Question

By measuring the go-and-return time for a laser pulse to travel from an Earth-bound observatory to a reflector on the Moon, it is possible to measure the separation between these bodies. (a) What is the predicted value of this time? (b) The separation can be measured to a precision of about 15 cm. To what uncertainty in travel time does this correspond? (c) If the laser beam forms a spot on the Moon 3 km in diameter, what is the angular divergence of the beam?

Short answer

(a) The predicted value of the time is $2.55s$.

(b) This precision correspond upto$5.0\times {10}^{-10}s$ of uncertainty in travel time.

(c) The angular divergence of the beam is $(7.84\times {10}^{-6})°$.

Step-by-step solution

The given data:

a) Distance of the Earth from the Moon’s surface,$d_{em}=3.82\times {10}^{8}m$

b) The precision of the separation is about $\delta d_{em}=15\mathrm{cm}$.

c) The diameter of the spot formed on the Moon,$D=3km$

Understanding the concept of laser pulse:

Pulses with pulsed lasers emit light not in continuous mode, but in the form of optical pulses (light intensity).

The distance between the Earth and the Moon helps to get the time that may the laser take to reach the Moon's surface from the Earth's observatory. Now, using the precision value and the uncertainty condition of time-distance, the uncertainty in time can be calculated. Again use the basic formula of tangent angle to find the angular divergence of the beam.

Formulas:

The time is taken by the light to travel the given distance back and forth,

$t=\frac{2d}{c}$ ….. (1)

The uncertainty condition of distance-time as

$2d_{em}\propto t$

$\frac{\delta t}{t}=\frac{\delta d_{e}m}{2d_{e}m}$ ….. (2)

The angular divergence of a beam,

$\theta =2ta{n}^{-1}\left(\frac{Radiusofthespot}{d_{e}m}\right)$ ….. (3)

(a) Calculation of the predicted time:

The time value that is predicted to be taken by the laser light to reach Moon from Earth and back to Earth can be calculated using equation (1) as follows:

$$\begin{gathered} t=\frac{2\left(3.82\times {10}^{8}m\right)}{3\times {10}^{8}m/s} \\ =2.55s \end{gathered}$$

Hence, the value of the time is 2.55 s.

(b) Calculation of the uncertainty in travel time

Using the given relation of equation (2) and the given data, the value of the uncertainty in travel time as follows:

$$\begin{gathered} \delta t=\frac{t\delta d_{em}}{2d_{em}} \\ =\frac{\left(2.55s\right)\left(0.15m\right)}{2\left(3.82\times {10}^{8}m\right)} \\ =5.0\times {10}^{-10}s \end{gathered}$$

Hence, the value of uncertainty in time is $5.0\times {10}^{-10}s$.

(c) Calculation of the angular divergence

The radius of the spot can be given as:

$$\begin{gathered} Radius=\frac{D}{2} \\ =\frac{3\mathrm{km}}{2} \\ =1.5\mathrm{km} \end{gathered}$$

Now, using this radius value in equation (3), the angular divergence of the beam is calculated as follows:

$$\begin{gathered} \theta =2{\tan }^{-1}\left(\frac{1.5\times {10}^{3}m}{3.82\times {10}^{8}m}\right) \\ =2{\tan }^{-1}\left(0.00000392\times {10}^{-5}\right) \\ =\left(7.84\times {10}^{-6}\right)° \end{gathered}$$

Hence, the value of angular divergence is $\left(7.84\times {10}^{-6}\right)°$.