When a u–v table contains one impossible row
A student determines the focal length of a thin convex lens by the u–v method on an optical bench. The object is a pin on a stand, the image is located with a second pin and a translucent screen, and all distances are recorded as positive magnitudes measured from the lens. Her data, written as (u, v), are: (20.0 cm, 60.0 cm), (25.0 cm, 37.5 cm), (30.0 cm, 30.0 cm) and (12.0 cm, 60.0 cm). For the last pair she notes that the screen had to be removed and the image could only be seen by looking through the lens from the far side, but she records v = 60.0 cm anyway. She then plots 1/v against 1/u. The teacher asks her to identify which pair is inconsistent with the others, to use the graph to obtain f and to comment on the slope and intercepts, and to predict what she would see if the lower half of the lens were covered with black paper.
1. Use the three consistent pairs to find the focal length, then show why the fourth pair cannot belong to a real-image measurement.
Read the worked answer
For a real image formed by a thin convex lens, 1/f = 1/v + 1/u using magnitudes.
- 1/60.0 + 1/20.0 = 0.0167 + 0.0500 = 0.0667 cm−1 → f = 15.0 cm
- 1/37.5 + 1/25.0 = 0.0267 + 0.0400 = 0.0667 cm−1 → f = 15.0 cm
- 1/30.0 + 1/30.0 = 0.0333 + 0.0333 = 0.0667 cm−1 → f = 15.0 cm
For u = 12.0 cm, which is less than f = 15.0 cm, the lens formula gives 1/v = 1/15.0 − 1/12.0 = (4 − 5)/60 = −1/60, so v = −60 cm. The image is virtual, erect, on the same side as the object, and therefore cannot be thrown on a screen. The magnitude 60 cm happens to match her figure, but the sign and the claim of a screen image are wrong. If the +60 cm were accepted, 1/60 + 1/12 = 0.100 cm−1 would give f = 10 cm, contradicting the other three rows.
2. Explain how the graph of 1/v against 1/u exposes the bad point, and what its slope and intercepts should be.
Read the worked answer
Rearranging the lens formula gives 1/v = −(1/u) + 1/f, which is a straight line in 1/v versus 1/u with slope −1 and intercept 1/f on both axes.
With f = 15.0 cm, both intercepts should be 0.0667 cm−1. The good points are (0.0500, 0.0167), (0.0400, 0.0267) and (0.0333, 0.0333): each satisfies y + x = 0.0667. The suspect point is (1/12.0, 1/60.0) = (0.0833, 0.0167), and 0.0833 + 0.0167 = 0.100, not 0.0667, so it lies far above the line.
A graph is the right tool because a single arithmetic or sign slip shows up as an outlier, whereas simply averaging f from all four rows would bias the result. The slope being exactly −1 is also a check on sign conventions: if a student plots 1/v against 1/u with mixed signs, the line will not be straight with slope −1.
3. Predict what the student would observe with the lower half of the lens covered, and explain the reasoning.
Read the worked answer
The full image still forms at the same position, but it is dimmer - roughly half as bright.
Covering the lower half removes a bundle of rays but does not move the convergence point: every point of the object still sends rays through the uncovered upper half, and those rays are refracted to the same image point. So the image is complete, not half an image, and its position and magnification are unchanged to first order.
The brightness falls because the effective aperture is halved, and the depth of focus increases slightly. The common wrong answer is that half the image disappears; that would require blocking one half of the object, not one half of the lens. The same logic explains why stopping down a lens darkens the image without cropping it.