Question

28. Consider a slide projector that consists of a single converging lens. The slide has an illuminated region of 2cm x 2cm. The screen is 10 m from lens. (a) Should the slides be put into the projector upside down or right side up (i.e. is the image inverted) ? (b) How far from the lens should the slide be placed so that the image has size of 2 m x 2 m (assume the focal length of the lens can be chosen to make the image in focus)? (c) If the focal length of the lens is 0.099 m (9.9 cm) will the slide be in focus on the screen? Explain. (d) If the focal length of the lens remains fixed at 0.099 m how far from the lens should the slide be placed in order to make a sharp image if the screen is moved to be only 2 m away. (e) How big will the new image be in this case?

Answer 1

Part a The converging lens will create an inverted real image, so the slide must be upside down. Part b: In this case: m=Y Yo -2 0.02 = -100 Remember that: s' 10 -100 m = -- = -- = -100 SS s= 0.1 m = 10 cm Part c: Let's calculate the s' for s = 0.1 m 1 1 1 1 1 1 5*5*f0. 15 0.099 →s' = 9.9 m The image is not focus on the screen. Part d: In this case: s =? s' = 2m Then: 1 1 1 1 1 1 + == +=+ = S= 0.104 m s's' f '2 0.099 Part e: y=yo ( 3 ) = (2 cm) (-2.104) = 38.5 cm