Two Lens Positions on a Fixed Optical Bench
A student is asked to find the focal length of a converging lens using a pin and a screen fixed 90.0 cm apart on an optical bench. With the lens at one position a sharp inverted image of the pin is formed on the screen; moving the lens 30.0 cm towards the screen gives a second sharp image. In the first position the image height is 24 mm while the pin height is 12 mm. The student uses the two lens positions to calculate f and then repeats the experiment with a second lens labelled f = 25 cm, keeping the pin and screen at the same 90.0 cm separation. She is surprised to find that no sharp image can be obtained at any lens position, and concludes that the second lens is defective. The bench scale is read to the nearest millimetre and the lens centres are aligned.
1. Calculate the focal length from D = 90.0 cm and d = 30.0 cm, find the two object distances and magnifications, and check consistency with the lens equation.
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The Bessel result for a fixed object-screen separation D with two lens positions separated by d is:
f = (D² − d²)/(4D) = (90.0² − 30.0²)/(4 × 90.0) = (8100 − 900)/360 = 7200/360 = 20.0 cm.
The two positions are u1 = (D − d)/2 = 30.0 cm, v1 = 60.0 cm and u2 = 60.0 cm, v2 = 30.0 cm.
Lens equation for the first: 1/30.0 + 1/60.0 = 3/60.0 = 1/20.0, so f = 20.0 cm, consistent.
Magnifications: m1 = v1/u1 = 2.0 and m2 = v2/u2 = 0.50. The image height 24 mm for a 12 mm pin confirms m1 = 2.0.
2. Explain why no sharp image is formed with the f = 25 cm lens when D = 90.0 cm. Calculate the minimum separation needed.
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A converging lens forms a real image on a screen only if the object-screen separation is at least 4f. This follows from u + v = D with 1/u + 1/v = 1/f; eliminating v gives u² − Du + Df = 0, whose discriminant is D² − 4Df.
For f = 25 cm and D = 90 cm: D² − 4Df = 8100 − 4 × 90 × 25 = 8100 − 9000 = −900 < 0, so there is no real object distance and no real image at any lens position.
The minimum separation is D_min = 4f = 100 cm. At exactly 100 cm there is one position (u = v = 50 cm) with magnification 1. The lens is not defective; the bench arrangement simply cannot produce a real image with that focal length.
3. The student says the two sharp images are 'the same image'. Use magnification to refute this. Which measurement limits the focal length most, and what is the resulting uncertainty?
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The images are not the same: m1 = 2.0 (enlarged, 24 mm) and m2 = 0.50 (reduced, 6 mm), and they are formed at different lens positions. Both are sharp because both satisfy the lens equation, but they are different images.
For the uncertainty, f = D/4 − d²/(4D), so ∂f/∂D = 1/4 + d²/(4D²) ≈ 0.25 + 0.028 = 0.278 and ∂f/∂d = −d/(2D) = −0.167. With δD = δd = 1 mm (bench read to the nearest mm), δf ≈ √[(0.278 × 1)² + (0.167 × 1)²] mm ≈ 0.32 mm.
So f = (20.0 ± 0.3) cm, about 1.6%. The D measurement contributes slightly more than d, and both matter more than the image-height measurement, which is not used in f.