55, 57 53 Thin lenses. Object $\mathit{O}$stands on the central axis of a thin symmetric lens. For this situation, each problem in Table 34-6 gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance $\mathit{i}$ and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real $\mathbf{\left(}\mathit{R}\mathbf{\right)}$or virtual $\mathbf{\left(}\mathit{V}\mathbf{\right)}$ , (d) inverted $\mathbf{\left(}\mathit{I}\mathbf{\right)}$ from object $\mathit{O}$ or noninverted $\left(NI\right)$, and (e) on the same side of the lens as object $\mathit{O}$ or on the opposite side.

The object distance is $P=+2\mathrm{cm}$

The given lens is diverging $\left(D\right)$

The distance between a focal point and the lens is $f=14\mathrm{cm}$.

We can use the Len’s formula. The diverging lens can only form a virtual image.

Formula:

$\frac{1}{f}=\frac{1}{P}+\frac{1}{i}$

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

The given lens is a diverging lens, and thus the focal length value should be negative.

$f=-14\mathrm{cm}$

For an object in front of the lens, the object distance $P$and image distance $i$ are related to the focal length of the lens.

$$\begin{gathered} \frac{1}{f}=\frac{1}{P}+\frac{1}{i} \\ \frac{1}{i}=\frac{1}{f}-\frac{1}{P} \end{gathered}$$

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+22cm\right)\left(-14cm\right)}{\left(+22cm\right)-\left(-14cm\right)} \\ =-8.6cm \end{gathered}$$

The image distance $i=-86\mathrm{cm}$.

The lateral magnification is the ratio of the object distance $P$ to the image distance $i$. It is given by

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(-8.6cm\right)}{+22cm} \\ =+0.39 \end{gathered}$$

The lateral magnification of the object is $m=+039$

Whether the image is real$R$or virtual$V$:

The image distance is negative; hence the image is virtual.

Whether the image is inverted from object or non -inverted

The value of magnification is positive; hence the image is not inverted $NI$.

The position of the image:

The value image is negative; hence the image is on the same side as the object.