cm}\).
03

Substitute the Values into the Lens Formula

Substituting \(u = -46\) and \(v = 17\) into the lens formula gives:\[\frac{1}{f} = \frac{1}{17} - \frac{1}{-46}\]
04

Calculate Each Term

Calculate \(\frac{1}{v} = \frac{1}{17} = 0.0588\) and \(\frac{1}{u} = \frac{1}{-46} = -0.0217\).
05

Solve for \(\frac{1}{f}\)

Substitute the calculated values into the equation:\[\frac{1}{f} = 0.0588 - (-0.0217) = 0.0588 + 0.0217 = 0.0805\]
06

Calculate the Focal Length \(f\)

Find \(f\) by taking the reciprocal of \(0.0805\):\[ f = \frac{1}{0.0805} \approx 12.42 \mathrm{cm} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Focal Length Calculation
The focal length of a lens is a crucial parameter in optics. It defines how strongly the lens converges or diverges light. To calculate it, we use the lens formula, which connects the focal length (\( f \)) to the object distance (\( u \)) and the image distance (\( v \)). The lens formula is stated as:
  • \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
In this formula:
  • \( f \) is the focal length,
  • \( u \) represents the object distance,
  • \( v \) is the image distance.
By rearranging these terms and substituting known values, we can solve for \( f \), understanding how an object and its image position affect it. This focal length aids in characterizing the power of the lens.
Object Distance
The object distance (\( u \)) is the distance from the object to the lens. In the context of lenses, distances measured against the light's direction of travel (towards the left) are considered negative, while distances in the direction of the light's travel (towards the right) are positive.
When solving problems, it's vital to accurately determine whether the object distance should be positive or negative based on these conventions. In our example, the object is positioned 46 cm to the left of the lens, hence the object distance is \( u = -46 \mathrm{cm} \).
Understanding this concept is crucial in applying the lens formula efficiently, as it affects the calculation of the focal length as well as the determination of image characteristics.
Image Distance
The image distance (\( v \)) is vital in understanding where the image forms in relation to a lens. It is the distance from the lens to the image itself. Conventionally, an image formed on the light's path (right side of the lens) carries a positive sign.
Considerations of positive and negative distances help to determine the direction and positioning of the image, which further affects the focal length calculation. In our example, the distance is described as 17 cm on the right side of the lens, meaning \( v = 17 \mathrm{cm} \).
Such distinctions reinforce correct usage of the lens formula and help predict whether images are real or virtual, and their nature concerning the lens’ curvature.
Optics
Optics, the branch of physics dealing with light, involves understanding how lenses interact with light to form images. It includes reflections, refractions, and the study of lenses and mirrors. Both convergent and divergent lenses operate using the principles of optics to manipulate light paths.
In optics, lenses play a significant role through:
  • Converging light rays to a point, forming either real or virtual images based on the object’s position relative to the focal point.
  • Determining the nature and size of images formed, which are consequential in applications like cameras, glasses, and telescopes.
Hence, mastering optics requires a comprehensive understanding of focal length, object, and image distances to thoroughly analyze and design optical systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use ray diagrams to show whether the image formed by a convex mirror increases or decreases in size as an object is brought closer to the mirror's surface.

Standing \(2.3 \mathrm{m}\) in front of a small vertical mirror, you see the reflection of your belt buckle, which is \(0.72 \mathrm{m}\) below your eyes. (a) What is the vertical location of the mirror relative to the level of your eyes? (b) What angle do your eyes make with the horizontal when you look at the buckle? (c) If you now move backward until you are \(6.0 \mathrm{m}\) from the mirror, will you still see the buckle, or will you see a point on your body that is above or below the buckle? Explain.

(a) Find the two locations where an object can be placed in front of a concave mirror with a radius of curvature of \(39 \mathrm{cm}\) such that its image is twice its size. (b) In each of these cases, state whether the image is real or virtual, upright or inverted.

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?

You view a nearby tree in a concave mirror. The inverted image of the tree is \(3.8 \mathrm{cm}\) high and is located \(7.0 \mathrm{cm}\) in front of the mirror. If the tree is \(23 \mathrm{m}\) from the mirror, what is its height?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.