Question

Reflections in a Mirror A highly reflective mirror reflects \(95 \%\) of the light that falls on it. In a light box having walls made of the mirror, the light reflects back-and-forth between the mirrors. (a) If the original intensity of the light is \(I_{0}\) before it falls on a mirror, write the \(n\) th term of the sequence that describes the intensity of the light after \(n\) reflections. (b) How many reflections are needed to reduce the light intensity by at least \(98 \% ?\)

Short answer

a) \( I_n = I_0 \times (0.95)^n \ b) At least 77 reflections

Step-by-step solution

Describe the problem

A highly reflective mirror reflects 95% of the light that falls on it. We need to find the intensity of the light after several reflections and determine how many reflections are needed to reduce the light intensity by at least 98%.

Define the reflection factor

Given that the mirror reflects 95% of the light, the reflection factor can be defined as 0.95.

Write the expression for the nth reflection

If the original intensity of the light is denoted by \( I_0 \), after each reflection, the intensity is multiplied by the reflection factor (0.95). Therefore, after \( n \) reflections, the intensity \( I_n \) can be expressed as \ I_n = I_0 \times (0.95)^n \.

Solve for the number of reflections needed to reduce intensity by 98%

To reduce the intensity by at least 98%, we need the remaining intensity to be less than or equal to 2% of the original intensity \ I_0 \). We can set up the equation: \ I_n \leq 0.02 \ I_0 \Substituting for \ I_n \: \ I_0 \times (0.95)^n \leq 0.02 \ I_0 \We can divide both sides by \ I_0 \(0.95)^n \leq 0.02 \To solve for \( n \), we take the natural logarithm of both sides: \ \ln((0.95)^n) \leq \ln(0.02) \Applying the power rule for logarithms: \ n \ln(0.95) \leq \ln(0.02) \Solve for \( n \): \ n \geq \frac{\ln(0.02)}{\ln(0.95)} \Using a calculator, we get: \ n \geq \frac{-3.912}{-0.0513} \ n \geq 76.24 \Since \ n \ must be an integer, \ n \ is at least 77.

Key concepts

Exponential Decay

In this problem, the light intensity decreases as it reflects off the mirror. The concept behind this is called exponential decay. Unlike linear decay, where something decreases by the same amount each time, exponential decay decreases by a percentage each time.
In our problem, each reflection reduces the light intensity by 5% (since 95% is reflected). So, the intensity after each reflection is 95% of what it was before. This can be expressed mathematically as:
\[ I_n = I_0 \times (0.95)^n \]
Here, \( I_0 \) is the initial intensity, and \( 0.95^n \) shows how the intensity is being reduced with each reflection. This exponential function shows the rapid decrease over time.
Understanding exponential decay is essential not only in physics but in finance, biology, and many other fields. Whenever you see something that shrinks by a constant percentage (like light here), it's likely to be exponential decay.

Reflection Factor

In the context of this problem, the reflection factor is a crucial part. It represents the percentage of light that is reflected back after hitting the mirror.
Given that the mirror reflects 95% of the light, the reflection factor is 0.95. This means for every hit, 95% of the light intensity remains. Mathematically, we show this as:
\[ I_{n+1} = I_n \times 0.95 \]
This reflection factor applies repeatedly for each reflection. So, after \( n \) reflections, the formula becomes:
\[ I_n = I_0 \times (0.95)^n \]
We continuously multiply by 0.95, showing the intensity at each subsequent step. Understanding and using the reflection factor is key to solving problems involving repeated reflections or bounces, like in optics and even some economic models.

Logarithmic Equations

To determine the number of reflections needed to reduce the light intensity by at least 98%, we use logarithmic equations.
From our exponential decay formula:
\[ I_n = I_0 \times (0.95)^n \]
We know that the remaining intensity should be 2% or less of the original to achieve a 98% reduction. Setting up our inequality:
\[ I_0 \times (0.95)^n \leq 0.02 \times I_0 \]
We start by dividing both sides by \( I_0 \):
\[ (0.95)^n \leq 0.02 \]
To solve for \( n \), we take the natural logarithm on both sides:
\[ \text{ln}((0.95)^n) \leq \text{ln}(0.02) \]
Using the power rule for logarithms, it simplifies to:
\[ n \times \text{ln}(0.95) \leq \text{ln}(0.02) \]
Solving for \( n \):
\[ n \geq \frac{\text{ln}(0.02)}{\text{ln}(0.95)} \]
With values:
\[ n \geq \frac{-3.912}{-0.0513} \approx 76.24 \]
This tells us at least 77 reflections are needed. Logarithms are very useful for reverse calculations in exponential problems. They help in situations where we need to find the number of times an event must happen to reach a certain threshold.