Chapter 24, Problem 83GP

To determine

The width of the beam when it reaches the Moon.

Answer to Problem 83GP

  $47.9\text{km}$

Explanation of Solution

Given data:

The width of slit, $\text{D}=0.010\text{m}$

The wavelength, $\lambda =6.30\times {10}^{-9}\text{m}$

Formula used:

  $\sin \theta =\frac{\lambda }{\text{D}}$ and $\text{y}=\text{L}\text{tanθ}$

Where,

  $D$ is width of the slit.

  $\lambda$ is wavelength.

  $y$ is distance.

  $\theta$ is an angle.

Calculation:

The angular half-width $\text{θ}$ : $\sin \theta =\frac{\lambda }{\text{D}}$

  $\sin \theta =\frac{\lambda }{\text{D}}=\frac{630\times {10}^{-9}\text{m}}{0.010\text{m}}=6.30\times {10}^{-5}$

Since for small angles, $\sin \text{θ}\approx \text{tanθ}$

  $\text{y}=\text{L}\sin \text{θ}=3.80\times {10}^8\text{m}\times \text{6}\text{.30}\times {\text{10}}^{-5}\times 2.394\times {10}^4\text{m}$

Width of the central maximum is

  $\text{2y}=4.788\times {10}^4\text{m}=\text{47}\text{.9}\text{km}$

Conclusion: The width of the beam will be $\text{47}\text{.9}\text{km}$ .