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Chapter 39: Problem 34

An opera glass has an objective lens of focal length \(+3.60 \mathrm{~cm}\) and a negative eyepiece of focal length \(-1.20 \mathrm{~cm} .\) How far apart must the two lenses be for the viewer to see a distant object at \(25.0 \mathrm{~cm}\) from the eye?

Short Answer

Expert verified
The lenses must be 4.80 cm apart.

Step by step solution

01

Understand the Lens Formula

The lens formula for the focal length of the combined system is given by: \( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \), where \( f_1 \) and \( f_2 \) are the focal lengths of the objective and the eyepiece respectively. This formula helps us understand how the two lenses work together as a system.
02

Identify the Focal Lengths

Identify the given focal lengths: \( f_1 = +3.60 \text{ cm} \) for the objective lens and \( f_2 = -1.20 \text{ cm} \) for the eyepiece. These values will be used in the calculations.
03

Calculate the Focal Length of the System

Substitute the focal lengths into the lens formula to find the effective focal length of the system: \[ \frac{1}{f} = \frac{1}{3.60} + \left(-\frac{1}{1.20}\right) \]. Calculate this to find \( f \).
04

Calculate the Effective Focal Length

Evaluate \( \frac{1}{3.60} = 0.278 \) and \( -\frac{1}{1.20} = -0.833 \). Adding these gives an effective focal length formula \( \frac{1}{f} = -0.555 \). Thus, \( f = -1.80 \text{ cm} \).
05

Find Lens Separation for Distant Object

To see a distant object at 25 cm from the eye, the distance \(d\) between the lenses should be such that the image formed by the objective is at the focal point of the eyepiece. Therefore, for a distant object, the separation \(d\) is the sum of the focal lengths: \( d = | f_1 | + |f_2| = 3.60 + 1.20 = 4.80 \text{ cm} \).

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Key Concepts

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

Lens Formula
The lens formula is an essential concept in optics, used to understand how multiple lenses interact to focus light. When using two lenses in combination, as in an opera glass, the lens formula helps determine the system's overall focal length. The formula is given by:\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]This equation states that the reciprocal of the effective focal length \( f \) is equal to the sum of the reciprocals of the focal lengths of the individual lenses \( f_1 \) and \( f_2 \).
  • \( f_1 \) is the focal length of the objective lens
  • \( f_2 \) is the focal length of the eyepiece
Negative focal lengths, as seen with the eyepiece here, indicate a diverging lens.
Applying this formula allows us to consider how each lens contributes to the focusing power of the entire system.
Focal Length Calculation
Calculating the effective focal length of a lens system requires substituting the individual focal lengths into the lens formula. For the opera glass in our example, let's calculate:1. Identify the given focal lengths: \( f_1 = +3.60 \text{ cm} \) (objective lens) and \( f_2 = -1.20 \text{ cm} \) (eyepiece).
2. Substitute these values into the formula:
\[ \frac{1}{f} = \frac{1}{3.60} + \left(-\frac{1}{1.20} \right) \]3. Compute each reciprocal:
- \( \frac{1}{3.60} = 0.278 \) - \(-\frac{1}{1.20} = -0.833 \)
Adding these values results in:
\[ \frac{1}{f} = 0.278 - 0.833 = -0.555 \]
4. Finally, calculate the effective focal length:
- \( f = -1.80 \text{ cm} \)
This negative value tells us that the overall system behaves as a diverging lens system.
Lens Separation
Lens separation refers to the physical distance between two lenses in an optical device like opera glasses. For the viewer to see a distant object clearly, the image formed by the first lens needs to land at the focal point of the second lens.
For the exercise, we consider the effective distance \( d \) needed:
- The objective lens focuses distant objects at its focal length: \( 3.60 \text{ cm} \).
- The eyepiece requires more alignment, needing alignment at \( 1.20 \text{ cm} \).
Hence, the distance or separation \( d \) between the lenses is simply the sum of the absolute values of the focal lengths:
\[ d = |f_1| + |f_2| = 3.60 + 1.20 = 4.80 \text{ cm} \]
This precise separation ensures that the image is clear and aligned for the viewer's eye, facilitating proper focus.

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Most popular questions from this chapter

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?

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 $$ \text { Power }=\frac{1}{0.375 \mathrm{~m}}=2.7 \text { diopters } $$

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.

A nearsighted man cannot see objects clearly that are beyond 50 \(\mathrm{cm}\) from his eye. Determine the focal length and power of the glasses that will enable him to see distant objects clearly.

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 \(s_{i o}=16 \mathrm{~cm}\), $$ \text { Magnification }=\left(\frac{d_{n}}{f_{E}}+1\right)\left(\frac{s_{i o}}{f_{o}}-1\right)=\left(\frac{25}{2.5}+1\right)\left(\frac{16}{0.8}-1\right)=2.1 \times 10^{2} $$
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