69 through 79 76, 78 75, 77 More lenses. Objectstands 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 or diverging , (b) the focal distance , (c) the object distance p, (d) the image distance , and (e) the lateral magnification . (All distances are in centimetres.) It also refers to whether (f) the image is real or virtual , (g) inverted or non-inverted from , and (h) on the same side of the lens asor 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. Object distance,$p=+8.0cm$
2. The focal length,$f=20cm$
3. The lateral magnification,role="math" localid="1662984996239" $m<1.0$
4. The image is non-inverted $\left(NI\right)$

Here, we need to use the concept of image formation by the thin lens. We can use equation 34.9 to solve for the image distance. The magnification of the lens can be calculated using equation 34.7. By using the values of image distance and magnification, and comparing the value of object distance and focal length we can determine whether the image is real or virtual, whether it is inverted or non-inverted, and whether it is on the same side as the object or on the opposite side.

(a)

Since, the image is non-inverted (NI) and the lateral magnification is$m<1.0$.

Hence, the lens is a diverging lens.

(b)

As the lens is diverging, the focal length will be negative.

Thus, the value of focal length is given as:$f=-20cm$

Hence, the focal distance is $-20cm$.

(c)

The object distance is$p=8.0cm$, as given in the table.

Hence, the object distance isrole="math" localid="1662985300858" $8.0cm$ .

(d)

Now, using the above data in equation (i), we can get the image distance as follows:

$$\begin{gathered} \frac{1}{i}=\frac{1}{-20\text{cm}}-\frac{1}{8.0\text{cm}} \\ =-0.175\text{cm} \\ i=-5.7\text{cm} \end{gathered}$$

Hence, the image distance is role="math" localid="1662985352773" $-5.7cm$.

(e)

Using the given data in equation (ii), we can get the lateral magnification of the lens as follows:

$$\begin{gathered} m=-\frac{-5.7}{8.0} \\ =+0.71 \end{gathered}$$

Hence, the value of magnification is$+0.71$.

(f)

The value of image distance is negative.

Hence, the image is virtual $\left(V\right)$.

(g)

The value of lateral magnification is positive.

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

(h)

From the above data, it is given that$p<f$and the image is diverging.

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