50 through 57 55, 57 53 Thin lenses. Object $\mathbf{O}$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 $\mathbf{i}$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 $\mathbf{O}$or noninverted $\left(\mathrm{NI}\right)$, and (e) on the same side of the lens as object $\mathbf{O}$or on the opposite side.

The lens is converging

Focal length,$f=4.0\mathrm{cm}$

Object distance, role="math" localid="1663068552749" $p=+16\mathrm{cm}$

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

Thin lens equation

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

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

(a)

Since the lens is converging, the focal length value should be positive, i.e.$f=+4.0cm$

Thin lens equation is

$$\begin{gathered} \frac{1}{f}=\frac{1}{p}+\frac{1}{i} \\ \frac{1}{4}=\frac{1}{16}+\frac{1}{i} \\ \frac{1}{i}=\frac{1}{4}-\frac{1}{16} \\ \frac{1}{i}=0.1875 \\ i=5.3cm \\ imagedis\tan cei=5.3cm \end{gathered}$$

(b)

Magnification is

$$\begin{gathered} m=-\frac{i}{p} \\ m=-\frac{5.33}{16} \\ m=-0.33 \\ Lateralmagnificationm=-0.33 \end{gathered}$$

(c)

As the image distance $i$is positive, the image is real (R).

(d)

As the magnification is negative, the image is inverted$\left(l\right)$

(e)

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

Since the image is inverted, it forms on the opposite side of the object.