Problem 42 - College Physics

Question

A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50 \(\mathrm{cm} .\) Looking through this lens, you observe that it forms an image of a very distant tree, at a distance of 1.87 \(\mathrm{cm}\) from the lens. What is the index of refrac- tion of the lens?

Step-by-Step Solution

Answer

The index of refraction of the lens is approximately 1.67.

Step 1: Understand the Lensmaker's Equation

The Lensmaker's Equation is given by \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), where \( f \) is the focal length, \( n \) is the index of refraction, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. For a double-convex lens with equal radii, \( R_1 = R \) and \( R_2 = -R \).

Step 2: Recognize Focal Length Calculation

Since the object is very distant, the converging lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) simplifies to \( \frac{1}{f} = \frac{1}{d_i} \), where \( d_o \) is the object distance and \( d_i \) is the image distance. So, \( f = d_i = 1.87\, \text{cm} \).

Step 3: Substitute into Lensmaker's Equation

Substitute \( f = 1.87 \text{ cm}, R_1 = 2.50 \text{ cm}, R_2 = -2.50 \text{ cm} \) into the Lensmaker's Equation: \( \frac{1}{1.87} = (n-1) \left( \frac{1}{2.50} - \frac{1}{-2.50} \right) \).

Step 4: Simplify the Equation

Simplify the term \( \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) to \( \frac{1}{2.50} + \frac{1}{2.50} = \frac{2}{2.50} = \frac{4}{5} \). This gives \( \frac{1}{1.87} = (n-1) \times \frac{4}{5} \).

Step 5: Solve for the Index of Refraction \( n \)

Rearrange the equation to solve for \( n \): \( n = 1 + \left( \frac{1}{1.87} \times \frac{5}{4} \right) \). Calculate \( \frac{1}{1.87} \approx 0.5355 \), then \( 0.5355 \times \frac{5}{4} \approx 0.6694 \). Finally, \( n = 1 + 0.6694 = 1.6694 \).

Key Concepts

Lensmaker's Equation

In optics, the Lensmaker's Equation is a crucial tool for understanding how lenses focus light. It relates the focal length of a lens to its material properties and its shape. The equation is given by: \[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where:

• \( f \) is the focal length of the lens.
• \( n \) is the index of refraction of the lens material.
• \( R_1 \) is the radius of curvature of the lens surface closest to the light source.
• \( R_2 \) is the radius of curvature of the lens surface farthest from the light source.

For a double-convex lens with equal radii, often you find that \( R_1 = R \) and \( R_2 = -R \), which simplifies the equation further. This makes it easier to solve for the index of refraction when you know the radii of curvature and the focal length.

Focal Length

The focal length \( f \) of a lens is the distance over which parallel rays of light are brought to a focus. In simpler terms, it tells you where the lens will bring light into focus after it passes through. For very distant objects, like a far-off tree, the light rays essentially travel parallel to each other.
In this scenario, the focal length can typically be measured by the distance from the lens to the clear image formed, known as the image distance \( d_i \). Given that objects far away make their light appear parallel, the formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) simplifies to \( \frac{1}{f} = \frac{1}{d_i} \) because the object distance \( d_o \) is infinitely large and hence negligible.

Index of Refraction

The index of refraction \( n \) helps us understand how much light slows down when it enters a material compared to its speed in a vacuum. It's a basic way of describing how much the path of light is bent, or refracted, as it passes through different materials.
Materials with a higher index of refraction bend light more significantly. The index's formula, derived earlier, is usually calculated using known values of focal length and radii of curvature. For this problem, you simplify and solve using the Lensmaker's Equation, since you are given the image distance and equal radii, resulting in the final equation: \[ n = 1 + \left( \frac{1}{f} \times \frac{R}{R_1 - R_2} \right) \] This lets you substitute and solve for \( n \), understanding light's behavior within that specific lens material.

Radius of Curvature

The radius of curvature of a lens surface is an imaginary line that describes the curve of the lens. It's crucial for determining how strongly the lens will bend light. In the case of a double-convex lens like the one in this problem, both surfaces have a curvature, each with its radius.
For our lens, both radii are equal, simplifying calculations. The positive radius \( R_1 = 2.50 \text{ cm} \) is for the side facing the incoming light, and \( R_2 = -2.50 \text{ cm} \) is for the opposite side. This means the first surface curves towards the light source, while the second curves away. Such symmetry allows for straightforward use of the Lensmaker's Equation.
The simplification involved is critical in understanding and calculating the lens's optical power, as light encounters two boundary changes when passing through the lens, making precise knowledge of each curvature important.