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A refracting telescope has an objective lens of focal length 16.0 in. and eyepieces of focal lengths \(15 \mathrm{~mm}, 22 \mathrm{~mm}, 35 \mathrm{~mm},\) and \(85 \mathrm{~mm}\) What are the largest and smallest angular magnifications you can achieve with this instrument?

Short Answer

Expert verified
Largest magnification: 27.093; smallest magnification: 4.781.

Step by step solution

01

Convert Units for Objective Lens

The focal length of the objective lens is given as 16.0 inches. To work with this in SI units, convert inches to millimeters. Since 1 inch equals 25.4 mm, multiply 16.0 inches by 25.4 to convert.\[ 16.0 \text{ inches} \times 25.4 \text{ mm/inch} = 406.4 \text{ mm} \]
02

Understand Angular Magnification Formula

The angular magnification \(M\) of a telescope is calculated using the formula:\[ M = \frac{f_o}{f_e} \]where \(f_o\) is the focal length of the objective lens and \(f_e\) is the focal length of the eyepiece.
03

Calculate Angular Magnifications

Substitute the focal length of the objective lens (406.4 mm) and each focal length of the eyepieces into the formula.- For \(f_e = 15\, \text{mm}\):\[ M = \frac{406.4}{15} = 27.093 \]- For \(f_e = 22\, \text{mm}\):\[ M = \frac{406.4}{22} = 18.473 \]- For \(f_e = 35\, \text{mm}\):\[ M = \frac{406.4}{35} = 11.611 \]- For \(f_e = 85\, \text{mm}\):\[ M = \frac{406.4}{85} = 4.781 \]
04

Identify Largest and Smallest Angular Magnifications

The largest angular magnification is 27.093, achieved with the 15 mm eyepiece. The smallest angular magnification is 4.781, achieved with the 85 mm eyepiece.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Magnification
When it comes to observing distant objects with telescopes, angular magnification is a crucial concept. **Angular magnification** refers to how much larger (or smaller) an object appears when viewed through the telescope compared to the naked eye. The angular magnification of a telescope is calculated using the formula: \[ M = \frac{f_o}{f_e} \] where:
  • \(M\) is the angular magnification,
  • \(f_o\) is the focal length of the objective lens,
  • \(f_e\) is the focal length of the eyepiece.
The telescope's ability to magnify depends on these two lenses: the objective and the eyepiece. A higher magnification means that the object appears bigger, helping astronomers and stargazers see more details of the celestial objects they're observing. Always remember: as magnification increases, the brightness of the viewed image might decrease.
Focal Length
**Focal length** is a fundamental property of lenses and optical systems. It's the distance between the lens and the point where it focuses light. In a refracting telescope, there are two critical focal lengths to consider: - **Objective Lens Focal Length**: This is usually longer and plays a significant role in gathering light. - **Eyepiece Focal Length**: This is shorter and is crucial for magnifying the image. The focal length affects both the magnification and the field of view. A longer focal length in the objective lens increases the system’s ability to magnify distant objects. Meanwhile, a shorter focal length in the eyepiece results in higher magnification. Proper understanding and adjustment of focal lengths are key to achieving desired observation outcomes.
Objective Lens
The **objective lens** is an essential component of a refracting telescope. It is the primary lens that first collects and bends light from the observed celestial objects. The focal length of the objective lens is a critical factor in determining the capability of the telescope. Generally: - A longer objective lens focal length allows the telescope to have a higher magnification potential. - It also helps in accumulating more light, which is vital for viewing faint objects. The quality, material, and design of the objective lens significantly impact the clarity and sharpness of the image. In the original problem, the objective lens has a focal length of 16 inches, converted to 406.4 mm in SI units. This consideration is essential for compatibility with the rest of the optical system.
Eyepiece
The **eyepiece** of a telescope is where you place your eye to view the magnified image. It is responsible for further increasing the magnification of the image created by the objective lens. The eyepiece focal length plays a direct role in determining the overall magnification factor: - Shorter focal length eyepieces result in higher magnification. - Longer focal length eyepieces offer wider fields of view but lower magnification. Choosing the right eyepiece allows the observer to optimize their viewing experience based on the object being observed. In a set up with multiple eyepieces, varying the eyepiece can be used to achieve different magnifications, as demonstrated in the solution.
SI Units
When dealing with scientific work of any kind, **SI units** (International System of Units) are the universal language. They ensure uniformity and clarity in measurements worldwide. In the context of optical instruments like telescopes: - The focal lengths of lenses are typically measured in **millimeters** (mm), an SI unit. - Converting measurements from other units, like inches, into SI units, as done in this exercise, allows easier comparison and calculation. Using SI units aids in reducing errors and streamlining equations, especially when handling high-precision scientific tools. For instance, the focal length of the objective lens was originally given in inches and converted to millimeters to make calculations using standard scientific formulas simpler.

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Most popular questions from this chapter

While Galileo did not invent the telescope, he was the first known person to use it astronomically, beginning around \(1609 .\) Five of his original lenses have survived (although he did work with others). Two of these have focal lengths of 1710 \(\mathrm{mm}\) and \(980 \mathrm{~mm}\). (a) For greatest magnification, which of these two lenses should be the eyepiece and which the objective? How long would this telescope be between the two lenses? (b) What is the greatest angular magnification that Galileo could have obtained with these lenses? (Note: Galileo actually obtained magnifications up to about \(30 \times\), but by using a diverging lens as the eyepiece.) (c) The moon subtends an angle of \(0.5^{\circ}\) when viewed with the naked eye. What angle would it subtend when viewed through this telescope (assuming that all of it could be seen)?

A camera has a lens with an aperture diameter of \(8.00 \mathrm{~mm}\). It is used to photograph a pet dog. What aperture diameter would correspond to an increase in the intensity of the dog's image on the film by a factor of \(2 ?\)

The objective lens and the eyepiece of a microscope are \(16.5 \mathrm{~cm}\) apart. The objective lens has a magnification of \(62 \times,\) and the eyepiece has a magnification of \(10 \times .\) Assume that the image of the objective lies very close to the focal point of the eyepiece. Calculate (a) the overall magnification of the microscope and (b) the focal length of each lens.

A camera is focused on an object that is \(1.2 \mathrm{~m}\) from the lens. The distance between the CCD image sensor and the lens is \(5 \mathrm{~cm}\). What is the focal length of the lens?

A compound microscope has an objective lens of focal length \(10.0 \mathrm{~mm}\) with an eyepiece of focal length \(15.0 \mathrm{~mm},\) and it produces its final image at infinity. The object to be viewed is placed \(2.0 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) How far from the objective lens is the first image formed? (b) What is the overall magnification of this microscope?

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