91影视

• The set value of horizontal scale is $p_s=10c\mathrm{m}$,
• The distance of the object from the mirror is$p=21cm$,

The expression for the magnification is as follows,

$\mathit{m}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\mathbf{p}}$

Here,is the object distance and is the image distance.

The expression for the mirror formula is as follows,

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{p}}$

Here, is the focal length.

Rearrange the mirror formula.

$$\begin{gathered} \frac{\text{1}}{i}=\frac{\text{1}}{f}-\frac{\text{1}}{p} \\ \frac{p}{i}=\frac{p-f}{f} \\ \frac{i}{p}=\frac{f}{p-f} \end{gathered}$$

Substitute the value of for$-m$and$\frac{i}{p}$simplify.

$-m=\frac{f}{p-f}$ 鈥�(颈)

Simplify the above expression.

$$\begin{gathered} m=-\frac{f}{p-f} \\ \frac{\text{1}}{m}=-\frac{p-f}{f} \\ \frac{\text{1}}{m}=-\frac{p}{f}+\text{1} \\ \frac{p}{f}=\text{1}-\frac{\text{1}}{m} \\ f=\frac{p}{\text{1}-\frac{\text{1}}{m}} \end{gathered}$$

It can be observed from the given graph that whenlocalid="1662984691028" $p=10\mathrm{cm}$,localid="1662984535050" $m=+0.5$. So, substitute all the values in the above expression.

$\begin{array}{l}f=\frac{100\mathrm{cm}}{1-\frac{1}{0.5}}\\ =-10.0\mathrm{cm}\end{array}$

Determine the magnification for the object distance $p=21\mathrm{cm}$by substituting all the values in equation (i).

$$\begin{gathered} m=\frac{-(-1000\mathrm{cm})}{21.0\mathrm{cm}-(-1000\mathrm{cm})} \\ =0.32 \end{gathered}$$,

Thus, the magnification of the object when the object islocalid="1662984721190" $21\mathrm{cm}$from the mirror is localid="1662984385670" $+0.32$.