91影视

To draw the principal rays, we have three principal rays coloured as shown in the figure below. The diagram shows that the image is real and inverted.

Given that the radius of curvature is$R=22cm$ and $S=16.5cm$.

The ratio between the distance of object and the distance of image is called lateral magnification and is given by

$m=鈥�\frac{s\text{'}}{s}=鈥�1$

The relation between the object鈥檚 distance and the image distance for the spherical mirrors is given by

$\frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}=\frac{1}{\mathrm{f}}$

Now, the focal length is half the radius of curvature.

$$\begin{gathered} \frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}=\frac{1}{\mathrm{f}} \\ \mathrm{s}\text{'}=\frac{\mathrm{sf}}{\mathrm{s}鈥�\mathrm{f}} \\ \mathrm{s}\text{'}=\frac{\mathrm{s}{\displaystyle \frac{\mathrm{R}}{2}}}{\mathrm{s}鈥�{\displaystyle \frac{\mathrm{R}}{2}}} \\ \mathrm{s}\text{'}=\frac{\mathrm{sR}}{2\mathrm{s}鈥�\mathrm{R}} \end{gathered}$$

Put the values in the equation and find the distance between the image and mirror.

$$\begin{gathered} \mathrm{s}\text{'}=\frac{22脳16.5}{2\left(16.5\right)鈥�22} \\ \mathrm{s}\text{'}=33\mathrm{cm} \end{gathered}$$

Positive sign indicates that the image is to the left of the mirror.

Calculate the lateral magnification

$\mathrm{m}=鈥�\frac{\mathrm{s}\text{'}}{\mathrm{s}}=鈥�\frac{33}{16.5}=鈥�2$

Also, the lateral magnification is given by$\mathrm{m}=鈥�\frac{\mathrm{y}\text{'}}{\mathrm{y}}$ . So,

$$\begin{gathered} \left|\mathrm{y}\text{'}\right|=\left|\mathrm{my}\right| \\ \left|\mathrm{y}\text{'}\right|=\left|鈥�2脳0.6\right| \\ \left|\mathrm{y}\text{'}\right|=1.2\mathrm{cm} \end{gathered}$$