50 through 57 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 localid="1662982946717" $\mathit{i}$and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real $\left(R\right)$ or virtual $\left(V\right)$ , (d) inverted $\left(I\right)$ from object $\mathit{O}$ or non inverted $\left(NI\right)$, and (e) on the same side of the lens as object $\mathit{O}$or on the opposite side.

The object distance is $P=+10\mathrm{cm}$.

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

The distance between a focal point and the lens is $f=60\mathrm{cm}$.

We can use the Lens 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.

$f=-60\mathrm{cm}$.

The image distance:

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

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

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+10cm\right)\left(-6.0cm\right)}{\left(+10cm\right)-\left(-6.0cm\right)} \\ =-3.8cm \end{gathered}$$

The image distance $i=-38\mathrm{cm}$.

The lateral magnification of the object:

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

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(-3.8cm\right)}{10cm} \\ =0.38 \end{gathered}$$

The lateral magnification of the object is$m=+038$.

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$I$ or not inverted:

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

The position of the image:

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