Question

The projection lens in a certain slide projector is a single thin lens. A slide 23.1 mm high is to be projected so that its image fills a screen 1.73 m high. The slide-to-screen distance is 3.01 m. (a) Determine the focal length of the projection lens. 38.5 x What is the relationship between the slide-to-screen distance and the object and image distances? mm (b) How far from the slide should the lens of the projector be placed to form the image on the screen? 40.19 X Make certain to keep a fairly large number of significant figures in your calculations to avoid round-off errors. mm

Answer 1

23.1 1.73

3.01

Let the distance of the slide from the lens be u and that of the screen be v

Then the magnitude of magnification

m = v/u = I/O

here I = height of the image

O = height of the object

v/u = 1.73/(23.1 x 10^-3)

v = 74.89*u

But the slide to screen distance is 3.01 m, hence

u + v = 3.01

u + 74.89*u = 3.01

u = 3.01/75.89

u = 0.039662 m

and

v = 2.9703 m

a)

The focal length of the lens is

1/f = 1/v – 1/u

1/f = 1/2.9703 – 1/(-0.039662)

f = 1/25.5497166

f = 0.03913 m.

f = 39.13 mm

b) The distance between the lens and slide is

u = 0.039662 m

u = 39.66 mm