Tutorial Exercise

image. Is the image inverted or upright? Is the image real or virtual?

Step $1$

The $1\mathrm{.}00-$cm$-$high object serves as a real object, located at a distance $p_1=3\mathrm{.}38cm$ in front of the converging lens $L_1,$ as shown in the diagram.

Figure

Upper Diagram

The converging lens is near the middle of the diagram on an Optical axis. The diverging lens is $6\mathrm{.}00$ cm $(6$ grid units$)$ right of the converging lens; and is grayed$-$out.

An arrow labeled $I_1$ is on the left of the diagram at a distance $|q_1|(8$ grid units$)$ left of the converging lens. The arrow extends vertically up from the Principal axis and has a height of $2$ grid units.

An arrow labeled $O_1$ is a distance $p_1(4$ grid units$)$ left of the converging lens. The arrow extends vertically up from the Principal axis and has a height of $1$ grid unit.

A point labeled $F_1$ is $2$ grid units right of the diverging lens.

Two rays extend from the tip of $O_1$

One ray extends horizontally from the tip of $O_1$ until it reaches the horizontal midpoint of the converging lens, whereupon it angles down to the right and ends at $F_1.$ A dashed line, parallel to the part of the ray that is to the right of the converging lens, starts from the point on the converging lens at which the ray bends, and then extends up to the left to the tip of $I_1.$

The other ray extends down and to the right from the tip of $O_1$ at a constant slope. It passes through the very center of the converging lens and maintains the same slope. A dashed line, parallel to the ray, starts at the tip of $O_1,$ and extends up and to the left to the tip of $I_1.$

Let $f_1$ be the focal length of the converging lens. The thin lens equation gives the image distance for this lens as

$q_1=\frac{{}_1{f}_1}{p_1-f_1}$

$=-\frac{(,m^2)}{(,m)-(,m)}$

$=-,$

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