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 $\mathit{i}$ 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 $\mathbf{\left(}\mathit{I}\mathbf{\right)}$ from object or non-inverted $\mathbf{\left(}\mathit{N}\mathit{I}\mathbf{\right)}$ , and (e) on the same side of the lens as object $\mathit{O}$or on the opposite side.

The object distance is $P=+12cm$

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

The distance between a focal point and the lens is $f=31\mathrm{mm}$.

We can use the concept of the lens formula. The diverging lens can only form a virtual image.

Formula:

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

(a)

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

$f=-31\mathrm{mm}$

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

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

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+12cm\right)\left(-31cm\right)}{\left(+12cm\right)-\left(-31cm\right)} \\ =-8.7cm \end{gathered}$$

The image distance $i=-87\mathrm{mm}$.

(b)

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

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(-8.7cm\right)}{+12cm} \\ =+0.72 \end{gathered}$$

The lateral magnification of the object is $m=+072$

(c)

Whether the image is real$\left(R\right)$or virtual$\left(V\right)$:

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

(d)

Whether the image is inverted from object$\left(I\right)$or non -inverted$\left(NI\right)$:

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

(e)

The position of the image:

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