Question

A canvas tent has a single, tiny hole in its side. On the opposite wall of the tent, \(2.0 \mathrm{~m}\) away, you observe a dot (due to sunlight incident upon the hole) of width \(2.0 \mathrm{~mm}\), with a faint ring around it. What is the size of the hole in the tent? (Assume a wave length of \(570 \mathrm{nm}\) for the sunlight.)

Short answer

Answer: The size of the hole in the tent is approximately 1.14 mm.

Step-by-step solution

Recall the formula for the angular width of the central maximum

The formula for the angular width of the central maximum in a single-slit diffraction pattern is given by: $$\theta = \frac{2\lambda}{b}$$ where \(\theta\): angular width of the central maximum, \(\lambda\): wavelength of the light, and \(b\): size of the hole (or slit).

Convert the given information into appropriate units

We are given the following information: Width of the sunlight dot: \(2.0 \ \text{mm}\), Distance between the hole and the opposite wall: \(2.0 \ \text{m}\), and Wavelength of the sunlight: \(570 \ \text{nm}\). We need to convert the width of the sunlight dot and the wavelength of the sunlight into meters to have consistent units. Width of the sunlight dot: \(2.0 \ \text{mm} = 2.0 \times 10^{-3} \ \text{m}\), Wavelength of the sunlight: \(570 \ \text{nm} = 570 \times 10^{-9} \ \text{m}\)

Compute the angular width of the central maximum

Knowing that the width of the sunlight dot on the opposite wall is \(2.0 \times 10^{-3} \ \text{m}\) and the distance between the hole and the wall is \(2.0 \ \text{m}\), we can compute the angle using the following formula: $$\tan(\theta) \approx \theta = \frac{\text{Width of the sunlight dot}}{\text{Distance between the hole and the wall}} = \frac{2.0 \times 10^{-3} \ \text{m}}{2.0 \ \text{m}} = 10^{-3} \ \text{rad}$$

Substitute the values into the formula and solve for the size of the hole

Now we can use the formula for the angular width of the central maximum to find the size of the hole: $$10^{-3} \ \text{rad} = \frac{2(570 \times 10^{-9} \ \text{m})}{b}$$ We can now solve for \(b\): $$b = \frac{2(570 \times 10^{-9} \ \text{m})}{10^{-3} \ \text{rad}} = 1.14 \times 10^{-3} \ \text{m}$$

Convert the size of the hole back to millimeters and state the answer

Finally, we can convert the size of the hole back to millimeters: Size of the hole: \(1.14 \times 10^{-3} \ \text{m} = 1.14 \ \text{mm}\) The size of the hole in the tent is approximately \(1.14 \ \text{mm}\).