The Lens Focal Length Practical
A student uses a converging lens to form an image of a distant window on a screen and measures the lens-to-screen distance as 15.2 cm. They then place a candle 25.0 cm from the lens and move the screen to find a sharp image, measuring 38.5 cm. A classmate says the second measurement contradicts the first because the image distance should equal the focal length. Another student measures the lens diameter as 4.0 cm and claims the image must be brighter than the object. The group must decide how to interpret the two measurements and what the image looks like.
1. Use the two measurements to find the focal length by each method and judge whether they are consistent.
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For a distant object, the image distance approximates the focal length, so f ≈ 15.2 cm. For the candle at u = 25.0 cm with v = 38.5 cm, the lens equation gives 1/f = 1/u + 1/v = 1/25.0 + 1/38.5 ≈ 0.0400 + 0.0260 = 0.0660, so f ≈ 15.2 cm. The two methods agree closely. The classmate's objection is wrong: the image distance equals the focal length only when the object is at infinity. For a nearby object the image forms beyond the focal point, as observed. The consistency supports the value f ≈ 15.2 cm.
2. Describe the nature of the image formed in the second case and assess the claim about brightness.
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With u = 25.0 cm and f = 15.2 cm, the object lies beyond the focal point but less than 2f (30.4 cm), so the image is real, inverted, and magnified. Magnification m = v/u = 38.5/25.0 ≈ 1.54, so the image is about 1.5 times the object's size. The brightness claim is flawed: a lens spreads the same light over a larger image area, so the image is generally dimmer per unit area, not brighter, and the lens diameter affects light gathered but not the magnification. The judgement is that the image is real, inverted, magnified, and no brighter than the object.