The equation $\frac{\mathbf{1}}{\mathbf{p}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{r}}$for spherical mirrors is an approximation that is valid if the image is formed by rays that make only small angles with the central axis. In reality, many of the angles are large, which smears the image a little. You can determine how much. Refer to Fig. 34-22 and consider a ray that leaves a point source (the object) on the central axis and that makes an angle a with that axis. First, find the point of intersection of the ray with the mirror. If the coordinates of this intersection point are x and y and the origin is placed at the center of curvature, then y=(x+p-r)tan a and x²+ y²= r²where pis the object distance and r is the mirror’s radius of curvature. Next, use tan$\mathbf{\beta }\mathbf{=}\frac{\mathbf{y}}{\mathbf{x}}$to find the angle b at the point of intersection, and then use$\mathbf{\alpha }\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{\beta }$tofind the value of g. Finally, use the relation$\mathbf{tan}\mathbf{}\mathbf{y}\mathbf{=}\frac{\mathbf{y}}{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{i}\mathbf{-}\mathbf{r}\mathbf{)}}$to find the distance iof the image. (a) Suppose r=12cmand r=12cm. For each of the following values of a, find the position of the image — that is, the position of the point where the reflected ray crosses the central axis:$\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0100}\mathbf{}\mathbf{rad}\mathbf{)}$. Compare the results with those obtained with theequation$\frac{\mathbf{1}}{\mathbf{p}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{r}}$.(b) Repeat the calculations for p=4.00cm.

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$\alpha$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 $\alpha$for r=12cm and p=20cm:

Given$\alpha =0.500rad,r=12cm,p=20cm.$

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,

$$\begin{gathered} y=d\tan \alpha \\ 0.500rad=28.65° \end{gathered}$$

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:

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

Solve as:

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

Then:

$\begin{array}{l}\gamma =2\beta -\alpha \\ =2\times 0.8233rad-0.500rad\\ =1.151rad\\ =65.{98}^0\end{array}$

Solve further as:

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

In a similar way, solve as:

For$\alpha =0.100rad,i=8.544cm$

For $\alpha =0.0100rad,i=8.571cm$

Now, the mirror equation relates an object distance p, mirror’s 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$\alpha =0.500rad,i=7.799cm$

For $\alpha =0.100rad,i=8.544cm$

For $\alpha =0.0100rad,i=8.571cm$

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

Repeat calculation for p=4cm:

For$\alpha =0.500rad,r=12cm,p=4cm.$

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:

$$\begin{gathered} y=d\tan \alpha \\ 0.500rad=28.65° \end{gathered}$$

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

From x² +y²=r²solve as:

$$\begin{gathered} x^2+(-4.3704+0.5463x{)}^2=(12cm{)}^2 \\ x=-8.1398\hspace{0.33em}cm \end{gathered}$$

Solve further as:

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

Solve for the beta as:

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

Solve further as:

$\begin{array}{ll}\gamma =2\beta -\alpha & \\ =2\times -0.8253rad-0.500& rad\\ =2.1506rad& \\ =132.{28}^0& \end{array}$

Consider the formula as:

$\tan \gamma =\frac{y}{(x+i-r)}$

Substitute the value and solve as:

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

i=-13.56cm

In a similar way, the obtained values are:

For$\alpha =0.100rad,i=-12.05cm$

For$\alpha =0.0100rad,i=-12cm$

Now, the mirror equation relates an object distance p, mirror’s 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”

For $\alpha =0.500rad,i=-13.56cm$$\alpha =0.500rad,i=-13.56cm$

For $\alpha =0.100rad,i=-12.05cm$

For$\alpha =0.0100rad,i=-12cm$

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