Question

15. (3 points) A convex lens of focal length of 20 cm is used to form an image of an object on a screen. (The setup is somewhat similar to what you used in the lab however it is not exactly the same.) The height of the object is 20 mm and the height of the image on the screen is 40 cm. (Note UNITS in the previous sentence!) How far from the lens must the screen be placed? Report your answer in meters.

Answer 1

The magnification of the system is thus,

$\large m=\frac{Image \;height}{Object\; height}=\frac{400}{20}=20$

Therefore, $\large m=\frac{Image \; distance}{Object\; distance}=\frac{v}{u}=20$

Thus, $\large v=20u$

The question asks the value of v.

By lens equation we know,

$\large \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Multiplying the above equation by v we get,

$\large 1+\frac{v}{u}=\frac{v}{f}$

Or, $\large 1+20=\frac{v}{20}$

Hence, $\large v=20\times 21\;cm=420\;cm$