69 through 79 76, 78 75, 77 More lenses. Object Ostands 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 (C)or diverging (D), (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(R) or virtual (V), (g) inverted (I)or non-inverted (NI)from O, and (h) on the same side of the lens as Oor on the opposite side. Fill in the missing information, including the value of mwhen only an inequality is given, where only a sign is missing, answer with the sign.

1. The object distance, p=+10cm
2. The lateral magnification, m=0.50
3. The image is not inverted.

Formula:

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

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

(a)

The image formed is smaller than the object as the magnification m<1.0. Also the image is not inverted. So the image will be virtual. A divergent lens always forms a reduced image, i.e., m<1.0 .

Hence, the type of lens is diverging.

(b)

The image distance is given using the given data in equation (ii) as follows:

$$\begin{gathered} i=-mp \\ =-0.5\times 10 \\ =-5.0cm \end{gathered}$$

As the lens used is divergent, the focal distance should be taken as negative. Now, using the given data in equation (i), the focal length of the lens can b given using the data in equation (i) as follows:

$$\begin{gathered} \frac{1}{f}=\frac{1}{10}+\frac{1}{-5.0} \\ =\frac{-1}{10} \\ =-10cm \end{gathered}$$

Hence, the focal length is -10cm.

(c)

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

(d)

The image distance is given using the given data in equation (ii) as follows:

$$\begin{gathered} i=-mp \\ =-0.5\times 10 \\ =-5.0cm \end{gathered}$$$$\begin{gathered} i=-mp \\ =-0.5\times 10 \\ =-5.0cm \end{gathered}$$

The image distance is -5.0cm.

(e)

From the given data in table, the lateral magnification is +0.50.

(f)

The value of image distance is negative.

Hence, the image is virtual (V)

(g)

The value of lateral magnification is positive.

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

(h)

From the above data, it is given that p<f.

Hence, the image is on the same side as object.