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 $\left(I\right)$ 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 lens is converging

Focal length,$f=16.0\mathrm{cm}$

Object distance, $p=+12\mathrm{cm}$

By using the thin lens equation and the formula for magnification, we can find all the required quantities.

Formula:

Thin lens equation, $\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{p}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{i}}$

Magnification, $\mathbf{m}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\mathbf{p}}$

Since the lens is converging, the focal length value should be positive, i.e.

$f=+16.0\mathrm{cm}$

Thin lens equation is

$$\begin{gathered} \frac{1}{f}=\frac{1}{p}+\frac{1}{i} \\ \frac{1}{16}=\frac{1}{12}+\frac{1}{i} \\ \frac{1}{i}=\frac{1}{16}-\frac{1}{12} \\ \frac{1}{i}=-0.021 \\ i=-47.61 \\ \approx -48cm \end{gathered}$$

Image distance =$-48cm$

Magnification is,

$\begin{array}{l}m=\frac{i}{p}\\ m=-\left(\frac{-48.0}{12}\right)\\ m=+4.0cm\end{array}$

Lateral magnification $m=+4.0$

As the image distance $i$is negative, the image is virtual$\left(V\right)$.

As the magnification is positive, the image is non-inverted $\left(NI\right)$.

For thin lens, the real images forms on the opposite side as the object and virtual images form on the same side as the object.

Since the image is non-inverted, it forms on the same side of the object.