Answer to Question #183793 in Optics for Tino

Question

1. The film in Seymour Cinema’s movie projector is placed 25.0 cm away from the lens which has a focal length of 20.0 cm.  
2. How far away from the lens is the image on the movie screen?
3. If the film (object) has a height of 5.0 cm, what is the height of the projected image?
4. What is the magnification?

Answer 1

1.) Object distance u= -25.0 cm

Focal length of convex lens = +20.0 cm

2.) Image distance v= ?

Using lens formula,

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\ \\\\ v=\dfrac{f\times u}{f+u}=\dfrac{-500}{-5}=100.0\ cm$

Movie screen will present at 100cm from the lens.

3.) Now, $\dfrac{h_i}{h_o}=\dfrac{v}{u}\\$

So, height of image $h_i=\dfrac{v\times h_o}{u}=\dfrac{100\times 5}{-25}=-20.0 \ cm$

So, Height of image projected image 20.0 cm but it will be inverted.

4.) Magnification $m= \dfrac{h_i}{h_o}=\dfrac{-20.0}{5.0}=-4$