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(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 centimetres.) It also refers to whether (f) the image is real $\left(R\right)$or virtual $\left(V\right)$, (g) inverted 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. Image distance$p=+16.0cm$
2. The lateral magnification, $m=+0.25$

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.

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 can be given as follows:

$\begin{array}{l}i=-mp\\ =-0.25\times 16\\ =-4.0\mathrm{cm}\end{array}$

Now, using the above values in equation (i), the focal length of the lens is given as follows:

$$\begin{gathered} \frac{1}{f}=\frac{1}{-4.0}+\frac{1}{16} \\ =-\frac{3}{16} \\ f=-5.3cm \end{gathered}$$

Here, the focal length and the image distance both are negative.

Hence, the lens is a diverging lens.

From the calculations of above part (a), the value of focal distance is$-5.3cm$ .

The object distance is $p=+16cm$ as given in the table.

Hence, the object distance islocalid="1663075423208" $+16cm$.

From the calculations of above part (a), the value of image distance is$-4.0cm$

The lateral magnification is $p=+0.25$as given in the table.

Hence, the lateral magnification is $+0.25.$

The value of image distance is negative.

Hence, the image is virtual (V).

The image is non-inverted (NI) because the magnification is positive.

The lens is diverging.

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