1.1 Pen and Paper

约 3323 个字 预计阅读时间 17 分钟

Homogeneous Coordinates

a) Proof that in homogeneous coordinates, the intersection point \(\tilde{x}\) of the two lines \(\tilde{l}_1\) and \(\tilde{l}_2\) is given by \(\tilde{x} = \tilde{l}_1 \times \tilde{l}_2\).

Proof:

假设两条直线分别为\(\tilde{l}_1 = (a_1, b_1, c_1)^T\)和\(\tilde{l}_2 = (a_2, b_2, c_2)^T\)，则

\[ \begin{aligned} & \tilde{l}_1 \times \tilde{l}_2 \\ =& \begin{pmatrix} a_1 \\ b_1 \\ c_1 \end{pmatrix} \times \begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix} \\ =& \begin{pmatrix} b_1c_2 - b_2c_1 \\ a_2c_1 - a_1c_2 \\ a_1b_2 - a_2b_1 \end{pmatrix} \end{aligned} \]

由于\(\tilde{x}\)是直线\(\tilde{l}_1\)和\(\tilde{l}_2\)的交点，因此\(\tilde{x}\)必然同时满足直线\(\tilde{l}_1\)和\(\tilde{l}_2\)的方程，即：

\[ \begin{aligned} a_1x + b_1y + c_1 &= 0 \\ a_2x + b_2y + c_2 &= 0 \end{aligned} \]

解这个方程组，得到：

\[ x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}, y = \frac{a_2c_1 - a_1c_2}{a_1b_2 - a_2b_1} \]

所以\(\tilde{x} = (x, y, 1)^T = \tilde{l}_1 \times \tilde{l}_2\)。

b) Similarly, proof that the line that joins two points \(\tilde{x}_1\) and \(\tilde{x}_2\) is given by \(\tilde{l} = \tilde{x}_1 \times \tilde{x}_2\).

Proof:

假设两个点分别为\(\tilde{x}_1 = (x_1, y_1, 1)^T\)和\(\tilde{x}_2 = (x_2, y_2, 1)^T\)，则

\[ \begin{aligned} & \tilde{x}_1 \times \tilde{x}_2 \\ =& \begin{pmatrix} x_1 \\ y_1 \\ 1 \end{pmatrix} \times \begin{pmatrix} x_2 \\ y_2 \\ 1 \end{pmatrix} \\ =& \begin{pmatrix} y_1 - y_2 \\ x_2 - x_1 \\ x_1y_2 - x_2y_1 \end{pmatrix} \end{aligned} \]

由于\(\tilde{l}\)是点\(\tilde{x}_1\)和\(\tilde{x}_2\)的连线，因此\(\tilde{l}\)必然同时满足点\(\tilde{x}_1\)和\(\tilde{x}_2\)的方程，即：

\[ \begin{aligned} x_1x + y_1y + 1 &= 0 \\ x_2x + y_2y + 1 &= 0 \end{aligned} \]

解这个方程组，得到：

\[ x = \frac{y_1 - y_2}{x_2y_1 - x_1y_2}, y = \frac{x_2 - x_1}{x_2y_1 - x_1y_2} \]

所以\(\tilde{l} = (x, y, 1)^T = \tilde{x}_1 \times \tilde{x}_2\)。

c) You are given the following two lines:\(\\ \mathbf{l}_1=\{(x,y)|x+y+3=0\}\\\mathbf{l}_2=\{(x,y)|-x-2y+7=0\}\\\). First, find the intersection point of the two lines by solving the system of linear equations. \(\\\)Next write the lines using homogeneous coordinates and calculate the intersection point using the cross product. Do you obtain the same intersection point?

Solution:

• First: 解方程组,解得\(\mathbf{l}_1\)和\(\mathbf{l}_2\)的交点为\((-13,10)\)。

• Next: 用齐次坐标表示直线\(\mathbf{l}_1\)和\(\mathbf{l}_2\)，并计算交点. \((1,1,3)^T \times (-1,-2,7)^T = (13,-10,-1)\sim(-13,10,1)\)，与上面的结果一致。

d) Write down the line whose normal vector is pointing into the direction \((3, 4)^T\) and which has a distance of 3 from the origin.

Note: normal vector的意思是法向量。

Solution:

单位化\((3,4)^T\)得到\((\frac{3}{5},\frac{4}{5})^T\)，所以直线的vector为\((\frac{3}{5},\frac{4}{5},3)^T\)，直线的方程为\(\frac{3}{5}x + \frac{4}{5}y + 3 = 0\)。

e) What distance from the origin and what (normalized) normal vector does the homogeneous line \(\tilde{l} = (2,5,\frac{\sqrt{29}}{5})^T\) have?

Solution:

\[\tilde{l} = (2,5,\frac{\sqrt{29}}{5})^T = (\frac{2}{\sqrt{29}},\frac{5}{\sqrt{29}},\frac{1}{5})^T\]

So the distance from the origin is \(\frac{1}{5}\), and the normal vector is \((\frac{2}{\sqrt{29}},\frac{5}{\sqrt{29}})^T\).

Transformations

a) Write down the \(2 × 3\) translation matrix which maps \((1, 2)^T\) onto \((0, 3)^T\).

Solution:

根据Translation的表达式：

\[\mathbf{x}'=\mathbf{x}+\mathbf{t} \Leftrightarrow \begin{pmatrix}x'\\y'\\1\end{pmatrix}=\begin{bmatrix}1&0&t_x\\0&1&t_y\\0&0&1\end{bmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix} \Leftrightarrow \mathbf{\overline{x}}'=\begin{bmatrix}\mathbf{I}&\mathbf{t}\\\mathbf{0}^T&1\end{bmatrix}\mathbf{\overline{x}}\]

容易得出该矩阵为：

\[\begin{bmatrix}1&0&-1\\0&1&1\\0&0&1\end{bmatrix}\]

写成\(2\times3\)的形式为：

\[\begin{bmatrix}1&0&-1\\0&1&1\end{bmatrix}\]

b)Let’s assume that you are given \(N\) 2D correspondence pairs \((\mathbf{x}_i,\mathbf{y}_i)=\begin{pmatrix}{\begin{pmatrix}x^i_1\\x^i_2\end{pmatrix}},{\begin{pmatrix}y^i_1\\y^i_2\end{pmatrix}}\end{pmatrix}\),where \(i = 1, . . . , N\).Find the \(2 × 3\) translation matrix mapping \(x_i\) onto \(y_i\) which is optimal in the least square sense.

Solution:

构造函数:

\[E(\mathbf{T})=\sum_{i=1}^N\|\mathbf{y}_i-\mathbf{T}\mathbf{x}_i\|^2_2\]

现在就是要找出使得\(E(\mathbf{T})\)最小的\(\mathbf{T}\)。

因为题干说的是translation，所以\(\mathbf{T}\)的形式为：

\[\mathbf{T}=\begin{bmatrix}1&0&t_x\\0&1&t_y\end{bmatrix}\]

代入\(E(\mathbf{T})\)，得到：

\[\begin{aligned} E(\mathbf{T}) &= \sum_{i=1}^N\|\mathbf{y}_i-\begin{bmatrix}1&0&t_x\\0&1&t_y\end{bmatrix}\mathbf{x}_i\|^2_2 \\ &= \sum_{i=1}^N\|\begin{pmatrix}y^i_1-t_x-x^i_1\\y^i_2-t_y-x^i_2\end{pmatrix}\|^2_2 \\ &= \sum_{i=1}^N(y^i_1-t_x-x^i_1)^2+(y^i_2-t_y-x^i_2)^2 \end{aligned}\]

分别对\(t_x\)和\(t_y\)求导，令导数为0，得到：

\[\begin{cases} \frac{\partial E(\mathbf{T})}{\partial t_x} = 2\sum_{i=1}^N(y^i_1-t_x-x^i_1) = 0 \\ \frac{\partial E(\mathbf{T})}{\partial t_y} = 2\sum_{i=1}^N(y^i_2-t_y-x^i_2) = 0 \end{cases}\]

所以：

\[\begin{cases} t_x = \frac{1}{N}\sum_{i=1}^N(y^i_1-x^i_1) \\ t_y = \frac{1}{N}\sum_{i=1}^N(y^i_2-x^i_2) \end{cases}\]

可见，最优的\(\mathbf{T}\)就是：

\[\mathbf{T}=\begin{bmatrix}1&0&\frac{1}{N}\sum_{i=1}^N(y^i_1-x^i_1)\\0&1&\frac{1}{N}\sum_{i=1}^N(y^i_2-x^i_2)\end{bmatrix}\]

也就是说，最优的\(\mathbf{T}\)就是所有\(\mathbf{T}_i\)的平均值，其中\(\mathbf{T}_i\)是将\(\mathbf{x}_i\)映射到\(\mathbf{y}_i\)的translation matrix。

c) You are given the following three correspondence pairs: \(\begin{pmatrix}{\begin{pmatrix}0\\1\end{pmatrix},\begin{pmatrix}3\\-5\end{pmatrix}}\end{pmatrix}\begin{pmatrix}{\begin{pmatrix}5\\7\end{pmatrix},\begin{pmatrix}7\\6\end{pmatrix}}\end{pmatrix}\begin{pmatrix}{\begin{pmatrix}4\\1\end{pmatrix},\begin{pmatrix}5\\-4\end{pmatrix}}\end{pmatrix}\). Using your derived equation, calculate the optimal 2 × 3 translation matrix \(\mathbf{T}^*\).

Solution:

由上一问的结论，可知：

\[\mathbf{T}^*=\begin{bmatrix}1&0&\frac{3+2+1}{3}\\0&1&\frac{-6-1-5}{3}\end{bmatrix}=\begin{bmatrix}1&0&2\\0&1&-4\end{bmatrix}\]

Camera Projections

a) Calculate the full rank \(4 × 4\) projection matrix \(\tilde{P}\) for the following scenario:\(\\\)- The camera pose consists of a \(90°\) rotation around the \(x\) axis and translation of \((1, 0, 2)^T\).\(\\\)- The focal lengths \(f_x, f_y\) are \(100\).\(\\\)- The principal point \((c_x, c_y)^T\) is \((25, 25)^T\).

Note:

• Full rank projection matrix是\(\begin{bmatrix}\mathbf{K}&\mathbf{0}\\\mathbf{0}^T&1\end{bmatrix}\begin{bmatrix}\mathbf{R}&\mathbf{t}\\\mathbf{0}^T&1\end{bmatrix}\)的形式
• \(\mathbf{K}\)是内参矩阵，\(\mathbf{K}=\begin{bmatrix}f_x&s&c_x\\0&f_y&c_y\\0&0&1\end{bmatrix}\)
• \(\mathbf{R}\)是旋转矩阵
• 绕\(x\)轴旋转的矩阵形式为\(\begin{bmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{bmatrix}\)
• 绕\(y\)轴旋转的矩阵形式为\(\begin{bmatrix}\cos\theta&0&\sin\theta\\0&1&0\\-\sin\theta&0&\cos\theta\end{bmatrix}\)
• 绕\(z\)轴旋转的矩阵形式为\(\begin{bmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{bmatrix}\)
• \(\mathbf{t}\)是平移向量

Solution:

依题意，可知：

\[\mathbf{K}=\begin{bmatrix}100&0&25\\0&100&25\\0&0&1\end{bmatrix}\]

\[\mathbf{R}=\begin{bmatrix}1&0&0\\0&0&-1\\0&1&0\end{bmatrix}\]

\[\mathbf{t}=\begin{bmatrix}1\\0\\2\end{bmatrix}\]

所以：

\[\begin{aligned} \tilde{P} &= \begin{bmatrix}\mathbf{K}&\mathbf{0}\\\mathbf{0}^T&1\end{bmatrix}\begin{bmatrix}\mathbf{R}&\mathbf{t}\\\mathbf{0}^T&1\end{bmatrix} \\ &= \begin{bmatrix}100&0&25&0\\0&100&25&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\begin{bmatrix}1&0&0&1\\0&0&-1&0\\0&1&0&2\\0&0&0&1\end{bmatrix} \\ &= \begin{bmatrix}100&25&0&150\\0&25&-100&50\\0&1&0&2\\0&0&0&1\end{bmatrix} \end{aligned}\]

b) For the previously defined projection, find the world point in inhomogeneous coordinates \(\mathbf{x}_w\) which corresponds to the projected homogeneous point in screen space \(\tilde{\mathbf{x}}_s = (25, 50, 1, 0.25)^T\).

Solution:

我们有：\(\tilde{\mathbf{x}}_s = \tilde{P}\tilde{\mathbf{x}}_w\)，所以：

\[\tilde{\mathbf{x}}_w = \tilde{P}^{-1}\tilde{\mathbf{x}}_s\]

经计算，得到\(\tilde{P}^{-1}\)为：

\[\tilde{P}^{-1}=\frac{1}{100}\begin{bmatrix}1&0&-25&-100\\0&0&100&200\\0&-1&25&0\\0&0&0&100\end{bmatrix}\]

所以：

\[\tilde{\mathbf{x}}_w = \frac{1}{100}\begin{bmatrix}1&0&-25&-100\\0&0&100&200\\0&-1&25&0\\0&0&0&100\end{bmatrix}\begin{pmatrix}25\\50\\1\\0.25\end{pmatrix}=\begin{pmatrix}0.25\\1.5\\-0.25\\0.25\end{pmatrix}\]

由\(\tilde{\mathbf{x}}_w\)的最后一维进行归一化，得到\(\mathbf{x}_w\)为：

\[\mathbf{x}_w=\begin{pmatrix}1\\6\\-1\end{pmatrix}\]

c) Let’s perform our first projection of a geometric shape. We define \(\mathcal{C}_0\) as the cube centered at \(\mathbf{c}_c = (0, 0, 15)^T\) with equal side lengths \(s = 20\).

i) Project the \(8\) corners of the cube \(\mathcal{C}_0\) to the image plane for the pinhole camera with focal lengths \(f_x = f_y = 5\) and the principal point \((c_x, c_y)^T = (10, 10)^T\). Draw the projected points and edges of the cube in a coordinate system on a paper.

Solution:

依题意：\(\mathbf{K}=\begin{bmatrix}5&0&10\\0&5&10\\0&0&1\end{bmatrix}\)，所以：\(\mathbf{\tilde{x}}_s=\begin{bmatrix}\mathbf{K}&\mathbf{0}\end{bmatrix}\mathbf{\overline{x}}_c\)。

由于\(\mathcal{C}_0\)是一个正方体，所以它的8个顶点的坐标为：

\[\begin{aligned} \mathbf{\overline{x}}_{c_1} &= \begin{pmatrix}-10\\-10\\5\\1\end{pmatrix} \mathbf{\overline{x}}_{c_2} = \begin{pmatrix}-10\\10\\5\\1\end{pmatrix} \mathbf{\overline{x}}_{c_3} = \begin{pmatrix}10\\10\\5\\1\end{pmatrix} \mathbf{\overline{x}}_{c_4} = \begin{pmatrix}10\\-10\\5\\1\end{pmatrix} \\ \mathbf{\overline{x}}_{c_5} &= \begin{pmatrix}-10\\-10\\25\\1\end{pmatrix} \mathbf{\overline{x}}_{c_6} = \begin{pmatrix}-10\\10\\25\\1\end{pmatrix} \mathbf{\overline{x}}_{c_7} = \begin{pmatrix}10\\10\\25\\1\end{pmatrix} \mathbf{\overline{x}}_{c_8} = \begin{pmatrix}10\\-10\\25\\1\end{pmatrix} \end{aligned}\]

经由\(\mathbf{K}\)的变换，得到：

\[\begin{aligned} \mathbf{\tilde{x}}_{s_1} &= \begin{pmatrix}0\\0\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_2} = \begin{pmatrix}0\\20\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_3} = \begin{pmatrix}20\\20\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_4} = \begin{pmatrix}20\\0\\1\end{pmatrix} \\ \mathbf{\tilde{x}}_{s_5} &= \begin{pmatrix}8\\8\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_6} = \begin{pmatrix}8\\12\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_7} = \begin{pmatrix}12\\12\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_8} = \begin{pmatrix}12\\8\\1\end{pmatrix} \end{aligned}\]

ii) Let’s move the cube further away from the pinhole camera along the \(z\)-axis such that the distance of the center of the new cube \(\mathcal{C}_1\) is now \(20\). Further, let’s zoom in with our camera such that the focal lengths are \(f_x = f_y = 10\) and the principal point remains the same. Draw the projected points and edges of the cube in a coordinate system on a paper.

Solution:

依题意：\(\mathbf{K}=\begin{bmatrix}10&0&10\\0&10&10\\0&0&1\end{bmatrix}\)，所以：\(\mathbf{\tilde{x}}_s=\begin{bmatrix}\mathbf{K}&\mathbf{0}\end{bmatrix}\mathbf{\overline{x}}_c\)。

由于\(\mathcal{C}_1\)是一个正方体，所以它的8个顶点的坐标为：

\[\begin{aligned} \mathbf{\overline{x}}_{c_1} &= \begin{pmatrix}-10\\-10\\10\\1\end{pmatrix} \mathbf{\overline{x}}_{c_2} = \begin{pmatrix}-10\\10\\10\\1\end{pmatrix} \mathbf{\overline{x}}_{c_3} = \begin{pmatrix}10\\10\\10\\1\end{pmatrix} \mathbf{\overline{x}}_{c_4} = \begin{pmatrix}10\\-10\\10\\1\end{pmatrix} \\ \mathbf{\overline{x}}_{c_5} &= \begin{pmatrix}-10\\-10\\30\\1\end{pmatrix} \mathbf{\overline{x}}_{c_6} = \begin{pmatrix}-10\\10\\30\\1\end{pmatrix} \mathbf{\overline{x}}_{c_7} = \begin{pmatrix}10\\10\\30\\1\end{pmatrix} \mathbf{\overline{x}}_{c_8} = \begin{pmatrix}10\\-10\\30\\1\end{pmatrix} \end{aligned}\]

经由\(\mathbf{K}\)的变换，得到：

\[\begin{aligned} \mathbf{\tilde{x}}_{s_1} &= \begin{pmatrix}0\\0\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_2} = \begin{pmatrix}0\\20\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_3} = \begin{pmatrix}20\\20\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_4} = \begin{pmatrix}20\\0\\1\end{pmatrix} \\ \mathbf{\tilde{x}}_{s_5} &= \begin{pmatrix}\frac{20}{3}\\\frac{20}{3}\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_6} = \begin{pmatrix}\frac{20}{3}\\\frac{40}{3}\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_7} = \begin{pmatrix}\frac{40}{3}\\\frac{40}{3}\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_8} = \begin{pmatrix}\frac{40}{3}\\\frac{20}{3}\\1\end{pmatrix} \end{aligned}\]

iii) Let’s move it even further away along the \(z\)-axis while zooming in. Now, the distance of the center of cube \(\mathcal{C}_2\) is \(100\) and the focal lengths are \(f_x = f_y = 90\). Draw the projected points and edges of the cube in a coordinate system on a paper.

Solution:

依题意：\(\mathbf{K}=\begin{bmatrix}90&0&10\\0&90&10\\0&0&1\end{bmatrix}\)，所以：\(\mathbf{\tilde{x}}_s=\begin{bmatrix}\mathbf{K}&\mathbf{0}\end{bmatrix}\mathbf{\overline{x}}_c\)。

由于\(\mathcal{C}_2\)是一个正方体，所以它的8个顶点的坐标为：

\[\begin{aligned} \mathbf{\overline{x}}_{c_1} &= \begin{pmatrix}-10\\-10\\90\\1\end{pmatrix} \mathbf{\overline{x}}_{c_2} = \begin{pmatrix}-10\\10\\90\\1\end{pmatrix} \mathbf{\overline{x}}_{c_3} = \begin{pmatrix}10\\10\\90\\1\end{pmatrix} \mathbf{\overline{x}}_{c_4} = \begin{pmatrix}10\\-10\\90\\1\end{pmatrix} \\ \mathbf{\overline{x}}_{c_5} &= \begin{pmatrix}-10\\-10\\110\\1\end{pmatrix} \mathbf{\overline{x}}_{c_6} = \begin{pmatrix}-10\\10\\110\\1\end{pmatrix} \mathbf{\overline{x}}_{c_7} = \begin{pmatrix}10\\10\\110\\1\end{pmatrix} \mathbf{\overline{x}}_{c_8} = \begin{pmatrix}10\\-10\\110\\1\end{pmatrix} \end{aligned}\]

经由\(\mathbf{K}\)的变换，得到：

\[\begin{aligned} \mathbf{\tilde{x}}_{s_1} &= \begin{pmatrix}0\\0\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_2} = \begin{pmatrix}0\\20\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_3} = \begin{pmatrix}20\\20\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_4} = \begin{pmatrix}20\\0\\1\end{pmatrix} \\ \mathbf{\tilde{x}}_{s_5} &= \begin{pmatrix}\frac{20}{11}\\\frac{20}{11}\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_6} = \begin{pmatrix}\frac{20}{11}\\\frac{200}{11}\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_7} = \begin{pmatrix}\frac{200}{11}\\\frac{200}{11}\\1\end{pmatrix} \mathbf{\tilde{x}}_{s_8} = \begin{pmatrix}\frac{200}{11}\\\frac{20}{11}\\1\end{pmatrix} \end{aligned}\]

iv) Project the \(8\) corners of the first cube \(\mathcal{C}_0\) using an orthographic projection and add the principal point \(\mathbf{c}_c = (10, 10)^T\) onto the obtained pixel coordinates to be in the same coordinate system as before. Draw the projected points and edges of the cube in a coordinate system on a paper.

Solution:

v) When is the perspective projection most similar to the orthographic projection?

Solution:

当焦距\(f_x\)和\(f_y\)趋近于无穷大时，透视投影就趋近于正交投影。

Photometric Image Formation

a) Write down the thin lens formula and calculate the focus distance \(z_c\) in meters for focal length \(f = 100mm\) and the distance to the image plane \(z_s = 102mm\).

Solution:

the thin lens formula:

\[\frac{1}{f}=\frac{1}{z_c}+\frac{1}{z_s}\]

所以：

\[z_c=\frac{z_sf}{z_s-f}=\frac{102\times100}{102-100}=5100mm=5.1m\]

b) Write the diameter of the circle of confusion \(c\) as a function of the focal length \(f\), the image plane distance \(z_s\) as well as the distance \(\Delta z_s\) and the f-number \(N\).

Solution:

We have:

\[\frac{c}{\Delta z_s}=\frac{d}{z_s},\quad N=\frac{f}{d}\]

So

\[c=\frac{d\Delta z_s}{z_s}=\frac{f\Delta z_s}{Nz_s}\]

c) Using your derived formula, calculate the diameter of the circle of confusion for the setting \(f = 35mm, N = 1.4, z_s = 40mm\) when \(\Delta z_s = 0.1mm\) as well as when \(\Delta z_s = 0.03mm\). Assuming the camera uses a sensor of size \(64mm^2\) and a pixel resolution of \(400 × 400\) with squared pixels, are the calculated projections sharp or not?

Solution:

\[c_1=\frac{f\Delta z_s}{Nz_s}=\frac{35\times0.1}{1.4\times40}=0.0625mm \]

\[ c_2=\frac{f\Delta z_s}{Nz_s}=\frac{35\times0.03}{1.4\times40}=0.01875mm \]

每个像素的面积为\(\frac{64}{400\times400}=0.0004mm^2\)，所以像素的边长为\(\sqrt{0.0004}=0.02mm\)。对比可知 \(c_1\)比像素的边长大，\(c_2\)比像素的边长小，所以\(c_1\)对应的投影不清晰，\(c_2\)对应的投影清晰。