Question

A telescope is advertised as providing a magnification of magnitude 41 using an eyepiece of focal length \(4.0 \cdot 10^{1} \mathrm{~mm}\). What is the focal length of the objective?

Short answer

Answer: The focal length of the objective is 1640 mm.

Step-by-step solution

Write down the given information

We are given the magnification (M) as 41 and the focal length of the eyepiece (f_e) as \(4.0 \cdot 10^{1} \mathrm{~mm}\).

Write the formula for magnification

The formula for magnification in a telescope is given by: M = \(\frac{f_o}{f_e}\) where M is the magnification, \(f_o\) is the focal length of the objective, and \(f_e\) is the focal length of the eyepiece.

Substitute the given values into the formula

We now substitute the given values of magnification (M) and the focal length of the eyepiece (f_e) into the formula: 41 = \(\frac{f_o}{4.0 \cdot 10^{1} \mathrm{~mm}}\)

Solve for the focal length of the objective (f_o)

Now we need to solve for \(f_o\). To do this, multiply both sides of the equation by the focal length of the eyepiece (\(4.0 \cdot 10^{1} \mathrm{~mm}\)): \(f_o\) = 41 \(\times\) \(4.0 \cdot 10^{1} \mathrm{~mm}\)

Calculate the focal length of the objective (f_o)

Perform the multiplication to find the focal length of the objective: \(f_o\) = 41 \(\times\) \(4.0 \cdot 10^{1} \mathrm{~mm}\) = 1640 mm The focal length of the objective is 1640 mm.