69 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 $\left(I\right)$ 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 lens is a converging type of lens.
2. The object distance, $p=+20\text{cm}$
3. The focal length of the lens, $f=10\text{cm}$
4. The lens is converging.

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 type of image is decided by the type of lens used, the focal length of the lens, and the distance of the object from the lens.

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)

From the given data in table, the lens is converging.

From the data in table, the focal distance is +10cm .

The object distance is +20 cm as given in the problem.

Using the given data in equation (i), the image distance can be given as follows:

$$\begin{gathered} \frac{1}{i}=\frac{1}{10}-\frac{1}{20} \\ =\frac{1}{20} \\ i=+20\text{cm} \end{gathered}$$

Hence, the image distance is $+20\mathrm{cm}$and positive.

Now, using the given object distance and the above image distance, we can get the lateral magnification of the lens as follows:

$$\begin{gathered} m=\frac{-20}{20} \\ =-1.0\text{cm} \end{gathered}$$

Hence, the lateral magnification is $-1.0\mathrm{cm}$.

The image is real because the image distance is positive.

Hence, the image is real (R).

The image is inverted as the lens is converging.

Hence, the image is inverted (I).

The image is on the opposite of the lens as O since the image distance is positive.

Hence, the image is on the opposite side of object.