Problem 81 - Prealgebra

Question

Diamonds Shining. The index of refraction \(n\) indicates the number of times slower that a light wave travels in a particular medium than it travels in a vacuum. A diamond has an index of refraction of \(2.4\). This is about one and onequarter times greater than the index of refraction of a zircon. What is the index of refraction of a zircon? Round your result to the nearest tenth.

Step-by-Step Solution

Answer

Zircon's index of refraction is 1.9.

Step 1: Understanding the Problem

We are given the index of refraction of a diamond ( =2.4) and told that this is about one and one-quarter times greater than that of zircon. Our task is to find the index of refraction for zircon and round it to the nearest tenth.

Step 2: Express the Relationship

The problem states that the diamond's index of refraction is one and one-quarter times larger than that of zircon's. Mathematically, this can be expressed as: \[ n_{diamond} = 1.25\, n_{zircon} \]

Step 3: Substitute the Given Values

Substitute the given value for diamond's index of refraction into the equation: \[ 2.4 = 1.25\, n_{zircon} \].

Step 4: Solve for Zircon's Index of Refraction

To find the index of refraction for zircon, divide both sides of the equation by 1.25: \[ n_{zircon} = \frac{2.4}{1.25} \].

Step 5: Perform the Division

Calculate the division to find zircon's index of refraction: \[ n_{zircon} = 1.92 \].

Step 6: Round the Result

Round 1.92 to the nearest tenth to obtain the final answer: \[ n_{zircon} = 1.9 \].

Key Concepts

Understanding the Index of Refraction

The index of refraction is a crucial concept when studying how light behaves as it moves through different materials. It tells us how much slower light travels in a particular medium compared to its speed in a vacuum. For example, the speed of light in a vacuum is about 300,000 kilometers per second. The index of refraction of a material, such as a diamond, shows how much light slows down when passing through it.

In this exercise, we're given that a diamond's index of refraction is 2.4. This means light travels 2.4 times slower in a diamond than in a vacuum. Understanding this value helps us predict how much light will bend, or refract, as it enters or exits a diamond. Comparing it with other materials, like zircon in this problem, allows us to see how similar or different their refractive properties are. By knowing the index of refraction, we can understand the basic physics behind how and why materials look the way they do when interacting with light.

Fractions and Decimals: Simplifying the Relationship

In this problem, we're dealing with the relationship between the diamond's and zircon's indices of refraction. We're told that the diamond's index is "one and one-quarter times greater" than the zircon's. This phrase "one and one-quarter" is a fractional number, which is another way to express a quantity using fractions.

To work with this relationship mathematically, it's helpful to express it as a decimal. The phrase "one and one-quarter" translates to the fraction \(\frac{5}{4}\), which is equivalent to the decimal 1.25. This allows us to build a simple mathematical equation to express the given relationship:

• Diamond's Index: \(n_{diamond} = 1.25 \, n_{zircon}\)

Using fractions and decimals interchangeably can simplify complex mathematical relationships, enabling easier calculation and problem-solving. In this case, converting the description into a mathematical expression helps us solve for zircon's index more easily.

Developing Problem-Solving Skills

Problem-solving skills are essential for mathematics and analyzing real-world situations. In this exercise, our focus was on determining the index of refraction for zircon, given certain conditions about a diamond. Here's a breakdown of steps that enhance our problem-solving abilities:

• Understanding the problem: Clearly define what's given and what's required. Here, the diamond's index is given, and the zircon's needs to be found.
• Expressing relationships mathematically: Convert word problems into algebraic expressions, such as the equation \(2.4 = 1.25 \, n_{zircon}\).
• Solving equations: Use algebra to isolate the variable and solve for it. Divide both sides by 1.25 to find zircon's index.
• Accuracy and precision: Compute the division and ensure rounding is applied when necessary, ensuring results are practical and understandable.

By breaking tasks into smaller steps, we approach problems systematically, leading to better accuracy and boosting our confidence in handling diverse mathematical challenges. Working through examples such as this enhances our overall problem-solving toolkit.