Question

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

Short answer

The fraction is reflected at ${\mathrm{P}}_{\mathrm{re}\mathrm{f}}=\frac{2}{3}$.

Step-by-step solution

Explanation

The integral of the "reflection fraction function" over the square gives the total fraction of reflected light.

In physics, radiant energy intensity is defined as the amount of energy transferred per unit area measured on a plane perpendicular to the energy's propagation direction.

Finding the Fraction Reflected.

Consider the given information,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{ref}}=\int {\int }_{\mathrm{A}}{\mathrm{f}}_{\mathrm{ref}}\mathrm{dxdy} \\ ={\int }_{-1}^1{\int }_{-1}^1\frac{{\left(\mathrm{x}-\mathrm{y}\right)}^2}{4}\mathrm{dxdy} \\ =\frac{1}{4}{\int }_{-1}^1{\int }_{-1}^1({\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2)\mathrm{dxdy} \end{gathered}$$

Further Solving

Now,

$$\begin{gathered} \mathrm{l}=\frac{1}{4}{\int }_{-1}^1{\overline{)\left(\frac{1}{3}{\mathrm{x}}^3-{\mathrm{x}}^2\mathrm{y}+{\mathrm{y}}^2\mathrm{x}\right)}}_{-1}^1\mathrm{dy} \\ =\frac{1}{4}{\int }_{-1}^1\left(\frac{2}{3}+2{\mathrm{y}}^2\right)\mathrm{dy} \\ =\frac{1}{4}{\overline{)\left(\frac{2}{3}\mathrm{y}+\frac{2}{3}{\mathrm{y}}^3\right)}}_{-1}^1 \\ =\frac{2}{3} \end{gathered}$$

Therefore, the required fraction is reflected atrole="math" localid="1658897542491" ${\mathrm{P}}_{\mathrm{ref}}=\frac{2}{3}$