The projection lens in a certain slide projector is a single thin lens. A slide 24.2 mm high is to be projected so that its image fills a screen 1.74 m high. The slide-to-screen distance is 3.04 m.
(a) Determine the focal length of the projection lens.
mm
(b) How far from the slide should the lens of the projector be placed to form the image on the screen?
(a) Determine the focal length of the projection lens.
mm
(b) How far from the slide should the lens of the projector be placed to form the image on the screen?
-
u+v = 3.04m
magnification m = v / u = 1.74 / 0.0242 = 71.9
so v = 71.9*u
so u + 71.9*u = 3.04m
u = 3.04 / 72.9 = 0.04170 m
so v = 3.04000 - 0.04170 = 2.9983 m
(a) focal length f = u*v / (u+v) = 0.0417 * 2.9983 / 3.0400 = 0.04113 = 41.1 mm (3 sig. figs)
(b) from above, distance from slide to lens is u = 0.04170 m = 41.7 mm (3 sig. figs)
[because of the high magnification from one lens the distances are very critical]
magnification m = v / u = 1.74 / 0.0242 = 71.9
so v = 71.9*u
so u + 71.9*u = 3.04m
u = 3.04 / 72.9 = 0.04170 m
so v = 3.04000 - 0.04170 = 2.9983 m
(a) focal length f = u*v / (u+v) = 0.0417 * 2.9983 / 3.0400 = 0.04113 = 41.1 mm (3 sig. figs)
(b) from above, distance from slide to lens is u = 0.04170 m = 41.7 mm (3 sig. figs)
[because of the high magnification from one lens the distances are very critical]