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Chapter 24: Problem 71

A refracting telescope has an objective lens with a focal length of $2.20 \mathrm{m}\( and an eyepiece with a focal length of \)1.5 \mathrm{cm} .$ If you look through this telescope the wrong way, that is, with your eye placed at the objective lens, by what factor is the angular size of an observed object reduced?

Short Answer

Expert verified
Answer: The angular size of an observed object is reduced by a factor of approximately 0.0068.

Step by step solution

01

Write the formula for the magnification of a refracting telescope

The magnification M of a refracting telescope can be 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.
02

Substitute given values into the formula

The objective lens has a focal length of \(2.20 \mathrm{m}\), and the eyepiece has a focal length of \(1.5 \mathrm{cm}\). To be consistent with units, we will convert the focal length of the eyepiece to meters: \(f_e = 1.5 \mathrm{cm} * \frac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.015 \mathrm{m}\). Now, we can substitute these values into the magnification formula: M = \frac{2.20 \mathrm{m}}{0.015 \mathrm{m}}
03

Calculate the magnification

Now, we can calculate the magnification of the refracting telescope when used in the incorrect manner: M = \frac{2.20}{0.015} = 146.67
04

Determine the angular size reduction factor

The magnification of the telescope when used correctly is 146.67. However, since the telescope is being used incorrectly, the magnification is inverted (i.e., 1/M). Therefore, the angular size reduction factor is: Angular size reduction factor = \frac{1}{M} = \frac{1}{146.67} \approx 0.0068 So, when looking through this telescope the wrong way, the angular size of an observed object is reduced by a factor of approximately 0.0068.

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Most popular questions from this chapter

A converging lens with a focal length of \(3.00 \mathrm{cm}\) is placed $24.00 \mathrm{cm}\( to the right of a concave mirror with a focal length of \)4.00 \mathrm{cm} .\( An object is placed between the mirror and the lens, \)6.00 \mathrm{cm}\( to the right of the mirror and \)18.00 \mathrm{cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.
Angular Magnification and the Simple Magnifier Thomas wants to use his 5.5 -D reading glasses as a simple magnifier. What is the angular magnification of this lens when Thomas's eye is relaxed?
Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of \(25 \mathrm{cm} .\) The focal length of the magnifying glass is \(+5.0 \mathrm{cm} .\) Assume that Keesha's eye is close to the magnifying glass. (a) What should be the distance between the magnifying glass and the beetle? (b) What is the angular magnification? (tutorial: magnifying glass II).
A refracting telescope is used to view the Moon. The focal lengths of the objective and eyepiece are \(+2.40 \mathrm{m}\) and \(+16.0 \mathrm{cm},\) respectively. (a) What should be the distance between the lenses? (b) What is the diameter of the image produced by the objective? (c) What is the angular magnification?
Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.
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