Physics 1308 Quiz 6
Physics 1308 Quiz 6 Problem A slide projector is made by placing an illuminated slide slightly more than one focal length in front of a converging lens. The projection lens in a certain slide projector is a single thin lens. A slide 24.0 mm high is to be projected so that its image fills a screen 1.80 m high. The slidetoscreen distance is 3.0 m. (a) Determine the focal length of the projection lens. (b) How far from the slide should the lens of the projector be placed so as to form the image on the screen? Solution We first notice that with the size of the slide and the size of the screen we are given the intended magnification. Remember that the image formed is inverted: M=
h' h
=
−s' s
=
1.80m 24×10 m −3
=−75 .
We are also given the slidetoscreen distance, which is equal to the sum of the object's distance and the image's distance in the thin lens approximation, Thus ss '=3.0 m . With these two equations we can solve for s and s'. Then s = 4 cm and s'= 2.96 m. Using the thin lens equation we obtain: 1 0.04 m
1 2.96 m
=
with f = 3.95 cm.
Useful Equations: h ' −s' = h s 1 1 1 = s s' f
M=
1/1
1 f
.
h' h
=
−s' s
=
1.80m 24×10 m −3
=−75 .
We are also given the slidetoscreen distance, which is equal to the sum of the object's distance and the image's distance in the thin lens approximation, Thus ss '=3.0 m . With these two equations we can solve for s and s'. Then s = 4 cm and s'= 2.96 m. Using the thin lens equation we obtain: 1 0.04 m
1 2.96 m
=
with f = 3.95 cm.
Useful Equations: h ' −s' = h s 1 1 1 = s s' f
M=
1/1
1 f
.
