Problem 27 - Algebra and Trigonometry with Analytic Geometry

Question

Telescope resolution Two stars that are very close may appear to be one. The ability of a telescope to separate their images is called its resolution. The smaller the resolution, the better a telescope's ability to separate images in the sky. In a refracting telescope, resolution \(\theta\) (see the figure) can be improved by using a lens with a larger diameter \(D\). The relationship between \(\theta\) in degrees and \(D\) in meters is given by \(\sin \theta=1.22 \lambda / D\), where \(\lambda\) is the wavelength of light in meters. The largest refracting telescope in the world is at the University of Chicago. At a wavelength of \(\lambda=550 \times 10^{-9}\) meter, its resolution is \(0.00003769^{\circ}\). Approximate the diameter of the lens.

Step-by-Step Solution

Answer

The diameter of the lens is approximately 1.02 meters.

Step 1: Understand the Given Formula

The problem gives you the formula for resolution as \( \sin \theta = \frac{1.22 \lambda}{D} \). In this formula, \( \lambda \) is the wavelength of light, \( D \) is the diameter of the lens, and \( \theta \) is the resolution in degrees. We need to find \( D \).

Step 2: Convert Resolution to Radians

Since \( \sin \theta \) requires \( \theta \) in radians, convert the given resolution \( \theta = 0.00003769^{\circ} \) to radians using the conversion factor \( 1^{\circ} = \frac{\pi}{180} \) radians. Thus, \( \theta = 0.00003769 \times \frac{\pi}{180} \).

Step 3: Calculate \( \sin \theta \)

Use \( \theta \) in radians to find \( \sin \theta \). For very small angles, you can approximate \( \sin \theta \approx \theta \), so \( \sin \theta \approx 0.00003769 \times \frac{\pi}{180} \).

Step 4: Solve for Diameter \( D \)

Rearrange the formula \( \sin \theta = \frac{1.22 \lambda}{D} \) to solve for \( D \):\[ D = \frac{1.22 \lambda}{\sin \theta} \]Substitute the values \( \lambda = 550 \times 10^{-9} \) meters and the calculated value for \( \sin \theta \).

Step 5: Perform Calculation

Calculate the value of \( D \) using:\[ D = \frac{1.22 \times 550 \times 10^{-9}}{0.00003769 \times \frac{\pi}{180}} \]This gives the diameter of the lens approximately.

Key Concepts

Refracting Telescopes

Refracting telescopes use lenses to bend (or refract) light and bring it into focus. They're one of the earliest types of telescopes invented and rely on the principle of refraction, where light changes direction when it passes through a lens. These telescopes are great for viewing objects like the Moon or planets. However, they can be limited by the size and quality of the lens. Key Features of Refracting Telescopes:

• Primary Lens: This is the main lens that gathers light and focuses it to create an image.
• Magnification: The magnification depends on the combination of the primary lens and the eyepiece.
• Image Quality: Affected by factors like lens imperfections and the telescope's resolution.

Refractors are appreciated for their simplicity and the crisp images they can provide, especially of bright celestial objects.

Wavelength of Light

The wavelength of light is a measure of the distance between successive peaks of light waves. It determines the color of the light we see. In this context, the exercise uses a specific wavelength of light, 550 nm (nanometers), which lies in the visible spectrum and appears as green light to our eyes. The wavelength is vital in calculating the resolution of telescopes. Wavelength Insight:

• Visible Light Spectrum: Wavelengths from about 400 nm to 700 nm.
• Importance in Astronomy: Helps determine the type and quality of images captured by telescopes.
• Resolution Relation: Shorter wavelengths can provide better resolution.

In telescope calculations, knowing the wavelength allows you to predict the angular resolution the telescope can achieve.

Lens Diameter

The diameter of a telescope lens is crucial because it determines the telescope’s light-gathering power and resolution. Larger lenses can gather more light, leading to brighter and clearer images. In the given exercise, the diameter needs to be calculated to understand the telescope's capability. Significance of Lens Diameter:

• Light Gathering: Larger diameter means more light, improving image brightness.
• Resolution Enhancement: Bigger diameter reduces the diffraction limit, improving resolution.
• Telescope Type Impact: Different designs leverage lens sizes differently, with refractors using lenses as primary image formers.

Understanding this concept helps when optimizing telescope designs for best viewing experiences.

Sine Function

The sine function is a fundamental part of trigonometry, relating to the ratio of the length of the side opposite to an angle, to the hypotenuse in a right triangle. In the field of astronomy, it's commonly used due to its periodic nature and role in circle-related calculations.Sine in Telescope Calculations:

• Resolution Formula: Used in the formula \( \sin \theta = \frac{1.22 \lambda}{D} \).
• Small Angle Approximation: For small angles where \( \sin \theta \approx \theta \) in radians, simplifying calculations.
• Wave Nature of Light: Helps in understanding wave interference and patterns.

Having a grasp of sine functions is essential for solving many physics problems, including those involving telescopic observations.

Angle Conversion

Angle conversion is a mathematical process used to switch between different units of angles, typically degrees and radians. For astronomical calculations, angles are usually more conveniently expressed in radians.Understanding Angle Conversion:

• Degrees to Radians: Multiply the angle in degrees by \( \frac{\pi}{180} \).
• Radians to Degrees: Multiply the angle in radians by \( \frac{180}{\pi} \).
• Practical Applications: Crucial for calculations in physics and engineering contexts where trigonometric functions use radian measure.

Mastering this conversion is vital for students tackling problems in trigonometry and those aiming to understand the mathematics behind telescopes and their functionality.