A 2.00 cm object is 18.0 cm in front of a thin converging lens of focal length 12.0 cm. Determine the image location, magnification, image height and whether the image can be projected. The lens is then used in a classroom experiment with several object distances. Describe a graph that determines focal length without relying on a single trial, and identify two alignment controls.

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The image is 36.0 cm beyond the lens, magnification −2.00, height −4.00 cm; it is real, inverted and projectable. Plot 1/d_i against 1/d_o; intercept = 1/f. Use the thin-lens equation 1/f = 1/d_o + 1/d_i with the real-is-positive convention. Then 1/d_i = 1/12.0 − 1/18.0 = 1/36.0 cm⁻¹. Thus d_i = +36.0 cm. The positive image distance places a real image on the far side of the lens, so a screen can intercept it. Magnification m = −d_i/d_o = −36.0/18.0 = −2.00. The image height is mh_o = (−2.00)(2.00 cm) = −4.00 cm; the sign denotes inversion. Rearrange to 1/d_i = −1/d_o + 1/f. Plot y = 1/d_i against x = 1/d_o; an ideal gradient is −1 and the y-intercept is 1/f. Keep object, lens centre and screen centre on one horizontal optical axis, and keep each plane perpendicular to it. Repeat focus judgments in both directions to reduce subjective focusing bias.

Image formation and an experimental focal length

Choose your AP course · School year 2026–27 · May 2027 exam

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