A Convex Lens, a Covered Aperture and a Missing Screen Image
A student sets up a convex lens on an optical bench. A candle flame is placed 30.0 cm from the lens, and a sharp image is obtained on a screen placed 60.0 cm on the other side. The lens is then covered with an opaque card except for a central circular hole of about half its diameter. The student moves the screen slightly and finds a dimmer but still complete image of the flame. Next, the candle is moved to 15.0 cm from the lens, and the student tries to catch the image on the screen but fails. The teacher asks for the focal length, the nature of the first image, and an explanation of the second observation. The bench scale is marked in millimetres, and the flame is treated as an object on the principal axis.
1. Calculate the focal length of the lens and describe the first image using the lens formula and magnification.
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Using the Cartesian sign convention, object distance u = −30.0 cm and image distance v = +60.0 cm for the real image on the screen. Lens formula: 1/f = 1/v − 1/u = 1/60 − (−1/30) = 1/60 + 1/30 = 3/60 = 1/20. Therefore f = +20.0 cm. Magnification m = v/u = 60/(−30) = −2. The negative sign means the image is inverted, and the magnitude 2 means it is twice as tall as the object. Because v is positive and the image is caught on a screen, the image is real. So the first image is real, inverted and magnified, formed 60 cm from the lens on the opposite side.
2. Explain why the candle at 15.0 cm does not form an image on the screen. Include the position and nature of the image.
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When the candle is 15.0 cm from the lens, the object distance is less than the focal length (15.0 cm < 20.0 cm). Using u = −15.0 cm and f = +20.0 cm: 1/v = 1/f + 1/u = 1/20 − 1/15 = (3 − 4)/60 = −1/60, so v = −60.0 cm. The negative image distance means the image is on the same side of the lens as the object. It is virtual, erect and magnified (m = v/u = −60/−15 = +4). A virtual image cannot be caught on a screen because the light rays diverge after passing through the lens; they only appear to meet when traced backward. The student must look through the lens from the opposite side to see the image.
3. Why is a complete, though dimmer, image still seen when most of the lens is covered?
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Covering most of the lens reduces the amount of light passing through it, so the image becomes dimmer, but it does not change the directions of the rays that do pass through the uncovered part. Every point of the object still sends rays to the uncovered aperture, and those rays are refracted to the same image point determined by the lens equation. The image position and size therefore remain the same; only its brightness decreases. A common misconception is that half-covering the lens should cut the image in half. That would happen only if the aperture acted like a mask in contact with the image, not when it limits the rays entering the lens.