Tutorial Exercise

A transparent photographic slide is placed in front of a converging lens with a focal length of $2\mathrm{.}46$ cm $.$ An image of the slide is formed $13\mathrm{.}7$ cm from the slide.

$($a$)$ How far is the lens from the slide if the image is real?

$($b$)$ How far is the lens from the slide if the image is virtual?

Step $1$

$($a$)$ The real image case is shown in the ray diagram.

Figure

Notice that here $p+q=13\mathrm{.}7cm,$ or $q=13\mathrm{.}7cm-p.$ The thin$-$lens equation, with focal length $f=2\mathrm{.}46cm,$ gives

$\frac{1}{p}+\frac{1}{13\mathrm{.}7(cm)-p}=$

$1$

$$

$(\frac{rcm}{?})$

or

$p^2-(13\mathrm{.}7cm)p+$

$$

Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. $c{m}^2=0.$

Using the quadratic formula to solve the above equation gives two solutions. The smaller solution is $p=$

Your.response differs from the correct answer by more than $10\%.$ Double check your calculations. cm and the larger solution is $p=$ Your response differs from the correct answer by more than $10\%.$ Double check your calculations. cm $.$

Both are valid solutic