Question

In an experiment to monitor the Moon’s surface with a light beam, pulsed radiation from a ruby laser (λ= 0.69 µm) was directed to the Moon through a reflecting telescope with a mirror radius of 1.3 m. A reflector on the Moon behaved like a circular flat mirror with radius 10 cm, reflecting the light directly back toward the telescope on Earth. The reflected light was then detected after being brought to a focus by this telescope. Approximately what fraction of the original light energy was picked up by the detector? Assume that for each direction of travel all the energy is in the central diffraction peak.

Short answer

The fraction of original energy picked up by the detector is$4\times {10}^{-13}$

Step-by-step solution

Identification of given data

The given data can be listed below,

• The wavelength of the light is,$\lambda =0.69\mu m$
• The diameter of the reflecting telescope at earth is,$d_0=2\times 1.3\text{m}=2.6\text{m}$
• The radius of the refractor on the moon is,$d_2=2\times 10\text{cm}=0.20\text{m}$

Concept/Significance of reflecting telescope

One surface is the recipient of light that has been bent by a reflector (the mirror). Two surfaces are in contact with the light coming through the lens. The light is weakened slightly by each interaction.

Chromatic aberration is not an issue with pure reflecting telescopes. Different hues of light are bent at various angles by lenses, on the other hand.

Determination of the fraction of the original light energy was picked up by the detector

The energy of the beam of light that is projected onto the Moon is concentrated in a circular spot of diameter$d_1$ is given by,

$2{\theta }_R=\frac{2\times 1.22\lambda }{d_0}$

Here,$d_0$ is the diameter of the mirror on the telescope at earth,$\lambda$ is the wavelength of the light.

The fraction of energy picked up by the reflector of diameter$d_2$ is given by,

${\eta }_1={\left(\frac{d_2}{d_1}\right)}^2$

The light upon reaching earth satisfies an equation which is given by,

$$\begin{gathered} \frac{d_3}{L}=2{\theta }_R \\ =\frac{2\times 1.22\lambda }{d_2} \end{gathered}$$

The fraction of energy for this light is given by,

${\eta }_2={\left(\frac{d_0}{d_3}\right)}^2$

So, the fraction of original energy picked up by detector is given by,

$$\begin{gathered} {\eta }_0={\eta }_1{\eta }_2 \\ ={\left(\frac{d_0{d}_2}{2.44\lambda d_{cm}}\right)}^4 \end{gathered}$$

Here,$d_0$ is the diameter of the mirror on the telescope at earth,$d_2$ is the diameter of reflector at moon,$\lambda$ is the wavelength of the light and$d_{em}$ is the separation between earth and moon whose value is $3.82\times {10}^8\text{m}$.

Substitute all the values in the above,

$$\begin{gathered} {\eta }_0={\left(\frac{2.6\times 0.20{\text{m}}^2}{2.44\times 0.69\times {10}^{-6}\text{m}\left(3.83\times {10}^8\text{m}\right)}\right)}^4 \\ =4\times {10}^{-13} \end{gathered}$$

Thus, the fraction of original energy picked up by the detector is $4\times {10}^{-13}$