Problem 68 - Physics for Scientists and Engineers with Modern Physics

Question

(II) \(\mathrm{A} 120 \times\) astronomical telescope is adjusted for a relaxed eye when the two lenses are 1.25 \(\mathrm{m}\) apart. What is the focal length of each lens?

Step-by-Step Solution

Answer

The focal lengths are approximately 1240 mm for the objective and 10.33 mm for the eyepiece.

Step 1: Understand the Telescope Formula

An astronomical telescope is adjusted for the relaxed eye when the image is formed at infinity. The formula relating the focal lengths of the eyepiece \(f_e\) and the objective lens \(f_o\) with the magnification \(M\) is given by \(M = \frac{f_o}{f_e}\). Here, \(M = 120\) (since it's a 120× telescope), and the total distance \(d\) between the lenses (objective and eyepiece) is 1.25 m.

Step 2: Use the Lens Separation Equation

Since the lenses are 1.25 m apart, we can write the equation as \(f_o + f_e = 1.25\ m\). This equation will be combined with the magnification equation to solve for the focal lengths.

Step 3: Solve the System of Equations

We have two equations: 1. \(f_o + f_e = 1.25\ m\)2. \(\frac{f_o}{f_e} = 120\)From Equation 2, express \(f_o\) as \(f_o = 120 f_e\). Substitute this expression into Equation 1 to get \(120 f_e + f_e = 1.25\ m\).

Step 4: Simplify and Solve for \(f_e\)

Combine like terms: \(121 f_e = 1.25\ m\). Solve for \(f_e\) by dividing both sides by 121:\[f_e = \frac{1.25}{121} \approx 0.01033\ m\] which is \(10.33\ mm\).

Step 5: Solve for \(f_o\)

Use \(f_o = 120 f_e\) to solve for \(f_o\):\[f_o = 120 \times 0.01033\ m \approx 1.24\ m\] which is \(1240\ mm\).

Key Concepts

Focal Length

In astronomical telescopes, the focal length is a crucial element. It is the distance over which light converges into a point after passing through a lens or reflecting off a mirror. This definition is simple yet powerful as it determines the telescope's ability to magnify celestial objects.
To comprehend focal length, imagine a lens gathering light from a distant star. The distance from the lens to the point where light rays meet and come into focus is the focal length of that lens.

• Objective lens: This lens has a longer focal length because it gathers light from celestial objects.
• Eyepiece lens: This has a shorter focal length and it magnifies the image formed by the objective lens.

In the given problem, the telescope consists of a combination of these two lenses. Knowing the length of each lens helps calculate how effective the telescope is at enlarging images from space.

Magnification

Magnification in telescopes indicates how much larger an object appears when viewed through the telescope compared to the naked eye. For instance, a magnification of 120× means the object looks 120 times larger.
The formula for magnification in an astronomical telescope is given as:\[M = \frac{f_o}{f_e}\]where \(M\) is the magnification, \(f_o\) is the focal length of the objective lens, and \(f_e\) is the focal length of the eyepiece.

• If \(f_o\) increases, magnification increases.
• If \(f_e\) decreases, magnification increases.

This relationship tells us that for a given focal length of the eyepiece, increasing the objective lens's focal length will result in higher magnification. In our exercise, a magnification of 120× means the objective lens's focal length divided by the eyepiece's focal length equals 120. Solving this allows us to understand the power behind the telescopic system.

Lens Separation

Lens separation is the distance between the objective lens and the eyepiece lens in a telescope. This distance is vital for the telescope to function correctly, as it determines focus and clarity.
Astronomical telescopes are set for relaxed eye viewing when the image is at infinity. This state is achieved when the sum of the focal lengths of both lenses equals the lens separation.The equation is presented as:\[f_o + f_e = d\]where \(f_o\) and \(f_e\) are the focal lengths of the objective and eyepiece lenses respectively, and \(d\) is the lens separation.
In our example, the lenses are separated by 1.25 m, which informs us about how close or far apart the focal points need to be to ensure optimal viewing. Proper lens separation ensures that light is focused correctly to provide the sharpest images possible.