/*! 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 137 A cheese enchilada is \(4.00 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A cheese enchilada is \(4.00 \mathrm{~cm}\) in front of a converging lens. The magnification of the enchilada is \(-2.00 .\) What is the focal length of the lens?

Short Answer

Expert verified
The focal length of the lens is \(2.67\, \mathrm{cm}\).

Step by step solution

01

Recognize Given Information

We know the object distance \(d_o\) is \(4.00 \, \mathrm{cm}\) and magnification \(m\) is \(-2.00\). The goal is to find the focal length \(f\).
02

Understand Magnification Formula

The magnification formula for a lens is \(m = -\frac{d_i}{d_o}\), where \(d_i\) is the image distance. Plug in \(m = -2.00\) and \(d_o = 4.00\, \mathrm{cm}\) to find \(d_i\).
03

Find Image Distance \(d_i\)

Rearrange the magnification formula to solve for \(d_i\): \(-2.00 = -\frac{d_i}{4.00}\). Thus, \(d_i = 8.00\, \mathrm{cm}\).
04

Use the Lens Formula

The lens formula is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Substitute \(d_o = 4.00\, \mathrm{cm}\) and \(d_i = 8.00\, \mathrm{cm}\) into the formula.
05

Calculate Focal Length \(f\)

\[\frac{1}{f} = \frac{1}{4.00} + \frac{1}{8.00} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}\]Thus, \(f = \frac{8}{3} = 2.67\, \mathrm{cm}\).

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Key Concepts

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

Converging Lens
A converging lens, also known as a convex lens, is one that bends light rays inward. To visualize this, imagine light coming from a distant object. As it passes through the lens, the light converges at a point on the opposite side. This point is called the focal point of the lens. Converging lenses are commonly used in devices like cameras, eyeglasses, and microscopes to focus light and form clear images.

The focal length of a lens is the distance from the center of the lens to its focal point. For a converging lens, the focal length is positive, indicating that the light converges on the opposite side of the light source. Understanding how a converging lens works is crucial when studying optics and knowing how to measure and calculate focal lengths. The exercise discussed here requires finding the focal length given the object distance and magnification.
Magnification Formula
The magnification formula provides a straightforward way to relate the size of an object to the size of its image, as well as the positions of these elements relative to the lens. The formula is given by:
  • \( m = -\frac{d_i}{d_o} \)
Here, \( m \) represents magnification, \( d_i \) is the image distance (the distance from the lens to where the image is formed), and \( d_o \) is the object distance (the distance from the lens to the object). The negative sign in the formula indicates that if the image is on the opposite side of the object concerning the lens, it will be inverted.

In our original exercise, we know the magnification \( m \) is \(-2.00\), and the object distance \( d_o \) is \( 4.00\, \mathrm{cm} \). By solving the equation, we calculated that the image distance \( d_i \) is \( 8.00\, \mathrm{cm} \). This shows that the image is twice the size of the object but inverted due to the negative magnification value.
Lens Formula
The lens formula connects object distance, image distance, and focal length in a single equation. This formula is essential for finding the focal length of a lens, especially in lenses used for educational experiments or optical devices. The formula is given as:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
Where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. When you have the information for any two of these distances, you can calculate the third

In the given exercise, the values provided were \( d_o = 4.00\, \mathrm{cm} \) and \( d_i = 8.00\, \mathrm{cm} \). By substituting these into the lens formula, we get:\[ \frac{1}{f} = \frac{1}{4.00} + \frac{1}{8.00} = \frac{3}{8} \]Thus, the focal length \( f \) is calculated as \( 2.67\, \mathrm{cm} \). This positive focal length confirms the lens is converging, as expected in such lens systems.

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Most popular questions from this chapter

A double-convex lens is to be made of glass with an index of refraction of \(1.5 .\) One surface is to have twice the radius of curvature of the other and the focal length is to be \(60 \mathrm{~mm}\). What is the (a) smaller and (b) larger radius?

You look through a camera toward an image of a hummingbird in a plane mirror. The camera is \(4.30 \mathrm{~m}\) in front of the mirror. The bird is at camera level, \(5.00 \mathrm{~m}\) to your right and \(3.30 \mathrm{~m}\) from the mirror. What is the distance between the camera and the apparent position of the bird's image in the mirror?

(a) A luminous point is moving at speed \(v_{O}\) toward a spherical mirror with radius of curvature \(r\), along the central axis of the mirror. Show that the image of this point is moving at speed \(v_{I}=-\left(\frac{r}{2 p-r}\right)^{2} v_{O}\) where \(p\) is the distance of the luminous point from the mirror at any given time. Now assume the mirror is concave, with \(r=15 \mathrm{~cm}\), and let \(v_{O}=5.0 \mathrm{~cm} / \mathrm{s} .\) Find \(v_{I}\) when \((\mathrm{b}) p=30 \mathrm{~cm}\) (far outside the focal point \(),(\mathrm{c}) p=8.0 \mathrm{~cm}\) (just outside the focal point), and (d) \(p=10 \mathrm{~mm}\) (very near the mirror).

A peanut is placed \(40 \mathrm{~cm}\) in front of a two-lens system: lens 1 (nearer the peanut) has focal length \(f_{1}=+20 \mathrm{~cm},\) lens 2 has \(f_{2}=-15 \mathrm{~cm},\) and the lens separation is \(d=10 \mathrm{~cm} .\) For the image produced by lens \(2,\) what are (a) the image distance \(i_{2}\) (including sign), (b) the image orientation (inverted relative to the peanut or not inverted), and (c) the image type (real or virtual)? (d) What is the net lateral magnification?

Object \(O\) stands on the central axis of a spherical or plane mirror. For this situation, each problem in Table \(34-4\) refers to (a) the type of mirror, (b) the focal distance \(f,\) (c) the radius of curvature \(r\), (d) the object distance \(p,\) (e) the image distance \(i\), and (f) the lateral magnification \(m\). (All distances are in centimeters.) It also refers to whether \((\mathrm{g})\) the image is real \((\mathrm{R})\) or virtual (V), (h) inverted (I) or noninverted (NI) from \(O\), and (i) on the same side of the mirror as object \(O\) or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

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