A Pinhole Camera in the Classroom
Students make a pinhole camera from a cardboard box 20 cm long. A square hole of side 2.0 mm is punched in one face and tracing paper is stretched across the opposite face as a screen. The camera is pointed at a window 3.0 m away. A clear inverted image of the window appears on the screen. A student then enlarges the hole to 10 mm, expecting a brighter and sharper image. Instead the image becomes brighter but noticeably blurred. The students are asked to explain the trade-off and to predict what happens if the box is lengthened to 40 cm while keeping the 2.0 mm hole.
1. Explain why enlarging the hole brightens but blurs the image, using the idea of overlapping images.
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Each point on the window sends light through the hole to the screen. With a small hole, each point produces a small patch on the screen, so neighbouring patches do not overlap much and the image is sharp. The patch size is set by the hole size and the geometry: light from one point spreads through the hole and lands on the screen as a patch roughly the size of the hole (scaled by the distances involved).
When the hole is enlarged to 10 mm, each point on the window produces a larger patch on the screen. Adjacent patches now overlap, so detail is washed out and the image blurs. At the same time, more light passes through the larger hole, so the image is brighter. This is the trade-off: brightness improves, sharpness worsens.
There is a limit to how small the hole should be. If the hole is made too small, diffraction of light at the edges becomes significant and the image blurs again, and the image also becomes very dim. So there is an optimum hole size, and 2.0 mm is a reasonable compromise for this box.
2. Predict the effect of lengthening the box to 40 cm with the 2.0 mm hole, and explain the reasoning in terms of similar triangles.
Read the worked answer
The image is formed by light travelling in straight lines through the hole. For an object of height H at distance D from the hole, and a screen at distance d behind the hole, similar triangles give:
image height / object height = d / D
So image height = H × d/D.
With the original box, d = 20 cm = 0.20 m and D = 3.0 m, so the magnification is 0.20/3.0 ≈ 0.067. If the window is, say, 1.5 m tall, the image is about 0.10 m tall.
With the longer box, d = 0.40 m, so the magnification doubles to about 0.133 and the image is about 0.20 m tall. The image is larger and, because the same light is spread over a larger area, dimmer per unit area.
The sharpness also changes. The blur patch from a hole of size w scales roughly as w × (D + d)/D, so a longer box increases the blur slightly. The 2.0 mm hole may now be slightly too large for a sharp image, and a smaller hole might be better. The student should recognise that lengthening the box magnifies the image but does not improve resolution.