91影视

Radius of curvature r=12cm

Object distance p=20cm

Object distance p=4cm

Use the mirror equation to find the image distance. Using the method given in the problem, then calculate image distance for different$伪$values and for different object distance values.

Formula:

$\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}=\frac{1}{\mathrm{f}}=\frac{2}{\mathrm{r}}$

localid="1664258469754" $f=\frac{r}{2}$

Position of the image for the given values of $伪$for r=12cm and p=20cm:

Given$伪=0.500rad,r=12cm,p=20cm.$

The distance from the object to point x is as follows:

$\begin{array}{l}d=p-r+x\\ =20\mathrm{cm}-12\mathrm{cm}+x\\ =8\mathrm{cm}+x\end{array}$

Also,

$$\begin{gathered} y=d\tan 伪 \\ 0.500rad=28.65掳 \end{gathered}$$

Solve for the value of y as:

$$\begin{gathered} y=(8\mathrm{cm}+x)tan28.{65}^o \\ =4.3704+0.54630x \end{gathered}$$

From x² + y²= r²solve as:

$$\begin{gathered} x^2+(4.3704+0.5463x{)}^2=(12cm{)}^2 \\ x=8.1398cm \end{gathered}$$

Solve as:

$\begin{array}{l}y=dtan伪\\ =4.3704+0.5463(8.1398)\\ =8.8172\mathrm{cm}\end{array}$

Solve as:

$\begin{array}{l}尾=\left(\frac{8.8172}{8.1398}\right)\\ =0.8253rad\end{array}$

Then:

$\begin{array}{l}纬=2尾-伪\\ =2脳0.8233rad-0.500rad\\ =1.151rad\\ =65.{98}^0\end{array}$

Solve further as:

$$\begin{gathered} \tan 纬=\frac{y}{(x+i-r)} \\ \tan \tan 65.98=\frac{8.8172}{(8.1398+i-12)} \\ i=7.799cm. \end{gathered}$$

In a similar way, solve as:

For$伪=0.100rad,i=8.544cm$

For $伪=0.0100rad,i=8.571cm$

Now, the mirror equation relates an object distance p, mirror鈥檚 focal length f and the image distance i as:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=\frac{2}{r}$

$$\begin{gathered} \frac{1}{20cm}+\frac{1}{i}=\frac{2}{12cm} \\ =i=8.571cm \end{gathered}$$

Thus, the mirror equation yields the value i = 8.571cm.

While using the given method the values are:

For$伪=0.500rad,i=7.799cm$

For $伪=0.100rad,i=8.544cm$

For $伪=0.0100rad,i=8.571cm$

The position of the image for the given values of $伪(0.500rad,0.100rad,0.0100rad)$when r=12cm and p=20 cm is$7.799cm,8.544cm,8.571cm.$

Repeat calculation for p=4cm:

For$伪=0.500rad,r=12cm,p=4cm.$

The distance from the object to point x is

$\begin{array}{l}d=p-r+x\\ =4\mathrm{cm}-12\mathrm{cm}+x\\ =-8\mathrm{cm}+x\end{array}$

Solve as:

$$\begin{gathered} y=d\tan 伪 \\ 0.500rad=28.65掳 \end{gathered}$$

$$\begin{gathered} y=(8\mathrm{cm}+x)tantan28.{65}^5 \\ =-4.3704+0.54630x \end{gathered}$$

From x² +y²=r²solve as:

$$\begin{gathered} x^2+(-4.3704+0.5463x{)}^2=(12cm{)}^2 \\ x=-8.1398鈥�cm \end{gathered}$$

Solve further as:

$\begin{array}{l}y=dtan伪\\ =-4.3704+0.5463(-8.1398)\\ =-8.8172\mathrm{cm}\end{array}$

Solve for the beta as:

localid="1663050628081" $\begin{array}{l}尾=(-\frac{8.8172}{8.1388})\\ =-0.8253rad\end{array}$

Solve further as:

$\begin{array}{ll}纬=2尾-伪& \\ =2脳-0.8253rad-0.500& rad\\ =2.1506rad& \\ =132.{28}^0& \end{array}$

Consider the formula as:

$\tan 纬=\frac{y}{(x+i-r)}$

Substitute the value and solve as:

$\tan \tan 132.28掳=\frac{(-8.8172)}{(-8.1398+i-12)}$

i=-13.56cm

In a similar way, the obtained values are:

For$伪=0.100rad,i=-12.05cm$

For$伪=0.0100rad,i=-12cm$

Now, the mirror equation relates an object distance p, mirror鈥檚 focal length f and the image distance i as:

$\frac{1}{4cm}+\frac{1}{i}=\frac{2}{12cm}$

i = -12cm

Thus, the mirror equation yields the value i = -12cm

From the given method the obtained values are鈥�

For $伪=0.500rad,i=-13.56cm$$伪=0.500rad,i=-13.56cm$

For $伪=0.100rad,i=-12.05cm$

For$伪=0.0100rad,i=-12cm$

The position of the image for the given values of$伪(0.500rad,0.100rad,0.0100rad)$when r=12cm and p=4cm is$-13.56cm,-12.05cm,-12cm$