A high-powered laser beam $\mathbf{\lambda }\mathbf{=}\mathbf{600}\mathbf{}\mathbf{nm}$with a beam diameter of 12 cm is aimed at the Moon, $\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{km}$distant. The beam spreads only because of diffraction. The angular location of the edge of the central diffraction disk (see Eq. 36-12) is given by$\mathbf{sin\theta }\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{\lambda }}{\mathbf{d}}$ , where dis the diameter of the beam aperture. What is the diameter of the central diffraction disk on the Moon’s surface?

The wavelength of the laser beam, $\lambda =600\mathrm{nm}$

Diameter of the beam, $d=12\mathrm{cm}$

Distance of the Earth from the Moon,$R=3.8\times {10}^5\mathrm{km}$

The angular location of the edge of the central diffraction disk,$\sin \theta =\frac{1.22\lambda }{d}$

The brightest central region of the diffraction pattern is called the central maximum. The center maximum is the broadest and most intense.

The "cone" of light has an apex angle equal to 2. Thus, by making small-angle approximations as the angle value is almost equal to the sine value of the angle, the diameter of the spot by substituting the given diffraction formula in the diameter equation.

Formula:

If we make the small-angle approximation, then the diameter of the spot on the Moon is given by,

$D=2R\theta$ ….. (1)

From the given formula of the angular location of the edge of the central diffraction disk considering small-angle approximations,

$\theta =\frac{1.22\lambda }{d}$

Now, using this value in equation (1), the diameter of the spot on the Moon is as follow.

$$\begin{gathered} D=2R\left(\frac{1.22\lambda }{d}\right) \\ =\frac{2\left(3.8\times {10}^{8}m\right)\left(1.22\right)\left(600\times {10}^{-9}m\right)}{\left(0.12m\right)} \\ =4.7\times {10}^3\mathrm{m} \\ =4.7\mathrm{km} \end{gathered}$$

Hence, the value of the diameter is 4.7 km.