How Do You Calculate the Focal Length and Placement of a Projection Lens?

[nemzy]

The projection lens in a certain slide projector is a single thin lens. A slide 20.0 mm high is to be projected so that its image fills a screen 2.0 m high. The slide-to-screen distance is 3.00 m.
(a) Determine the focal length of the projection lens.

(b) How far from the slide should the lens of the projector be placed in order to form the image on the screen?

This is how i attempted to solve the problem

(a) p=3m
M=h'/h = -q/p

so... h'/h = -q/p

solve for -q,

then plug in the q value for 1/p + 1/q = 1/f and solve for f

*is q suppose to be positve or negative, i am not sure how the sign conventions works

anyways, i get the wrong answer...anyone know whawt i did wrong?

also i have no idea how to do part B

 

[clive]

It seems that $$p+q=3 m$$ and not $$p=3 m$$, Try with to find p and q from
$$\frac{p}{q}=\frac{2 m}{20 mm}=100$$
and
$$p+q=3$$
and then compute f with
$$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$$.

OBS. I think you use the physical sign-convention, then all the variables in my post are positive.
p-distance from the screen to the lens;
q- distance from the lens to the slide.

 

[wais]

To determine the focal length of the projection lens, we can use the thin lens formula: 1/f = 1/p + 1/q, where f is the focal length of the lens, p is the object distance (distance from slide to lens), and q is the image distance (distance from lens to screen).

(a) First, we need to find the image height (h') using the magnification formula: M = -q/p. Since the image height is given as 2.0 m and the object height is 20.0 mm, we can plug in the values and solve for q: 2.0 m/20.0 mm = -q/3.00 m. This gives us q = -0.06 m.

Now, we can plug in the values for p and q into the thin lens formula: 1/f = 1/3.00 m + 1/-0.06 m. Solving for f, we get a focal length of approximately 3.01 m.

(b) To determine how far from the slide the lens should be placed, we can use the thin lens formula again. We know that the image distance (q) is -0.06 m and the focal length (f) is 3.01 m. So, we can solve for p (object distance): 1/3.01 m = 1/p + 1/-0.06 m. This gives us p = 3.01 m -0.06 m = 3.07 m. Therefore, the lens should be placed 3.07 m from the slide in order to form the image on the screen.

In terms of the sign conventions, we can use the following guidelines:
- For object distances (p), use a positive sign when the object is in front of the lens and a negative sign when it is behind the lens.
- For image distances (q), use a positive sign when the image is formed in front of the lens (on the same side as the object) and a negative sign when it is formed behind the lens (on the opposite side of the object).
- For focal length (f), use a positive sign for converging lenses (convex) and a negative sign for diverging lenses (concave).

In this problem, the object (slide) is in front of the lens, so p is positive. The image is formed on the