Question

Sunrays pass through a pinhole in the roof of a hut and produce an elliptical spot on the floor. The minor and major axes of the spot are 6 cm and 12 cm respectively. The angle subtended by the diameter of the sun at our eye is 0.5°. Calculate the height of the roof.

• A. The height of the roof is $\frac{2160}{\pi}$ cm.
• B. The height of the roof is approximately 687.55 cm.
• C. The height of the roof is approximately 6.88 m.
• D. The height of the roof is $\frac{360}{\pi}$ cm.

Answer 1

Answer: (A), (B), (C)

The height of the roof is $\frac{2160}{\pi}$ cm. Numerically, this is approximately 687.55 cm (or 6.88 m).

Answer 2

The sun's angular diameter is given as $\theta = 0.5^\circ$. Converting to radians: $\theta = 0.5^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{360}$ radians.

In a pinhole camera, the size of the image ($d_{image}$) is related to the object's angular size ($\theta$) and the distance from the pinhole to the screen ($h$) by the formula: $d_{image} = h \times \theta$

When a circular spot is projected onto an inclined plane, it forms an ellipse. The minor axis of the ellipse is equal to the diameter of the original circular spot. Given: Minor axis = 6 cm Major axis = 12 cm

Therefore, the diameter of the circular spot ($d$) is equal to the minor axis, so $d = 6$ cm.

Now, using the pinhole camera formula: $d = h \times \theta$ $6 \text{ cm} = h \times \frac{\pi}{360}$

Solving for $h$: $h = \frac{6 \times 360}{\pi} \text{ cm}$ $h = \frac{2160}{\pi} \text{ cm}$

Using $\pi \approx 3.14159$: $h \approx \frac{2160}{3.14159} \text{ cm} \approx 687.55 \text{ cm}$ This is approximately 6.88 meters.