91影视

$f=+20cm$.

$p=+60cm$.

Here, the focal distance and object distance is given in the problem. Using that the radius of curvature and image distance can be found. Then, by using image distance and object distance, the magnification ratio can be found. Using all these values, it can be decided if the image is virtual or real and the position of the image.

The formula is as follows:

$\begin{array}{rcl}\mathit{r}& \mathbf{=}& \mathbf{2}\mathit{f}\\ \frac{\mathbf{1}}{\mathbf{f}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{p}}\\ \mathit{m}& \mathbf{=}& \frac{\mathbf{-}\mathbf{i}}{\mathbf{p}}\end{array}$

It is given that the image is on the same side as the object. Thus, the image is real. This implies that the mirror is concave.

As the mirror is concave, from the table 34-4, the focal distance is,

$f=+20cm$.

Use the following formula to find the radius of curvature,

$$\begin{gathered} r=2脳f \\ r=2脳20 \\ r=40cm \end{gathered}$$

The object distance is $p=+60cm$, as given in the table.

It is known,

$$\begin{gathered} \frac{1}{f}=\frac{1}{i}+\frac{1}{p} \\ \frac{1}{20}=\frac{1}{i}+\frac{1}{60} \\ \frac{1}{i}=\frac{1}{20}-\frac{1}{60}=\frac{1}{30} \\ i=30cm \end{gathered}$$

The image distance is, $i=+30cm$.

The magnification ratio is given as,

$$\begin{gathered} M=\frac{-i}{p} \\ M=\text{-}\frac{\text{30}}{\text{60}} \\ M=\text{- 0.50} \end{gathered}$$

Since the image distance is positive, the image is real.

As the magnification is negative, the image is inverted.

For spherical mirrors, the real image is formed on the same side as the object. Since the image is real here, so it is on the same side of the mirror as the object $O$.