through 79 76, 78 75, 77 More lenses. Object $o$stands on the central axis of a thin symmetric lens. For this situation, each problem in Table 34-8 refers to (a) the lens type, converging $\left(C\right)$or diverging $\left(D\right)$ , (b) the focal distance $f$, (c) the object distance p, (d) the image distance i, and (e) the lateral magnification m. (All distances are in centimeters.) It also refers to whether (f) the image is real $\left(R\right)$or virtual $\left(V\right)$, (g) inverted $\left(l\right)$or non-inverted (NI) from $O$, and (h) on the same side of the lens as $O$or on the opposite side. Fill in the missing information, including the value of m when only an inequality is given, where only a sign is missing, answer with the sign.

1. The focal distance of the lens, $f=10\mathrm{cm}$
2. The object distance, $p=+5.0\mathrm{cm}$
3. The lateral magnification, $m<1.0$
4. The image is on the same side as O.

An object, when placed in front of a lens, produces an image. It could be real or virtual, magnified or diminished, inverted or not inverted. The characteristics of the image are decided by the type of lens used, the focal length of the lens, and the distance of the object from the lens. A divergent lens always forms a virtual mage irrespective of the object's distance.

Formulae:

The lens formula, $\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ (i)

The magnification formula of the lens, $m=-\frac{i}{p}$ (ii)

The image formed will be smaller than the object as the magnification $m<1.0$. Also the object distance is less than the focal distance. Thus, the image will be virtual, non-inverted. It is given that it is on the same side of the lens as O. This type of image can be formed only with a divergent type of lens.

Hence, the type of lens is diverging.

The focal length and the object distance are given as and respectively. As the lens is divergent, the focal distance should be taken as negative.

Hence, the value of focal distance is $+10\mathrm{cm}$.

From the given data in table, the object distance is$+5\mathrm{cm}$ .

Now, using the given data in equation (i), the image distance is given as follows:

$\frac{1}{i}=\frac{1}{f}-\frac{1}{p}$

$\frac{1}{i}=\frac{1}{-10}-\frac{1}{5.0}$

role="math" $=\frac{-3}{10}$

$i=-3.3\text{cm}$

Hence, the image distance is $-3.3\mathrm{cm}$.

The lateral magnification of the lens is calculated using the given data in equation (ii) as follows:

$$\begin{gathered} m=\frac{-\left(-3.33\right)}{5.0} \\ =+0.67,\text{i.e.,}m<1.0 \end{gathered}$$

Hence, the lateral magnification is $+0.67$.

The image is virtual because the image distance is negative.

Hence, the image is virtual (V).

The value of lateral magnification is positive.

Hence, the image is non-inverted (NI).

It is given in the tabular data that the image is on the same side of the lens as O.