Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to 2000 km. One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of 1.40 nm. The element that emits light is the end of a wire with diameter 0.200 mm. (a) Calculate the diameter of the central beam at a target 2000 km away from the beam source. (b) What is the ratio of the beam intensity at the target to that at the end of the wire? (The laser is fired from space, so neglect any atmospheric absorption.)

Let the diameter of the central beam be D.

The expression to calculate the diameter of the central beam is given by,

$\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}}$

Here,$\mathit{\lambda }$ is the wavelength, d is separation, and$\mathit{\theta }$ is angle.

It is known that,

$$\begin{gathered} \mathit{L}\mathit{\theta }\mathbf{=}\mathit{D} \\ \mathit{\theta }\mathbf{=}\frac{\mathbf{D}}{\mathbf{L}} \end{gathered}$$

From the equation$\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}}$ ,

$$\begin{gathered} \mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}} \\ \frac{\mathbf{D}}{\mathbf{L}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}} \\ \mathit{D}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}}\mathit{L} \end{gathered}$$ …….. (1)

Substitute$2000\times {10}^3\text{m}$ for L,$0.2\times {10}^{-3}\text{m}$ for d , and$1.4\times {10}^{-9}\text{m}$ for $\lambda$in equation (1).

$$\begin{gathered} D=\frac{1.22\left(1.4\times {10}^{-9}\text{m}\right)\left(2000\times {10}^3\text{m}\right)}{0.2\times {10}^{-3}\text{m}} \\ \approx 17.1\text{m} \end{gathered}$$

Therefore, the diameter of the central beam is $17.1m$.

The expression to calculate the beam intensity is given by,

$\frac{I_1}{I_2}={\left(\frac{d}{D}\right)}^2$ …… (2)

Substitute $0.2\times {10}^{-3}\text{m}$for d and $17.1\text{m}$for D in equation (2).

$$\begin{gathered} \frac{I_1}{I_2}={\left(\frac{0.2\times {10}^{-3}\text{m}}{17.1\text{m}}\right)}^2 \\ =1.37\times {10}^{-10} \end{gathered}$$

Therefore, the ratio of the beam intensity is $1.37\times {10}^{-10}$.