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 object distance , $p=+16\text{cm}$
2. The lateral magnification,$m=-0.25$

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)

Using the given data in equation (ii), the image distance is given as follows:

$$\begin{gathered} i=-mp \\ =-\left(-0.25\right)\times 16 \\ =+4.0\text{cm} \end{gathered}$$

The positive value of i gives us the following information that the lens is converging.

Hence, the lens is converging as only this type of lens gives real image.

The focal length of the lens is calculated using the given data in equation (i) as follows:

$$\begin{gathered} \frac{1}{f}=\frac{1}{p}+\frac{1}{i} \\ \frac{1}{f}=\frac{1}{16}+\frac{1}{4.0} \\ =\frac{5.0}{16} \\ f=3.2\text{cm} \end{gathered}$$

Hence, the focal distance is$3.2\mathrm{cm}$ .

From the given table, the value of object distance is$+16\mathrm{cm}$ .

From part (a) calculations, the image distance is $+4.0\mathrm{cm}$ .

From the table data, the lateral magnification is found to be$+0.25$ .

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.

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.