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Chapter 6: Problem 3

Show that if two lenses of focal lengths \(f_{1}\) and \(f_{2}\) can be considered to have zero physical separation, then the effective focal length of the combination of lenses is $$\frac{1}{f_{\mathrm{eff}}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$$ Note: Assuming that the actual physical separation of the lenses is \(x\), this approximation is strictly valid only when \(f_{1} \gg x\) and \(f_{2} \gg x\)

Short Answer

Expert verified
The effective focal length for two lenses in zero separation is \( f_{\text{eff}} = \frac{1}{\frac{1}{f_1} + \frac{1}{f_2}} \).

Step by step solution

01

Understanding the Lens Formula

For each lens, the lens maker's formula is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. For two lenses in contact (zero separation), this formula can be applied to each lens separately.
02

Analyzing the First Lens

For the first lens with focal length \( f_1 \), let \( u_1 \) be the object distance and \( v_1 \) be the image distance. The relationship is \( \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \). Rearrange to express \( v_1 \): \( \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} \).
03

Analyzing the Second Lens

For the second lens, consider the image from the first lens to be the object for the second lens. Thus, \( u_2 = v_1 \). For the second lens with focal length \( f_2 \), the lens equation is \( \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \). Substitute \( u_2 \) from the first lens: \( \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{v_1} \).
04

Combining the Formulas

By rearranging from the second lens, we have \( \frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{v_1} \). From Step 2, we have \( \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} \). Substitute \( \frac{1}{v_1} \) into the expression for \( v_2 \): \( \frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{f_1} + \frac{1}{u_1} \).
05

Determining Effective Focal Length

For the combination of lenses, an effective object (u_{ ext\( ext{eff} \)}) and image distance (\( v_{ ext{eff}} \)) are used where \( \frac{1}{f_{\text{eff}}} = \frac{1}{v_{\text{eff}}} - \frac{1}{u_{\text{eff}}} \). In the limit where the physical separation is zero, and using result from Step 4, we get \( \frac{1}{f_{\text{eff}}} = \frac{1}{f_1} + \frac{1}{f_2} \).
06

Finalizing the Understanding

This derived equation shows that the combined focal length is based solely on the individual focal lengths of the lenses when zero separation is approximated. This simplified relationship holds when the distances involved are significantly smaller than the focal lengths of the lenses, ensuring that other factors such as aberration or separation factor do not affect the light paths substantially.

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Key Concepts

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

Lens Formula
When discussing optics, the lens formula is a crucial concept. It helps explain how lenses form images. The formula is expressed as: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Here, \( f \) represents the focal length of the lens, \( v \) is the distance from the lens to the image, and \( u \) is the distance from the lens to the object. By using this formula, we can determine where an image will appear when light passes through a lens. It arises from the lens-making process and underlies many calculations in optics. Understanding this formula is essential for solving more complex problems involving multiple lenses or different configurations.
Focal Length
The focal length of a lens is a central property that measures how strongly the lens converges or diverges light. It is the distance between the lens's center and its focus. A lens with a shorter focal length bends light more sharply, bringing it to focus closer to the lens. Conversely, a longer focal length lens bends light less and brings it to focus further away.
  • Short focal lengths are typically found in wide-angle lenses, which capture broader views.
  • Long focal lengths are in telephoto lenses, useful for magnifying distant objects.
This characteristic is foundational in determining how lenses affect light paths and produce images.
Effective Focal Length
When we combine two lenses, the effective focal length (EFF) comes into play. This term represents the overall focal length of the lens system. To determine the EFF, consider two close-placed lenses with respective focal lengths \(f_1\) and \(f_2\). The formula for determining their combined focal length is: \[ \frac{1}{f_{\text{eff}}} = \frac{1}{f_1} + \frac{1}{f_2} \] This equation is valid when physical separation between the lenses is negligible. It helps simplify calculating the focal length of lens combinations in complex optical systems. By using EFF, predicting how a lens system affects light becomes streamlined and intuitive.
Lens Combination
Combining lenses is a common practice in optics to achieve desired image properties or magnifications. When lenses are situated closely, their focal properties interact, creating a new effective focal length.
  • This combination enhances optical systems by providing flexibility in designing corrective lenses, telescopes, and microscopes.
  • It can correct aberrations found in single-lens systems.
  • When combined, lenses can either increase magnifying power or adjust the field of view.
Understanding how combinations work helps in creating various optical devices that meet specific requirements for different applications.
Physics Education
Physics education is crucial in developing problem-solving and analytical skills. In optics, students learn about lenses and light behavior, which is essential for understanding broader physics concepts and applications.
  • By studying optics, learners explore the intricacies of vision and design.
  • It fosters skills necessary for fields such as photography, astronomy, and engineering.
  • The study of lenses and light interactions is often foundational for students venturing into various science and technology disciplines.
Optics education equips students with practical insight into everyday phenomena and cutting-edge technologies, reinforcing the significance of physics in our understanding and manipulation of the natural world.

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Most popular questions from this chapter

Suppose that a radio telescope receiver has a bandwidth of \(50 \mathrm{MHz}\) centered at \(1.430 \mathrm{GHz}\) \((1 \mathrm{GHz}=1000 \mathrm{MHz}) .\) Assume that, rather than being a perfect detector over the entire bandwidth, the receiver's frequency dependence is triangular, meaning that the sensitivity of the detector is \(0 \%\) at the edges of the band and \(100 \%\) at its center. This filter function can be expressed as $$f_{v}=\left\\{\begin{array}{cc} \frac{v}{v_{m}-v_{\ell}}-\frac{v_{\ell}}{v_{m}-v_{\ell}} & \text { if } v_{\ell} \leq v \leq v_{m} \\ -\frac{v}{v_{u}-v_{m}}+\frac{v_{u}}{v_{u}-v_{m}} & \text { if } v_{m} \leq v \leq v_{u} \\ 0 & \text { elsewhere } \end{array}\right.$$ (a) Find the values of \(v_{\ell}, v_{m},\) and \(v_{u}\) (b) Assume that the radio dish is a \(100 \%\) efficient reflector over the receiver's bandwidth and has a diameter of \(100 \mathrm{m}\). Assume also that the source NGC 2558 (a spiral galaxy with an apparent visual magnitude of 13.8 ) has a constant spectral flux density of \(S=2.5 \mathrm{mJy}\) over the detector bandwidth. Calculate the total power measured at the receiver. (c) Estimate the power emitted at the source in this frequency range if \(d=100\) Mpc. Assume that the source emits the signal isotropically.

The light rays coming from an object do not, in general, travel parallel to the optical axis of a lens or mirror system. Consider an arrow to be the object, located a distance \(p\) from the center of a simple converging lens of focal length \(f,\) such that \(p>f\). Assume that the arrow is perpendicular to the optical axis of the system with the tail of the arrow located on the axis. To locate the image, draw two light rays coming from the tip of the arrow: (i) One ray should follow a path parallel to the optical axis until it strikes the lens. It then bends toward the focal point of the side of the lens opposite the object. (ii) A second ray should pass directly through the center of the lens undeflected. (This assumes that the lens is sufficiently thin.) The intersection of the two rays is the location of the tip of the image arrow. All other rays coming from the tip of the object that pass through the lens will also pass through the image tip. The tail of the image is located on the optical axis, a distance \(q\) from the center of the lens. The image should also be oriented perpendicular to the optical axis. (a) Using similar triangles, prove the relation $$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$$ (b) Show that if the distance of the object is much larger than the focal length of the lens \((p \gg f),\) then the image is effectively located on the focal plane. This is essentially always the situation for astronomical observations. The analysis of a diverging lens or a mirror (either converging or diverging) is similar and leads to the same relation between object distance, image distance, and focal length.

When operated in "planetary"mode, HST's WF/PC 2 has a focal ratio of \(f / 28.3\) with a plate scale of \(0.0455^{\prime \prime}\) pixel \(^{-1}\). Estimate the angular size of the field of view of one \(\mathrm{CCD}\) in the planetary mode

For some point \(P\) in space, show that for any arbitrary closed surface surrounding \(P\), the integral over a solid angle about \(P\) gives $$\Omega_{\mathrm{tot}}=\oint d \Omega=4 \pi$$

How much must the pointing angle of a two-element radio interferometer be changed in order to move from one interference maximum to the next? Assume that the two telescopes are separated by the diameter of Earth and that the observation is being made at a wavelength of \(21 \mathrm{cm} .\) Express your answer in arcseconds.
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