Problem 36 - College Physics

Question

\(\cdot\) The front, convex, surface of a lens made for eyeglasses has a radius of curvature of \(11.8 \mathrm{cm},\) and the back, concave, surface has a radius of curvature of 6.80 \(\mathrm{cm} .\) The index of refraction of the plastic lens material is 1.67 . Calculate the local length of the lens.

Step-by-Step Solution

Answer

The focal length of the lens is approximately 6.44 cm.

Step 1: Understand the Lensmaker's Formula

The lensmaker's formula is used to calculate the focal length \( f \) of a lens: \[\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\] where \( n \) is the index of refraction, \( R_1 \) is the radius of curvature of the front surface, and \( R_2 \) is the radius of curvature of the back surface. It's important to note that \( R_1 \) is positive for convex surfaces and \( R_2 \) is negative for concave surfaces.

Step 2: Identify and Substitute the Values

We have \( n = 1.67 \), \( R_1 = 11.8 \, \text{cm} \), and \( R_2 = -6.8 \, \text{cm} \). Substitute these values into the lensmaker's formula.\[\frac{1}{f} = (1.67 - 1)\left(\frac{1}{11.8} - \frac{1}{-6.8}\right)\]Calculate each component separately to prepare for the next step.

Step 3: Calculate the \(\frac{1}{f}\) Value

First, calculate the parts of the equation separately:\( (n-1) = 1.67 - 1 = 0.67 \).Then calculate \( \frac{1}{R_1} = \frac{1}{11.8} \approx 0.08475 \) and \( \frac{1}{R_2} = \frac{1}{-6.8} \approx -0.14706 \).Substitute these into the equation:\[\frac{1}{f} = 0.67 \left(0.08475 + 0.14706\right)\]

Step 4: Simplify and Solve for \( f \)

Add the terms inside the parentheses:\[ 0.08475 + 0.14706 = 0.23181 \]Multiply by \( 0.67 \):\[\\frac{1}{f} = 0.67 \times 0.23181 \approx 0.15531\]To find \( f \), take the reciprocal:\[ f \approx \frac{1}{0.15531} \approx 6.44 \, \text{cm}\]

Step 5: Verify the Solution

Review the calculations to ensure accuracy, especially with signs and units. Ensure that each step logically follows from the previous and check that \( f \) is in the correct units, which it is, as 6.44 cm.

Key Concepts

Radius of Curvature

The radius of curvature is a key concept in understanding the geometry of lenses and their effect on light. For a lens, it refers to the radius of the sphere from which a lens surface is a part. In simpler terms, if you imagine a perfectly round ball, the radius of curvature is the radius of that ball.

Lenses can have convex or concave surfaces:

• *Convex surface*: Curves outward, like the exterior of a ball. The radius of curvature here is considered positive.
• *Concave surface*: Curves inward, like creating a hollow. The radius of curvature is negative.

In the exercise, the convex front surface of the lens has a positive radius of curvature of 11.8 cm, and the concave back surface has a negative radius of curvature of -6.80 cm. Understanding these signs is essential in performing accurate lens calculations.

Index of Refraction

The index of refraction, denoted as "n," describes how light travels through a medium. It's calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium. It's a unique property for every material.

This index determines how much light bends, or refracts, when it enters the material. Materials with a higher index cause light to bend more sharply. For example:

• *For air*: The index of refraction is approximately 1.0
• *For the plastic lens in our problem*: The index is 1.67

This means light travels more slowly through the plastic lens material, bending at a significant angle, which is crucial for the lens's focusing ability.

Focal Length Calculation

The focal length of a lens, “f”, is a measurement that determines how strongly the lens converges or diverges light. The Lensmaker's Formula is used for calculating this: \[ \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] where:

• : "n" is the index of refraction.
• : "R₁" is the radius of curvature of the front surface.
• : "R₂" is the radius of curvature of the back surface.

For our lens, we substitute the known values:

• : The lens material's index is 1.67.
• : The front surface radius is 11.8 cm.
• : The back surface radius is -6.8 cm.

Using these in the formula helps us find the lens's focal length — a critical factor in determining its functionality in focusing light effectively.

Refraction in Lenses

Refraction occurs when light passes through different materials at an angle, bending the light. Lenses, such as eyeglasses or camera lenses, use this property to alter the pathway of light rays.

When light enters a lens:

• : The light slows down and bends towards the normal due to the lens's material index of refraction.
• : As light exits the lens, it speeds up again, bending away from the normal, focusing or diverging light rays to form an image.

Understanding how light refracts in lenses helps in designing lenses to correct vision or focus images properly. This is why calculations involving refraction indices and curvatures are critical in lens making, ensuring that lenses work effectively for their intended purpose.