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, and (h) on the same side of the lens asO 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=+16.0cm.
2. The lateral magnification, m=+1.25.

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 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.

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 is greater than the object as the magnification m>1.0. Also the magnification is positive. Hence the image will be virtual, not inverted. This type of image can be formed only with a convergent type of lens.

Hence, the lens used here is convergent.

As the lens used is convergent, the focal distance should be taken as positive. Now, using the given data in equation (i), the focal distance can be given as follows:

$$\begin{gathered} \frac{1}{f}=\frac{1}{16}+\frac{1}{-20} \\ =\frac{1}{80} \\ f=+80cm \end{gathered}$$

Hence, the focal distance is +80cm.

From the given data in table, the object distance is +16cm.

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

$i=-\frac{m}{p}$

$$\begin{gathered} =-1.25\times 16 \\ =-20cm \end{gathered}$$

From the given data in table, the lateral magnification is +1.25.

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).

The image is on the same side of the lens as O since the image distance is negative.

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