Question

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

The fraction of incident light reflected is \ \( \frac{2}{3} \).

Step-by-step solution

Understand the region of integration

The mirror covers a square area with vertices at \((\pm 1, \pm 1)\). This means we need to consider the region \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\).

Write down the reflection fraction function

The fraction of light reflected at point \((x, y)\) is given by \((x-y)^{2} / 4\).

Set up the integral

To find the total fraction of reflected light, we need to integrate \(\frac{(x-y)^2}{4}\) over the square area. The integral is \: \ \( \int_{-1}^{1} \int_{-1}^{1} \frac{(x-y)^2}{4} \, dy \, dx \)

Simplify the integral

Simplify the integrand: \(\frac{(x-y)^2}{4}\) can be separated into \ \( \frac{1}{4} \int_{-1}^{1} \int_{-1}^{1} (x^2 - 2xy + y^2) \, dy \, dx \)

Evaluate the integral for each part

Evaluate the integral in parts: \ \( \int_{-1}^{1} \int_{-1}^{1} x^2 \, dy \, dx = \int_{-1}^{1} x^2 (2) \, dx = 2 \int_{-1}^{1} x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{4}{3} \)

Evaluate for remaining parts

Similarly, for \( \int_{-1}^{1} \int_{-1}^{1} y^2 \, dy \, dx\): \ \( 2 \int_{-1}^{1} y^2 \, dy = \frac{4}{3} \)And for cross term \( \int_{-1}^{1} \int_{-1}^{1} -2xy \, dy \, dx \): \ \(-2 \int_{-1}^{1} x \left[ \int_{-1}^{1} y \, dy \right] dx = -2 \int_{-1}^{1} x \left[ 0 \right] dx = 0 \)

Combine results

Combine all results: \ \( \frac{1}{4} \left( \frac{4}{3} + \frac{4}{3} \right) = \frac{1}{4} \left( \frac{8}{3} \right) = \frac{2}{3} \)

Key concepts

Definite Integral

A definite integral is a powerful tool in calculus that allows us to compute the area under a curve within a specific interval. When you calculate a definite integral, you are essentially summing up small, infinitesimal slices of the area. In this exercise, the region of interest is a square with vertices at \((\pm 1, \pm 1)\). This means we are integrating over the range from \(-1\) to \(1\) for both the x and y coordinates. The function we’re integrating is the fraction of light reflected, given as \((x-y)^2/4\). To compute the total reflected light, we set up a double integral \(\int_{-1}^{1} \int_{-1}^{1} \frac{(x-y)^2}{4} \, dy \, dx \). Understanding the limits of integration and the function to be integrated is crucial to solving such problems.

Reflection Fraction

In this problem, the reflection fraction tells us how much light is reflected at any given point \((x, y)\) on the partially silvered mirror. The fraction is given by the function \((x-y)^2/4\). This function reflects the idea that the light reflection changes depending on the position on the mirror. For instance, at points where \(x = y\), the fraction of reflected light is zero because \((x-y)^2/4\) becomes zero. This emphasizes that reflections are not uniform and vary across the mirror’s surface. By integrating this reflection fraction over the entire square area \([-1, 1] \), we can find out the overall fraction of light that gets reflected.

Integrand Simplification

Simplifying the integrand is a critical step in solving the integral efficiently. In our exercise, the integrand \(\frac{(x-y)^2}{4}\) can be expanded to make the integration easier. This gives us \(\frac{1}{4}(x^2 - 2xy + y^2)\). Now, we can separate this into individual integrals: \(\frac{1}{4} \int_{-1}^{1} \int_{-1}^{1} \(x^2 - 2xy + y^2\) dy dx\). This simplification breaks the original problem into manageable parts. Each part can then be integrated separately, which simplifies the overall integration process. By doing this, we ensure that each term can be dealt with using basic integral properties.

Area of Integration

The area of integration is the region over which we calculate the integral. In this problem, the region of integration is the square with vertices at \((\pm 1, \pm 1)\). This implies that our integration boundaries for x and y are both from \(-1\) to \(1\). Understanding the area of integration helps us set up the correct limits for our integral. When dividing the area into smaller slices to sum up, these limits ensure we cover the entire square. This step is crucial as any error in defining the area can lead to an incorrect result. Once the boundaries are well understood, we proceed by summing up the contributions from each small rectangle within the defined area.