17 through 29 22 23, 29 More mirrors. Object $\mathit{O}$stands on the central axis of a spherical or plane mirror. For this situation, each problem in Table 34-4 refers to (a) the type of mirror, (b) the focal distance $\mathit{f}$, (c) the radius of curvature $\mathit{r}$, (d) the object distance $\mathit{p}$, (e) the image distance $\mathit{i}$, and (f) the lateral magnification $\mathit{m}$. (All distances are in centimeters.) It also refers to whether (g) the image is real localid="1662999140986" $\left(R\right)$or virtual $\mathbf{\left(}\mathit{V}\mathbf{\right)}$, (h) inverted $\mathbf{\left(}\mathit{I}\mathbf{\right)}$or non-inverted from $\mathbf{\left(}\mathit{N}\mathit{I}\mathbf{\right)}$from $\mathit{O}$, and (i) on the same side of the mirror as the object $\mathit{O}$or the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

$f=+20cm$.

$p=+60cm$.

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,

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

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 $O$.