Question

An astronomical telescope provides an angular magnification of 12. The two converging lenses are \(66 \mathrm{cm}\) apart. Find the focal length of each of the lenses.

Short answer

Answer: The focal length of the objective lens is 60.96 cm, and the focal length of the eyepiece lens is 5.08 cm.

Step-by-step solution

Identify the formula for angular magnification

The formula for the angular magnification (m) of a telescope is given by: $$m=\frac{f_\mathrm{o}}{f_\mathrm{e}}$$, where \(f_\mathrm{o}\) is the focal length of the objective lens, and \(f_\mathrm{e}\) is the focal length of the eyepiece lens. In this case, \(m=12\). Our goal is to find the values of \(f_\mathrm{o}\) and \(f_\mathrm{e}\).

Identify the formula for the lens tube length

The length of the telescope's tube is given by the distance between its objective and eyepiece lenses (\(L\)). This is related to the focal lengths of the lenses by the following formula: $$L = f_\mathrm{o} + f_\mathrm{e}$$ In this problem, the telescope's tube length is given as \(L=66\,\mathrm{cm}\).

Write the system of equations

We can use the two formulas obtained in steps 1 and 2 to write a system of equations: $$\begin{cases} 12 = \frac{f_\mathrm{o}}{f_\mathrm{e}} \\ 66 = f_\mathrm{o} + f_\mathrm{e} \end{cases}$$

Solve the system of equations

We can multiply the first equation by \(f_\mathrm{e}\) to eliminate the denominator: $$12f_\mathrm{e} = f_\mathrm{o}$$ Now, substitute this expression for \(f_\mathrm{o}\) in the second equation: $$66 = 12f_\mathrm{e} + f_\mathrm{e}$$ Combine terms and solve for \(f_\mathrm{e}\): $$66 = 13 f_\mathrm{e}$$ $$f_\mathrm{e} = \frac{66}{13}$$ $$f_\mathrm{e} = 5.08\,\mathrm{cm}$$ Now we can substitute this value of \(f_\mathrm{e}\) in the equation for \(f_\mathrm{o}\): $$f_\mathrm{o} = 12f_\mathrm{e}$$ $$f_\mathrm{o} = 12(5.08)$$ $$f_\mathrm{o} = 60.96\,\mathrm{cm}$$

Find the focal length of each lens

We have found the focal length of the objective lens (\(f_\mathrm{o}\)) to be \(60.96\,\mathrm{cm}\) and the focal length of the eyepiece lens (\(f_\mathrm{e}\)) to be \(5.08\,\mathrm{cm}\).