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Chapter 39: Problem 11

Compute the magnifying power of a telescope, having objective and eyepiece lenses of focal lengths \(+60\) and \(+3.0 \mathrm{~cm}\), respectively, when it is focused for parallel rays. The image is inverted.

Short Answer

Expert verified
The magnifying power of the telescope is 20.

Step by step solution

01

Understanding the Formula

The magnifying power of a telescope focused for parallel rays 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 lens.
02

Substitute the Focal Lengths

Substitute the given focal lengths into the formula: \( f_o = 60 \) cm and \( f_e = 3 \) cm. So we have: \[ M = \frac{60}{3} \]
03

Calculate the Magnifying Power

Perform the division: \[ M = 20 \] This means the telescope has a magnifying power of 20.

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Key Concepts

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

Focal Length
Focal length is a key concept in optics and plays a crucial role in understanding telescopes. It is the distance between the lens and the point where the light rays converge in focus. The focal length determines how strongly a lens concentrates light. It is usually denoted by the symbol \( f \). In telescopes, both the objective lens and the eyepiece lens have specific focal lengths. A longer focal length implies less bending of light, leading to stronger magnification potential. When using a telescope, you use the focal lengths of both lenses to calculate the magnifying power, which shows how much larger an object appears compared to the naked eye view.
Objective Lens
An objective lens is the primary lens in a telescope. Its main function is to gather light from a distant object and converge it to form an image. This lens is larger in size compared to the eyepiece lens, which helps in collecting more light and providing a brighter image.The objective lens in a telescope has a fixed focal length identified as \( f_o \). In the exercise, the focal length of the objective lens is given as \( 60 \) cm. This lens is crucial as it defines the quality and clarity of the image. A key feature is that a good objective lens can capture and bring parallel rays to a point, which is needed when observing distant celestial objects.
Eyepiece Lens
An eyepiece lens is found at the viewing end of a telescope and is tasked with magnifying the image formed by the objective lens. While it is smaller than the objective lens, its focal length, denoted as \( f_e \), plays a crucial role in determining the overall magnification.In our telescope exercise, the eyepiece lens has a focal length of \( 3 \) cm. The shorter focal length allows for higher magnification levels. The eyepiece takes the focused light from the objective lens and spreads it over the observer鈥檚 eye, creating a magnified and more discernible image through the eyepiece.
Parallel Rays
In the context of telescopes, understanding parallel rays is important for effective focusing. Parallel rays are those that travel in the same direction and never converge or diverge. When a telescope is adjusted for parallel rays, it means the telescope is optimized for viewing objects at a great distance, such as stars or planets. When you align a telescope for parallel rays, it assumes that incoming light is already parallel, which simplifies the calculation of magnification. It lets the objective and eyepiece lenses work together efficiently to form a clear, focused image on the retina of the observer, by converging these rays appropriately.

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Most popular questions from this chapter

In the compound microscope shown, the objective and eyepiece have focal lengths of \(+0.80\) and \(+2.5 \mathrm{~cm}\), respectively. The real intermediate image \(A^{\prime} B^{\prime}\) formed by the objective is \(16 \mathrm{~cm}\) from the objective. Determine the total magnification if the eye is held close to the eyepiece and views the virtual image \(A^{\prime \prime} B^{\prime \prime}\) at a distance of \(25 \mathrm{~cm}\). Method 1 Let \(s_{o} 0=\) Object distance from the objective \(s_{o} 0=\) Real-image distance from the objective $$ \frac{1}{s_{o o}}=\frac{1}{f_{o}}-\frac{1}{s_{i O}}=\frac{1}{0.80}-\frac{1}{16}=\frac{19}{16} \mathrm{~cm}^{-1} $$ and so the objective produces the linear magnification $$ M_{T O}=-\frac{s_{i O}}{s_{o o}}=-(16 \mathrm{~cm})=\left(\frac{19}{16} \mathrm{~cm}^{-1}\right)=-19 $$ The intermediate image is inverted. The magnifying power of the eyepiece is $$ M_{T E}=-\frac{s_{i E}}{s_{o E}}=-s_{i E}\left(\frac{1}{f_{E}}-\frac{1}{s_{i E}}\right)=-\frac{s_{i E}}{f_{E}}+1=-\frac{-25}{+2.5}+1=11 $$ The eyepiece does not flip the image: the intermediate image is inverted and the final image is inverted. Therefore, the magnifying power of the instrument is \(-19 \times 11=-2.1 \times 10^{2}\). Alternatively, under the conditions stated, the magnifying power of the eyepiece can be found as $$ \frac{25}{f_{E}}+1=\frac{25}{2.5}+1=11 $$ Method 2 From Eq. (39.2) with \(s_{i o}=16 \mathrm{~cm}\), $$ \text { Magnification }=\left(\frac{d_{n}}{f_{E}}+1\right)\left(\frac{s_{i o}}{f_{o}}-1\right)=\left(\frac{25}{2.5}+1\right)\left(\frac{16}{0.8}-1\right)=2.1 \times 10^{2} $$

An old camera produces a clear image of a distant landscape when the thin lens is \(8 \mathrm{~cm}\) from the film. What adjustment is required to get a good photograph of a map placed \(72 \mathrm{~cm}\) from the lens? When the camera is focused for distant objects (for parallel rays), the distance between lens and film is the focal length of the lens, namely, \(8 \mathrm{~cm}\). For an object \(72 \mathrm{~cm}\) distant: $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8}-\frac{1}{72} \quad \text { or } \quad s_{i}=9 \mathrm{~cm} $$ The lens should be moved farther away from the film a distance of \((9-8) \mathrm{cm}=1 \mathrm{~cm}\)

Reflecting telescopes make use of a concave mirror, in place of the objective lens, to bring the distant object into focus. What is the magnifying power of a telescope that has a mirror with \(250 \mathrm{~cm}\) radius and an eyepiece whose focal length is \(5.0 \mathrm{~cm}\) ? As it is for a refracting telescope (i.e., one with two lenses), \(M_{A}=\) \(-f_{O} / f_{E}\) again applies where, in this case, \(f_{O}=-R / 2=125 \mathrm{~cm}\) and \(f_{E}\) \(=5.0 \mathrm{~cm} .\) Thus, \(M_{A}=-25\).

A farsighted person named Amy cannot see clearly objects closer to the eye than \(75 \mathrm{~cm}\). Determine the power of the spectacle lenses which will enable her to read type at a distance of \(25 \mathrm{~cm}\). The image, which must be right-side-up, must be on the same side of the lens as the type (hence, the image is virtual and \(s_{i}=75 \mathrm{~cm}\) ), and farther from the lens than the type (hence, converging on positive lenses are prescribed). Keep in mind that for virtual images formed by a convex lens \(\left|s_{i}\right|>s_{0}\). We have $$ \frac{1}{f}=\frac{1}{25}-\frac{1}{75} \quad \text { or } \quad f=+37.5 \mathrm{~cm} $$ and $$ \text { Power }=\frac{1}{0.375 \mathrm{~m}}=2.7 \text { diopters } $$

A refracting astronomical telescope has a magnifying power of 150 when adjusted for minimum eyestrain. Its eyepiece has a focal length of \(+1.20 \mathrm{~cm} .\) (a) Determine the focal length of the objective lens. (b) How far apart must the two lenses be so as to project a real image of a distant object on a screen \(12.0 \mathrm{~cm}\) from the eyepiece?
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