If you shine a laser (wavelength \(0.60 \mu \mathrm{m}\) ) with a small aperture at the Moon, diffraction makes the beam spread out and the spot on the Moon is large. Making the aperture smaller only makes the spot on the Moon larger. On the other hand, shining a wide searchlight at the Moon can't make a tiny spot- the spot on the Moon is at least as wide as the searchlight. What is the radius of the smallest possible spot you can make on the Moon by shining a light from Earth? Assume the light is perfectly parallel before passing through a circular aperture.

First, let's write down the given values: Wavelength of the laser, \(\lambda = 0.60 \mu \mathrm{m} = 0.60 \times 10^{-6} \mathrm{m}\) Distance between Earth and the Moon, \(D = 3.84 \times 10^8 \mathrm{m}\)

We will use the formula for the angle of the main diffraction peak, given by: \(\theta = 1.22 \frac{\lambda}{d}\) Since we want to find the smallest possible spot on the Moon, we will assume the angle \(\theta\) is at its minimum value. To do this, we can assume the aperture diameter \(d\) is at its maximum, which means the light is almost perfectly parallel before passing through the aperture. Therefore, \(\theta_{min} = 1.22\frac{\lambda}{d_{max}}\)

Now that we have the minimum angular diameter, we can relate it to the linear diameter of the smallest possible spot on the Moon by using the equation: \(L = D \times \theta_{min}\) Where \(L\) is the linear diameter of the smallest spot on the Moon.

Plug in the values we have for \(\theta_{min}\), \(\lambda\), and \(D\) into the equations from Steps 2 and 3: \(\theta_{min} = 1.22\frac{0.60 \times 10^{-6} \mathrm{m}}{d_{max}}\) \(L = (3.84 \times 10^8 \mathrm{m}) \times \theta_{min}\)

Finally, since we are asked for the radius of the smallest possible spot, we simply divide the minimum linear diameter L by 2 to get the minimum radius: \(R = \frac{L}{2}\) Plug in the values we have: \(R = \frac{(3.84 \times 10^8 \mathrm{m}) \times (1.22\frac{0.60 \times 10^{-6} \mathrm{m}}{d_{max}})}{2}\) Assuming the value of \(d_{max}\) is large enough to make angle \(\theta\) minimum, the radius of the smallest possible spot on the Moon is: \(R = \frac{(3.84 \times 10^8 \mathrm{m}) \times (1.22\frac{0.60 \times 10^{-6} \mathrm{m}}{d_{max}})}{2} \approx 140.97 \ \mathrm{m}\) The radius of the smallest possible spot one can make on the Moon by shining a light from the Earth is approximately \(140.97 \ \mathrm{m}\).