Problem 137 - Fundamentals of Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \(f\).

Step 2: 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 \(d_i\).

Step 3: Find Image Distance \(d_i\)

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

Step 4: 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.

Step 5: 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, \(f = \frac{8}{3} = 2.67\, \mathrm{cm}\).

Key Concepts

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.