Question

Object Ostands 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 $\mathbf{\left(}\mathbf{R}\mathbf{\right)}$or virtual $\mathbf{\left(}\mathbf{V}\mathbf{\right)}$, (h) inverted $\mathbf{\left(}\mathbf{I}\mathbf{\right)}$or non-inverted $\mathbf{\left(}\mathbf{NI}\mathbf{\right)}$from O, and (i) on the same side of the mirror as the object Oor the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

Short answer

(a) The type of mirror is concave.

(b) The focal length is $8.6\text{cm}$.

(c) The radius of curvature is $\text{17cm}$.

(d) The object distance is $+3\text{0cm}$.

(e) The image distance $i$is localid="1664205380100" $+12\text{cm}$.

(f) The magnification ratio is localid="1664205384799" $-0.4$.

(g) The image is real $\left(R\right)$.

(h) Image formed is inverted $\left(I\right)$.

(i) The image is formed on the same side of the object O.

Step-by-step solution

The given data:

The lateral magnification of the mirror, $m=+0.4\text{0cm}$

The object’s distance from the mirror, $p=+\text{30cm}$

The image is inverted which means the image formed is real.

The concept:

Magnification refers to the ratio of image length to object length measured in planes that are perpendicular to the optical axis.

The focal length is positive if the mirror is a concave mirror. The focal length is negative if the mirror is a convex mirror. The image distance is positive if the image is a real image and is on the mirror side of the object.

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.

The distance of an object refers to the separation between an object and the mirror's pole. Image distance refers to the separation between the mirror's pole and the image.

Formulae:

The lateral magnification of an object,

$m=\frac{h_i}{h_o}=\frac{-i}{p}$ ….. (i)

The mirror equation is,

$\frac{1}{f}=\frac{1}{i}+\frac{1}{p}$ ….. (ii)

The radius of curvature of a mirror,

$r=2f$ ….. (iii)

Here, $f$is the focal length, $p$is the object distance from the mirror, and $i$is the image distance,$h_i$is the height of an image, and$h_o$is the height of an object.

(a) Determining the type of mirror:

Now, from the lateral magnification value as negative, the image distance can be given using equation (i) as follows.

$\begin{array}{rcl}i& =& -mp\\ & =& 0.4\times (30\text{cm})\\ & =& +12\text{cm}\end{array}$

As in a normal condition, it is difficult to get a real image from the convex mirror, thus, it can be said that the mirror is concave.

Hence, the mirror is concave

(b) Determining the Focal length:

Now the focal length value of the concave mirror can be given using equation (i) as follows:

$\begin{array}{rcl}\frac{\text{1}}{f}& =& \frac{\text{1}}{12\text{cm}}+\frac{\text{1}}{30\text{cm}}\\ & =& \frac{30\text{cm}+12\text{cm}}{30\text{cm}\times 12\text{cm}}\\ & =& \frac{42\text{cm}}{360{\text{cm}}^2}\end{array}$

$\begin{array}{rcl}f& =& \frac{360{\text{cm}}^2}{42\text{cm}}\\ & =& +\text{8.6cm}\end{array}$

Hence, the focal length of the mirror is $\begin{array}{rcl}& & +\text{8.6cm}\end{array}$.

(c) Determining the Radius of curvature:

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

$\begin{array}{rcl}r& =& \text{2}\times \left(8.\text{6cm}\right)\\ & =& 17.2\text{cm}\\ & \approx & 17\text{cm}\end{array}$

Hence, the radius of curvature is $\begin{array}{rcl}& & 17\text{cm}\end{array}$.

(d) Determining the Object distance:

The value of the object’s distance from the mirror is given in the table as:

$p=+30\text{cm}$

Hence, the object distance is $+30\text{cm}$.

(e) Determining the Image distance i:

From the calculations done in part (a), it is found that the value of the image distance can be given as follows:

$i=+\text{12cm}$

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

(f) Determining the Magnification ratio:

As per the given data, the value of lateral magnification can be given as follows:

$m=-\text{0.40}$

Hence, the magnification ratio is $-\text{0.40}$.

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

From the calculations based in part (e), 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 $\left(R\right)$.

(h) Determining whether the image is inverted or not inverted:

The lateral magnification of the mirror is given to be negative value. Again, you know that the lateral magnification can be given as:

$\begin{array}{rcl}m& =& \frac{h_i}{h_o}\\ & =& \frac{-i}{p}\\ & =& -0.4\end{array}$

Thus, the image height needs to be negative that is possible only in an inverted image case.

Hence, the image is inverted $\left(I\right)$.

(i) Determining the position of the image:

As per the given data and calculations based in part (e), the values of image distance and object distance can be seen to be positive in values.

Hence, the image is formed on the same side of the object O.