Question

A slide projector has a converging lens whose focal length is 104.00 mm. (a) How far (in meters) from the lens must the screen be located if a slide is placed 107.66 mm from the lens? If the slide measures 24.0 mm (width) x 36.0 mm (height), what are the magnitudes of the (b) width (in mm) and (c) height (in mm) of its image?

Answer 1

given
focal lenght oc lens f = 104mm = 0.104m
object distance p = 107.66mm = 0.10766m
from the lens formula
1/p + 1/q = 1/f
1/q = 1/f - 1/p
= 1/0.104m - 1/0.10766m =

q = 3.059 m

b) the dimensions of the image on the screen are
xi = - q/p(x0)
=- 3.059m/ 0.10766m)(36mm)
= - 1022.89mm

yi = -q/p( y0)
==- 3.059m/ 0.10766m)(24mm)
= - 681.93mm
the dimensions of image are 681.93mm x 1022.89mm

Answer 2

I have figured this out for you by just teaching myself how this works. I had no idea how to do this 5 minutes ago but now I know. You should try it yourself sometime and that way you will learn more! Good luck!

1/S1 + 1/S2 = 1/f

1/S2 = 1/f - 1/S1

1/S2 = 1/0.154 + 1/0.15835

S2 = 5.6 m

Scale slide accordingly.