Chapter 5: Problem 36

A partially silvered mirror covers the square area with vertices at \((\pm 1, \pm 1)\). The fraction of incident light which it reflects at \((x, y)\) is \((x-y)^{2} / 4\). Assuming a uniform intensity of incident light, find the fraction reflected.

Short Answer

Expert verified

The fraction of incident light reflected is \[ \frac{2}{3} \].

Step by step solution

- Understand the region of interest

The region of interest is a square with vertices at \( (\text{±}1, \text{±}1) \). This means the limits for both x and y are from -1 to 1.

- Set up the fraction of light reflection equation

Given the fraction of light reflected at any point \( (x, y) \) is \( \frac{(x-y)^2}{4} \), we need to calculate this expression over the entire square region.

- Integrate the reflection function over the area

To find the fraction of light reflected over the entire square, we integrate the reflection function \( \frac{(x-y)^2}{4} \) over the limits \( -1 \) to \( 1 \) for both x and y: \[ \int_{-1}^1 \int_{-1}^1 \frac{(x-y)^2}{4} \ dx \ dy \]

- Simplify the integrand

The integrand \( \frac{(x-y)^2}{4} \) can be expanded: \[ \frac{1}{4} \int_{-1}^1 \int_{-1}^1 (x^2 - 2xy + y^2) \ dx \ dy \]

- Separate the integral

We can break the integral into three separate integrals: \[ \frac{1}{4} \left( \int_{-1}^1 \int_{-1}^1 x^2 \ dx \ dy - 2 \int_{-1}^1 \int_{-1}^1 xy \ dx \ dy + \int_{-1}^1 \int_{-1}^1 y^2 \ dx \ dy \right) \]

- Evaluate the integrals

Evaluate each integral separately: \[ \int_{-1}^1 x^2 \ dx \] and \[ \int_{-1}^1 y^2 \ dy \] are both \[ \frac{2}{3} \] because of symmetry, and \[ \int_{-1}^1 xy \ dx \ dy \] is zero because it's an odd function. So, the expression becomes: \[ \frac{1}{4} \left( \frac{2}{3} \cdot 2 + \frac{2}{3} \cdot 2 - 0 \right) = \frac{1}{4} \cdot \frac{8}{3} = \frac{2}{3} \]

- Conclude the fraction of light reflected

Thus, the total fraction of incident light that is reflected is \[ \frac{2}{3} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

integration

Integration is a fundamental concept in calculus that helps us find areas under curves, volumes, and other quantities that are accumulated over a range. In this exercise, we use integration to determine the total fraction of light reflected by the mirror.
We start with an expression for the fraction of light reflected at any point \( (x, y) \) and integrate this function over the entire area of the mirror. We do this as follows:

The limits for both \( x \) and \( y \) are from -1 to 1 because the mirror covers a square with vertices at \( (\text{±}1, \text{±}1) \).
The reflection function \( \frac{(x-y)^2}{4} \) is set up and integrated over these limits.

This process of integration helps in summing up the small fractions of light reflected at each point on the mirror to find the total reflection.

mirror reflection

Mirror reflection is the phenomenon where light bounces off a reflective surface. For a partially silvered mirror, the amount of light reflected may vary from one point to another on the surface.
In this problem, the reflection at any point \( (x, y) \) depends on the formula \( \frac{(x-y)^2}{4} \). This means that at different points, the fraction of light reflected varies based on the \( x \) and \( y \) coordinates.

Higher values of \( (x-y)^2 \) result in more light being reflected.
Lower values mean less light is reflected.

Thus, to calculate the total fraction of light reflected by the entire mirror, we need to consider these variations across its whole surface by using integration.

uniform intensity

Uniform intensity assumes that the light hitting the mirror is uniform across its entire surface. This simplifies calculations because we don't need to account for variations in the incident light.
When the problem states 'assuming a uniform intensity of incident light,' it means:

The amount of light hitting every point on the mirror is the same.
We only need to focus on how the mirror's reflectivity varies across its surface.

This uniformity is essential for our calculations because it lets us use the provided reflection function \( \frac{(x-y)^2}{4} \) without worrying about changing light intensity.

multivariable calculus

Multivariable calculus extends the principles of calculus to functions of more than one variable. In this exercise, we deal with a function of two variables, \( x \) and \( y \).
This involves:

Setting up a double integral to sum the reflections over the mirror's surface.
Handling the cross-term \( xy \) in the reflection function.

We break the reflection function \( \frac{(x-y)^2}{4} \) down into simpler parts and integrate them separately:

The integrals of \( x^2 \) and \( y^2 \) are straightforward and result in \( \frac{2}{3} \).
The cross-term \( xy \) integrates to zero because it is an odd function over a symmetric interval.

Combining these results gives us the total fraction of light reflected by the mirror.