A Lens, a Screen, and a Claimed Image
A converging lens of focal length 15 cm is used to project an image of a small illuminated arrow onto a screen. The arrow is placed 25 cm from the lens. A student moves the screen until a sharp image appears and measures the image to be larger than the object. A second student argues that because the image is real and inverted, it must be smaller, 'like a camera.' Later the object is moved to 10 cm from the lens and the student looks for an image on the screen but finds none. The lens is thin and the setup is in air.
1. Use the thin-lens equation to locate the image and compute the magnification for the 25 cm object position. Resolve the disagreement about image size.
Read the worked answer
1/f = 1/d_o + 1/d_i → 1/15 = 1/25 + 1/d_i → 1/d_i = 1/15 − 1/25 = (5−3)/75 = 2/75 → d_i = 37.5 cm. Magnification m = −d_i/d_o = −37.5/25 = −1.5. The image is real (d_i > 0), inverted (m negative), and larger (|m| = 1.5). The second student conflates 'real and inverted' with 'reduced'; a camera is reduced only because its object distance is much greater than f. Here d_o is between f and 2f, so the image is magnified. The screen position (37.5 cm) and the measured enlargement are consistent.
2. Explain why no screen image forms when the object is 10 cm from the lens, and describe what the observer sees looking through the lens.
Read the worked answer
With d_o = 10 cm < f = 15 cm: 1/d_i = 1/15 − 1/10 = (2−3)/30 = −1/30 → d_i = −30 cm. A negative image distance means a virtual image on the same side as the object; light diverges after the lens and never converges on a screen, so no sharp image can be projected. Magnification m = −(−30)/10 = +3: upright and enlarged. The observer looking through the lens sees a magnified upright virtual image - the magnifying-glass regime. The judgment: a screen only catches real images (d_i > 0); absence of a screen image does not mean 'no image.'
3. The student now places the object exactly at the focal point (15 cm). Predict the outcome and explain the limiting behavior.
Read the worked answer
1/d_i = 1/15 − 1/15 = 0 → d_i → ∞. The rays emerge parallel, so the image is at infinity and cannot be focused on any screen; effectively no image forms at a finite distance. As d_o → f⁺, d_i → +∞ and |m| → ∞; as d_o → f⁻, d_i → −∞. The focal point is the boundary between real and virtual image formation for a converging lens. Recognizing this discontinuity (not a finite large number) is the conceptual point.