Problem 43 - College Physics

Question

(. A converging meniscus lens (see Fig. 24.31\()\) with a refrac- tive index of 1.52 has spherical surfaces whose radii are 7.00 \(\mathrm{cm}\) and 4.00 \(\mathrm{cm} .\) What is the position of the image if an object is placed 24.0 \(\mathrm{cm}\) to the left of the lens? What is the magnification?

Step-by-Step Solution

Answer

The image is 6.46 cm to the right of the lens and the magnification is -0.27.

Step 1: Understand Lens Formula

The lensmaker's formula for calculating the focal length \( f \) of a lens is given by: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( n \) is the refractive index, \( R_1 \) is the radius of the first surface, and \( R_2 \) is the radius of the second surface.

Step 2: Assign Variable Values

Assign the given values to the formula: \( n = 1.52 \), \( R_1 = 7.00 \) cm, and \( R_2 = -4.00 \) cm. Note that \( R_2 \) is negative because the surface is concave to the incoming light.

Step 3: Calculate Focal Length

Substitute the given values into the lensmaker's formula:\[ \frac{1}{f} = (1.52 - 1) \left( \frac{1}{7} - \frac{1}{-4} \right) \] Simplify and solve for \( f \). The calculation yields: \[ \frac{1}{f} = 0.52 \left( \frac{1}{7} + \frac{1}{4} \right) = 0.52 \left( \frac{4 + 7}{28} \right) = 0.52 \times \frac{11}{28} \approx 0.2043 \] Therefore, \( f \approx 4.89 \) cm.

Step 4: Use Lens Formula for Image Position

With the focal length \( f \approx 4.89 \) cm, use the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where \( d_o = 24.0 \) cm (object distance) and solve for \( d_i \) (image distance). Substitute \( f \) and \( d_o \) into the formula:\[ \frac{1}{4.89} = \frac{1}{24} + \frac{1}{d_i} \] Solve for \( d_i \) which gives approximately \( d_i = 6.46 \) cm.

Step 5: Calculate Magnification

The magnification \( m \) is given by: \[ m = -\frac{d_i}{d_o} \] Substitute \( d_i = 6.46 \) cm and \( d_o = 24.0 \) cm:\[ m = -\frac{6.46}{24.0} \approx -0.27 \] Magnification tells us that the image is inverted and smaller than the object.

Key Concepts

Lensmaker's Equation

The Lensmaker's Equation is an essential tool to understand how lenses focus light and form images. It helps us calculate the focal length of a lens. The formula is written as:\[\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]In this equation:

• \(f\) is the focal length.
• \(n\) is the refractive index of the lens material.
• \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.

This equation accounts for the geometry of the lens and the material's properties. For positive focal lengths, like those for converging lenses, the incoming parallel rays converge to a point. In practical use, knowing the focal length allows us to predict how lenses modify light paths and form images.

Refractive Index

The refractive index is a fundamental property of materials that describes how light travels through them. It is denoted by \(n\) and defined as the ratio of the speed of light in a vacuum to the speed of light in the material:\[ n = \frac{c}{v} \]Where:

• \(c\) is the speed of light in a vacuum (approximately \(3 \times 10^8\) m/s).
• \(v\) is the speed of light in the material.

In lens calculations, the refractive index determines how much light bends when entering or exiting the lens. A higher refractive index means the light will bend more sharply, greatly influencing the lens's ability to focus. For the given converging meniscus lens, the refractive index is 1.52, indicating relatively high bending power, crucial for focusing light efficiently.

Image Position

To find where an image forms when an object is placed in front of a lens, we use a variation of the Lens Formula:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]In this formula:

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

By rearranging this formula, we can solve for \(d_i\), finding where the image will appear. This allows for the detailed positioning of components in optical systems, from simple glasses to complex laser setups. For the exercise, calculating an image position at \(6.46\) cm demonstrates how lenses can magnify or minimize images.

Magnification

Magnification is a measure of how much larger or smaller an image is compared to the object. It is determined using the formula:\[ m = -\frac{d_i}{d_o} \]Where:

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

This negative sign indicates that the image is inverted. In the described exercise situation, the calculated magnification \(-0.27\) tells us that the image is smaller than the object and flipped upside down due to the lens's behavior. Understanding magnification is crucial in applications like microscopy, photography, and vision correction, where precise image size and orientation are essential.