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 approximately 2.67 cm.

Step 1: Understand the Relationship

The magnification (\(m\)) of a lens is related to the image distance (\(d_i\)) and the object distance (\(d_o\)) by the formula \[ m = -\frac{d_i}{d_o} \]In this problem, we know that \(d_o = 4.00 \, \mathrm{cm}\) and \(m = -2.00\).

Step 2: Solve for the Image Distance

Rearrange the magnification formula to solve for \(d_i\):\[ d_i = -m \cdot d_o \]Substituting the known values gives:\[ d_i = -(-2.00) \cdot 4.00 \, \mathrm{cm} = 8.00 \, \mathrm{cm} \]

Step 3: Apply the Lens Formula

The lens formula relates the object distance (\(d_o\)), the image distance (\(d_i\)), and the focal length (\(f\)) of the lens:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

Step 4: Substitute Known Values

Substitute \(d_o = 4.00 \, \mathrm{cm}\) and \(d_i = 8.00 \, \mathrm{cm}\) into the lens formula:\[ \frac{1}{f} = \frac{1}{4.00} + \frac{1}{8.00} \]

Step 5: Calculate the Focal Length

Simplify the equation: \[ \frac{1}{f} = \frac{1}{4.00} + \frac{1}{8.00} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8} \]Therefore, the focal length is:\[ f = \frac{8}{3} \, \mathrm{cm} \approx 2.67 \, \mathrm{cm} \]

Key Concepts

Lens Formula

The lens formula is a critical concept in optics that helps us understand the relationship between object distance, image distance, and the focal length of a lens. The formula is expressed as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where:

• \(f\) is the focal length of the lens.
• \(d_o\) is the distance from the object to the lens.
• \(d_i\) is the distance from the image to the lens.

This equation assumes that light travels through a thin lens, and both distances (object and image) are measured from the lens. It's key in lens calculations, helping you determine one of the variables if the other two are known. For example, in our exercise, knowing the object and image distance allows us to calculate the focal length.

Magnification

Magnification in optics is a measure of how much larger or smaller an image is compared to the object itself. The formula used to calculate magnification is:\[m = -\frac{d_i}{d_o}\]where:

• \(m\) is the magnification.
• \(d_i\) is the image distance.
• \(d_o\) is the object distance.

The negative sign indicates that if the image is upside down (inverted) relative to the object, the magnification is considered negative. A magnification of -2.00 means the image is twice the size of the object and inverted. This is common with converging lenses when the object is located closer to the lens than its focal length.

Focal Length

Focal length is an essential property of a lens that describes how strongly it converges or diverges light. It is the distance from the center of the lens to the focal point, where parallel rays of light either come together or appear to do so. A converging lens, like the one mentioned in our exercise, has a positive focal length. In this scenario, we found the focal length using the lens formula:\[f = \frac{8}{3} \, \mathrm{cm} \approx 2.67 \, \mathrm{cm}\]This tells us how the lens influences the path of light rays to form an image. A shorter focal length indicates a stronger lens power to bend light rays and bring them to focus. This is significant in applications like cameras and glasses, where precise control over image focus is needed.