Problem 81 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A planoconvex lens (Fig. 33-2a) has one flat surface and the other has \(R=15.3 \mathrm{~cm} .\) This lens is used to view a red and yellow object which is \(66.0 \mathrm{~cm}\) away from the lens. The index of refraction of the glass is 1.5106 for red light and 1.5226 for yellow light. What are the locations of the red and yellow images formed by the lens?

Step-by-Step Solution

Answer

The red image is located at approximately 45.9 cm, and the yellow image at 44.3 cm from the lens.

Step 1: Use the Lensmaker's Equation

To find the focal length of the lens for each color, we use the lensmaker's equation for a thin lens: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \). For a planoconvex lens, \( R_2 = \infty \) and \( R_1 = R \), and since the refractive index differs for red and yellow, calculate each separately.

Step 2: Calculate Focal Length for Red Light

Using the lensmaker's equation for red light where \( n = 1.5106 \): \[ \frac{1}{f_\text{red}} = (1.5106 - 1) \left( \frac{1}{15.3} \right) \] Simplifying, we find \( f_\text{red} \approx 31.1 \text{ cm} \).

Step 3: Calculate Focal Length for Yellow Light

Now, apply the lensmaker's equation for yellow light where \( n = 1.5226 \): \[ \frac{1}{f_\text{yellow}} = (1.5226 - 1) \left( \frac{1}{15.3} \right) \] Simplifying, we find \( f_\text{yellow} \approx 29.3 \text{ cm} \).

Step 4: Use the Lens Formula to Find Image Position for Red Light

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Insert values for red light with \( d_o = 66.0 \text{ cm} \) and \( f_\text{red} \approx 31.1 \text{ cm} \): \[ \frac{1}{31.1} = \frac{1}{66.0} + \frac{1}{d_{i,\text{red}}} \] Solving for \( d_{i,\text{red}} \) gives \( d_{i,\text{red}} \approx 45.9 \text{ cm} \).

Step 5: Use the Lens Formula to Find Image Position for Yellow Light

Now, apply the lens formula for yellow light with \( f_\text{yellow} \approx 29.3 \text{ cm} \): \[ \frac{1}{29.3} = \frac{1}{66.0} + \frac{1}{d_{i,\text{yellow}}} \] Solving for \( d_{i,\text{yellow}} \) gives \( d_{i,\text{yellow}} \approx 44.3 \text{ cm} \).

Key Concepts

Lensmaker's Equation

The lensmaker's equation is pivotal in understanding how lenses form images. It relates the focal length of a lens to the curvatures of its surfaces and the refractive index of its material. The equation is given as:

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

This formula helps calculate the focal length of lenses be they convex, concave, or a combination of both. For a planoconvex lens, the lens has one flat surface, meaning the radius of curvature \( R_2 \) is infinity, simplifying the equation to:

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

Here, \( n \) is the refractive index of the lens material, and \( R_1 \) is the radius of curvature of the lens's convex surface. This equation allows us to fine-tune lenses for specific optical requirements, considering material properties and design geometries.

Focal Length

The focal length of a lens is a crucial measure in optics, signifying the distance over which light rays converge after passing through the lens. It's an essential factor determining how lenses focus images:

• A shorter focal length means light converges quickly, producing a tighter image.
• A longer focal length results in a broader, more extensive image convergence.

In the provided problem, we calculate the focal lengths for red and yellow light using the lensmaker’s equation, taking into account the different refractive indices. For the planoconvex lens in question, the focal lengths differ slightly due to the variation in refractive indices, leading to values of approximately:

• 31.1 cm for red light
• 29.3 cm for yellow light

These differences affect where the respective images form, demonstrating how specific wavelengths of light impact focusing behavior.

Refractive Index

The refractive index of a material explains how much the material bends the light passing through it. It is a fundamental property in optics, crucial for designing lenses. It is signified by \( n \) in equations and is 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 vacuum, and \( v \) is the speed within the medium. The index can vary based on the wavelength of light, which explains why different colors (red and yellow in the problem) result in different focal lengths in lenses:

• Red light refractive index: 1.5106
• Yellow light refractive index: 1.5226

This concept allows us to comprehend how lenses can be designed to handle chromatic aberration by balancing the bending of different light wavelengths.

Lens Formula

The lens formula is a straightforward yet powerful tool that assists in calculating the position of an image formed by a lens. It is expressed as:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

Here, \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance from the lens. By rearranging the formula, we can calculate unknown quantities if the others are known:

• For red light, with \( f = 31.1 \text{ cm} \) and \( d_o = 66.0 \text{ cm} \), the image distance \( d_i \) is approximately 45.9 cm.
• For yellow light, with \( f = 29.3 \text{ cm} \), \( d_i \) works out to be around 44.3 cm.

This illustrates how the lens formula is invaluable in optical calculations, helping determine where the resulting images will form based on the focal properties of the lens and the positioning of objects with respect to it.