## Section48.8Optical Instruments Bootcamp

### Subsection48.8.5Miscellaneous

What is the angular size of the Moon if viewed from a binocular that has a focal length of $1.2\text{ cm}$ for the eyepiece and a focal length of $8\text{ cm}$ for the objective? Use the radius of the Moon to be $1.74 \times 10^6\text{ m}$ and the distance of the Moon from the observer to be $3.8 \times 10^8\text{ m}\text{.}$

Hint

Use the angle subtended by the moon itself and magnification of the binucular.

$6.1\times 10^{-2}\text{ rad} \text{.}$

Solution

We use $s=R\theta$ with $s=\text{diameter of moon}$ and $R=\text{distance to the Moon}$ to find the angle subtended by the Moon directly to be

\begin{equation*} \theta = \frac{s}{R} = \frac{2\times 1.74 \times 10^6\text{ m}}{3.8 \times 10^8\text{ m}} = 9.158\times 10^{-3}\text{ rad}. \end{equation*}

Multiplying by the magnification we will get the desired angle. So, we need the value for magnification. From the focal lengths of the objective and the eyepiece, we get the magnification of the binocular to be

\begin{equation*} M = -\frac{f_\text{o}}{f_\text{e}} = -\frac{8\text{ cm}}{1.2\text{ cm}} = -6.67. \end{equation*}

Therefore, the angle subtended by the image will me $|M|$ times

\begin{equation*} \theta_\text{image} = 6.67\times 9.158\times 10^{-3}\text{ rad} = 6.1\times 10^{-2}\text{ rad}. \end{equation*}

In a reflecting telescope the objective is a concave mirror of radius of curvature equal to $50\text{ cm}$ and an eyepiece is a convex lens of focal length $5\text{cm}\text{.}$ Find the apparent size of a $25\text{-m}$ tree at a distance of $10\text{ km}$ that you would perceive when looking through the telescope.

Hint

The apparent height will be multiple of the magnification.

$125\text{ m}$
We can think of $10\text{ km}$ as infinitely far away with virtual image forming that far as well, as is the case in telescopes. This happens in telescopes since the distance between objective and eyepiece is very close to the sum of their foci. Therefore, the height of the tree will be magnified $5$ fold. That is the apparent height will be $125\text{ m}\text{.}$