Question

17 through 29 22 23, 29 more mirror. 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 m. (All distances are in centimeters.) It also refers to whether (g) the image is real (R) or virtual (V), (h) inverted (I) or non-inverted (NI) from $\mathit{O}$, and (i) on the same side of the mirror as the object$\mathit{O}$or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

Short answer

1. The type of mirror is concave.
2. The focal length is $8\text{cm}$.
3. The radius of curvature is $16\text{cm}$.
4. The object distance is $+24\text{cm}$.
5. The image distance $i$is $+12\text{cm}$.
6. The magnification ratio is $-0.50$.
7. The image is real.
8. The image formed is inverted.
9. The position of the image is the same side.

Step-by-step solution

Step 1: The given data

1. The distance of the object from the mirror,$p=+24\text{cm}$
2. The lateral magnification of the mirror,$m=0.5$
3. The image is inverted.

Understanding the concept of the properties of the mirrors

The given image is inverted which implies the type of type is concave. Again, for an inverted image, the lateral magnification will be negative. This implies the object is at a distance greater than the focal length of the concave mirror. A concave mirror is a diverging mirror with its reflective surface bugling opposite of the light source. Thus, as per the properties, the images formed by the concave mirrors can be both real and virtual based on the position of the object. When the object is closer to the mirror, the image formed is found to be virtual and magnified in nature.

Formulae:

The radius of curvature of a mirror, $r=\text{2}f$ (i)

Where,$f$= focal length of the mirror

The mirror equation, $\frac{\text{1}}{f}=\frac{\text{1}}{i}+\frac{\text{1}}{p}$ (ii)

Where,$f$= focal length,$p$= object distance from the mirror,$i$= image distance.

The lateral magnification of an object, $m=\frac{-i}{p}\text{or}m=\frac{h_i}{h_0}$ (iii)

Where, $p$= object distance from the mirror, $i$= image distance, $h_i$is the height of the image, $h_o$is the height of the object.

(a) determining the type of mirror

Only a concave mirror can form an inverted image for the case that the object distance is greater than that of the focal length of the concave mirror, while plane and convex mirror produce upright images.

Therefore, the type of mirror is concave.

(b) determining the focal length of the mirror

As the image is inverted, thus its magnification value will be negative as per equation (iii), that is.$m=-0.5$

The image distance of the object can be given using the given data in equation (iii) as follows:

$\begin{array}{rcl}i& =& -mp\\ & =& -\left(-0.5\right)\times 24\text{cm}\\ & =& +12\text{cm}\\ & & \end{array}$

Now, the focal length of the mirror can be calculated using the data in equation (ii) as follows:

$\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{12\text{cm}}+\frac{1}{24\text{cm}}\\ & =& \frac{\left(24+12\right)\text{cm}}{\left(12\text{cm}\right)\left(24\text{cm}\right)}\\ & =& +8.0\text{cm}\\ & & \end{array}$

Therefore, the focal length is .$+8\text{cm}$

(c) Determining the radius of curvature

Thus, the radius of curvature of the mirror can be given using equation (i) as follows:

$\begin{array}{rcl}r& =& 2\times 8\text{cm}\\ & =& 16\text{cm}\\ & & \end{array}$

Therefore, the radius of curvature is $+16\text{cm}$.

(d) Determining the object distance

The value of the object distance from the mirror as per the given data is $p=+24\text{cm}$

Therefore, the object distance is $+24\text{cm}$.

(e) Determining the image distance

Based on the calculations in part (b), the value of the imagedistance from the mirror is $i=+12\text{cm}$that is on the reflective side of the mirror.

Therefore, the image distance is $+12\text{cm}$.

(f) Determining the magnification ratio

As per the discussion based in the part (b), it can be said that an inverted image will have its height in negative sign that implies a negative magnification of the concave system for the given position of the object.

The magnification ratio is thus$m=-0.50$as given in the problem.

Therefore, the magnification ratio is $-0.50$.

(g) Determining whether the image is virtual or real.

From the calculations based in part (a), it is found that the image distance is positive in value. Thus, for an image distance to be positive, the image can be concluded to be real.

Hence, the image formed by the mirror is real (R).

(h) Determining whether inverted or non-inverted

For the given case, the image is given to be inverted that is downward to the line of curvature of the mirror.

Therefore, the image is inverted.

(i) Determining the position of the image

For spherical mirrors, real images form on the side of the mirror where the object is located and virtual images form on the opposite side.

An image is formed on the same side as the mirror from the object because the image is real.

Therefore, the position of the image is the same side.