Projector Lens: Fixed Screen and Magnification
A student is assembling a simple projector for a 3.0 cm tall transparency. A thin convex lens of focal length 15 cm is fixed on an optical bench, and a screen is clamped 45 cm from the lens. The student first places the transparency 20 cm from the lens and sees a blurred patch on the screen instead of a sharp image. The transparency holder can slide along the bench, but the screen cannot be moved. A second thin convex lens of focal length 10 cm is available. The lens aperture is 4.0 cm in diameter. The room is darkened so that the image is visible, and the lens may be treated as thin and free of aberrations. The student wants a sharp image on the fixed screen and then wants to know how to obtain a larger image without moving the screen.
1. Using the lens formula, find the image distance for the initial 20 cm object distance. Estimate the diameter of the blurred patch on the screen and explain why the image is blurred.
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Use 1/f = 1/v − 1/u with u = −20 cm and f = +15 cm:
1/v = 1/15 + 1/(−20) = 1/60, so v = 60 cm.
The lens would focus the image 60 cm behind the lens, but the screen is only 45 cm away. The rays have not yet converged at the screen. Treating the aperture as a cone, the blur diameter is approximately aperture × (v − 45)/v = 4.0 × (60 − 45)/60 = 1.0 cm. That is a large blur compared with the 3.0 cm object, so the screen shows a blurred patch rather than a sharp image.
2. Find the object distance that puts a sharp image exactly on the 45 cm screen. Calculate the magnification and image height. Which way must the holder move?
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Now v = +45 cm. Use 1/u = 1/v − 1/f:
1/u = 1/45 − 1/15 = (1 − 3)/45 = −2/45, so u = −22.5 cm.
The object must be 22.5 cm from the lens, i.e. it must move 2.5 cm farther from the lens than the original 20 cm position. The magnification is m = v/u = 45/(−22.5) = −2.0. The image is real, inverted, and 2.0 × 3.0 cm = 6.0 cm tall. This is a usable sharp image on the fixed screen.
3. The student now wants a larger sharp image on the same fixed screen. Should the transparency be moved toward or away from the 15 cm lens? Justify your answer and state what actually must be changed.
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For a fixed screen distance v and a fixed lens, the lens equation fixes u. With f = 15 cm and v = 45 cm, u must remain 22.5 cm. Moving the transparency either way would only blur the image; it cannot increase magnification while keeping the image on the screen.
The magnification magnitude is |m| = v/|u|. Using 1/u = 1/v − 1/f gives |m| = v/f − 1. For v = 45 cm, f = 15 cm gives |m| = 2. To increase |m| while keeping v fixed, the focal length must be reduced. With the available f = 10 cm lens, 1/u = 1/45 − 1/10 = −7/90, so u ≈ −12.9 cm and |m| = 45/12.9 ≈ 3.5. The image becomes 10.5 cm tall. Thus the student must change to the shorter-focal-length lens, or move the screen farther away; sliding the object alone will not work.