1. ## lens projection

A slide projector has a coverging 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 108.70 mm from the lens?
m

(b) If the slide measures 24.0 mm 36.0 mm, what are the dimensions (in mm) of its image?
mm x mm

need help solving the lens problems please.

2. Originally Posted by rcmango
A slide projector has a coverging 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 108.70 mm from the lens?
m

(b) If the slide measures 24.0 mm 36.0 mm, what are the dimensions (in mm) of its image?
mm x mm
to (a): Use

$\frac1{d_O} + \frac1{d_I} = \frac1f$

Plug in the values you know:

$\frac1{108.7} + \frac1{d_I} = \frac1{104}$ . Solve for $d_I$ . I've got: $d_I \approx 2405.3\ mm$

So the distances of the screen from the lens should be 2.41 m

to (b) Use

Let O denote the length of the object (here: the measures of the slide) and I the length of the image:

$\frac IO = \frac{d_I}{d_O}$ . Plug in the values you know:

$\frac I{O} = \frac{2405.3}{108.7} \approx \underbrace{22.128}_{enlargement\ factor}$

Therefore $24\ mm \mapsto 531.1\ mm$ and $36\ mm \mapsto 796.6\ mm$

3. Okay, I kinda see how to plug them in now, but I was getting confused because I was reading d0 and d0- to mean the right or the left of the lens, and i didn't think I had that information. Even though its easily seen what the focal point is. Thanks for explaining the setup. Much more clear.