Problem 29 - University Physics with Modern Physics

Question

A converging meniscus lens (see Fig. 34.32a) with a refractive index of 1.52 has spherical surfaces whose radii are 7.00 cm and 4.00 cm. What is the position of the image if an object is placed 24.0 cm to the left of the lens? What is the magnification?

Step-by-Step Solution

Answer

The image is formed 4.07 cm to the right of the lens, with a magnification of 0.17.

Step 1: Understand the Lensmaker's Formula

A converging (convex) lens has a positive focal length. The Lensmaker's Formula is used to find the focal length \(f\) of a lens. It is given by:\[\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]where \(n\) is the refractive index of the lens material, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces. With the given values, \(R_1 = 7.0 \, \text{cm}\) and \(R_2 = -4.0 \, \text{cm}\) (noting that \(R_2\) is negative since it is in the opposite direction). The refractive index \(n = 1.52\).

Step 2: Calculate the Focal Length

Substitute the given values into the Lensmaker's Formula:\[\frac{1}{f} = (1.52 - 1) \left( \frac{1}{7.0} - \frac{1}{-4.0} \right)\]Simplify this expression:\[\frac{1}{f} = 0.52 \left( \frac{1}{7.0} + \frac{1}{4.0} \right)\]Calculate the result of the fractions:\[\frac{1}{f} = 0.52 \left( \frac{4 + 7}{28} \right) = 0.52 \times \frac{11}{28} = \frac{0.52 \times 11}{28} = \frac{5.72}{28}\]Thus, the focal length \(f\) is:\[f = \frac{28}{5.72} \approx 4.90 \, \text{cm}\]

Step 3: Apply the Lens Formula for Image Position

The lens formula relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\):\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]With the object distance \(d_o = -24.0 \, \text{cm}\) (negative since the object is to the left of the lens), substitute \(f = 4.90 \, \text{cm}\):\[\frac{1}{4.90} = \frac{1}{-24.0} + \frac{1}{d_i}\]Solving for \(d_i\):\[\frac{1}{d_i} = \frac{1}{4.90} - \frac{1}{-24.0} = \frac{24 + 4.90}{117.6}\]\[\frac{1}{d_i} = \frac{28.90}{117.6} \Rightarrow d_i \approx 4.07 \, \text{cm}\]

Step 4: Calculate the Magnification

The magnification \(m\) of the lens is given by the formula:\[m = -\frac{d_i}{d_o}\]Using \(d_i \approx 4.07 \, \text{cm}\) and \(d_o = -24.0 \, \text{cm}\):\[m = -\frac{4.07}{-24.0} = \frac{4.07}{24.0} \approx 0.17\]This means the image is upright and smaller than the object.

Key Concepts

Lensmaker's formula

To understand how lenses form images, we begin with the Lensmaker's formula. This important formula helps us determine the focal length of a lens based on its curvature and the material it is made from. The formula is:

• \[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

The variables mean the following:

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

It is crucial to remember that the radius of curvature is positive if the center of curvature is on the outgoing side of the light. Conversely, it is negative if the center is on the incoming side. By substituting the provided values into the formula, we can calculate the lens's focal length.

focal length

The focal length of a lens tells us how strongly it converges or diverges light. A positive focal length indicates a converging (convex) lens. To compute the focal length using the Lensmaker's formula, substitute the known values:

• Refractive index \( n = 1.52 \)
• Radius of curvature \( R_1 = 7.0 \, \text{cm} \)
• Radius of curvature \( R_2 = -4.0 \, \text{cm} \)

Fill in the Lensmaker's formula:\[ \frac{1}{f} = (1.52 - 1) \left( \frac{1}{7.0} - \frac{1}{-4.0} \right) \]Simplifying gives:\[ \frac{1}{f} = 0.52 \left( \frac{1}{7.0} + \frac{1}{4.0} \right) = \frac{5.72}{28} \]Thus, the focal length \( f \) is approximately \( 4.90 \, \text{cm} \). This tells us the lens is strong in converging light rays to a point.

image position

With knowledge of the focal length, we determine the position of the image formed by the lens. The lens formula bridges the object distance, image distance, and focal length:

• \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

In this equation:

• \( f \) is the focal length,
• \( d_o \) is the object distance,
• \( d_i \) is the image distance.

For the problem, assume the object is positioned to the left of the lens, giving it a negative distance \( d_o = -24.0 \, \text{cm} \). Using our focal length result \( f = 4.90 \, \text{cm} \), we substitute into the formula:\[ \frac{1}{4.90} = \frac{1}{-24.0} + \frac{1}{d_i} \]Solving for \( d_i \) gives:\[ d_i \approx 4.07 \, \text{cm} \]This result signifies where the image forms on the lens's opposite side when the object is on the left.

magnification

Understanding magnification helps us predict the size and orientation of an image produced by a lens. The formula for magnification \( m \) is:

• \[ m = -\frac{d_i}{d_o} \]

Where:

• \( d_i \) is the image distance
• \( d_o \) is the object distance

For this scenario, with an image distance \( d_i \approx 4.07 \, \text{cm} \) and an object distance \( d_o = -24.0 \, \text{cm} \), \[ m = -\frac{4.07}{-24.0} = \frac{4.07}{24.0} \approx 0.17 \]This magnification factor tells us the image is upright and reduced to 17% of the object's original size in height. Magnification being positive confirms the image's upright orientation.