91Ó°ÊÓ

The given data can be listed below,

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

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.

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,

$2{\mathrm{θ}}_R=\frac{2×1.22\mathrm{λ}}{d_0}$

Here,$d_0$ is the diameter of the mirror on the telescope at earth,$\mathrm{λ}$ is the wavelength of the light.

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

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

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

$$\begin{gathered} \frac{d_3}{L}=2{\mathrm{θ}}_R \\ =\frac{2×1.22\mathrm{λ}}{d_2} \end{gathered}$$

The fraction of energy for this light is given by,

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

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

$$\begin{gathered} {\mathrm{η}}_0={\mathrm{η}}_1{\mathrm{η}}_2 \\ ={\left(\frac{d_0{d}_2}{2.44\mathrm{λ}{d}_{cm}}\right)}^4 \end{gathered}$$

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

Substitute all the values in the above,

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

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