Chapter 25, Problem 52P

To determine

The diameter of the flash light beam when it reaches the Moon.

Answer to Problem 52P

Solution:

The diameter of the beam when it reaches the Moon is found to be $1.0\times {10}^4\text{m}$

Explanation of Solution

When a powerful flash light is shone towards the Moon, the beam spreads out on the surface of the Moon due to diffraction. The angular half width of the Airy’s disk formed on the surface of the Moon is given by,

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

Where, the wavelength of light is $\lambda$ and the diameter of the aperture is D.

If, the Earth- Moon distance is denoted by R and the diameter of the Airy’s disk is d then,

$2\theta =\frac{d}{R}$

From the two equations, the diameter of the beam on the surface of the beam is given by,

$d=\left(2\theta \right)R=2\left(\frac{1.22\lambda }{D}\right)R$

Given:

The diameter of the aperture $D=5.0\text{ cm}=5.0\times {10}^{-2}\text{m}$

Wavelength of light $\lambda =550\text{ nm}=5.5\times {10}^{-7}\text{m}$

Earth-Moon distance (from standard tables)$R=3.84\times {10}^8\text{m}$

Formula used:

The diameter of the disk of the beam on the surface of the Moon is given by,

$d=2\left(\frac{1.22\lambda }{D}\right)R$

Calculation:

Substitute the given values in the expression for diameter and calculate the value.

$\begin{array}{c}d=2\left(\frac{1.22\lambda }{D}\right)R\\ =2\left(\frac{1.22\left(5.5\times {10}^{-7}\text{m}\right)}{5.0\times {10}^{-2}\text{m}}\right)\left(3.84\times {10}^8\text{m}\right)\\ =1.0306\times {10}^5\text{m}\\ =1.0\times {10}^4\text{m }\left(2sf\right)\end{array}$