91影视

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.