The lens shown in FIGURE CP35.49 is called an achromatic doublet, meaning that it has no chromatic aberration. The left side is flat, and all other surfaces have radii of curvature R. Because of dispersion, either lens alone would focus red rays and blue rays at different points. Define ∆n₁ and ∆n₂ as n_blue - n_red for the two lenses. What value of the ratio ∆n₁ / ∆n₂ makes f_blue = f_red for the two-lens system? That is, the two-lens system does not exhibit chromatic aberration.

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Step 1: Begin by recalling the lensmaker's equation for thin lenses: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), where \( f \) is the focal length, \( n \) is the refractive index, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. For the achromatic doublet, we need to ensure that the focal lengths for blue and red light are equal.

Step 2: Define \( \Delta n_1 = n_{blue,1} - n_{red,1} \) and \( \Delta n_2 = n_{blue,2} - n_{red,2} \), which represent the differences in refractive indices for blue and red light for the two lenses. These differences cause chromatic aberration, which we aim to eliminate.

Step 3: The focal length of the system is determined by the combination of the two lenses. For the system to be achromatic, the contributions to chromatic aberration from each lens must cancel out. This requires that \( \Delta n_1 / \Delta n_2 \) be chosen such that the chromatic focal shifts of the two lenses are equal and opposite.

Step 4: Use the principle of superposition for the focal lengths of the two lenses. The effective focal length \( f_{eff} \) of the system is given by \( \frac{1}{f_{eff}} = \frac{1}{f_1} + \frac{1}{f_2} \). For achromatic behavior, the chromatic shifts \( \Delta f_1 \) and \( \Delta f_2 \) must satisfy \( \Delta f_1 + \Delta f_2 = 0 \). Substitute \( \Delta n_1 \) and \( \Delta n_2 \) into the lensmaker's equation to express this condition mathematically.

Step 5: Solve the resulting equation for \( \Delta n_1 / \Delta n_2 \). This ratio ensures that the chromatic aberration is eliminated, making \( f_{blue} = f_{red} \) for the two-lens system. The exact value of the ratio depends on the radii of curvature \( R \) and the refractive indices \( n_1 \) and \( n_2 \) of the lenses.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chromatic Aberration

Chromatic aberration occurs when a lens fails to focus all colors to the same convergence point due to varying refractive indices for different wavelengths of light. This results in colored fringes around images, as blue light is refracted more than red light. Achromatic lenses, like the doublet in the question, are designed to minimize this effect by combining two different types of glass to bring different colors into focus at the same point.

Refractive Index and Dispersion

The refractive index of a material indicates how much light slows down and bends when entering that material. Dispersion refers to the phenomenon where different wavelengths of light are refracted by different amounts, leading to a separation of colors. In the context of the achromatic doublet, the difference in refractive indices for blue and red light (∆n) is crucial for determining how to combine the two lenses to achieve a common focal point.

Focal Length and Lens Combination

The focal length of a lens is the distance from the lens to the point where parallel rays of light converge. When combining two lenses, the effective focal length can be calculated using the lens maker's formula, which takes into account the individual focal lengths and the distances between the lenses. For the achromatic doublet to function without chromatic aberration, the ratio of the differences in refractive indices (∆n<sub>1</sub> / ∆n<sub>2</sub>) must be such that the focal lengths for blue and red light are equal.

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Textbook Question

FIGURE CP35.50 shows a lens combination in which the lens separation is less than the focal length of the converging lens. The procedure for combination lenses is to let the image of the first lens be the object for the second lens, but in this case the image of the first lens—shown as a dot—is on the far side of the second lens. This is called a virtual object, a point that light rays are converging toward but never reach. The top half of Figure CP35.50 shows that the converging rays are refracted again by the diverging lens and come to a focus farther to the right. The procedure for combination lenses will continue to work if we use a negative object distance for a virtual object. Equation 35.1 defined the effective focal length f_eff of a lens combination, but we didn't discuss how it is used. Although an actual ray refracts twice, once at each lens, we can extend the output rays leftward to where they need to bend only once in a plane called the principal plane. The principal plane is similar to the lens plane of a single lens, where a single bend occurs, but the principal plane generally does not coincide with the physical lens; it's just a mathematical plane in space. The effective focal length is measured from the principal plane, so parallel input rays are focused at distance f_eff beyond the principal plane. Find the positions of the principal planes for lens separations of 5 cm and 10 cm. Give your answers as distances to the left of the diverging lens.

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Textbook Question

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Textbook Question

A beam of white light enters a transparent material. Wavelengths for which the index of refraction is n are refracted at angle θ₂. Wavelengths for which the index of refraction is n + δn, where δn << n, are refracted at angle θ₂ + δθ. A beam of white light is incident on a piece of glass at 30°. Deep violet light is refracted 0.28° more than deep red light. The index of refraction for deep red light is known to be 1.552. What is the index of refraction for deep violet light?

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The Hubble Space Telescope has a mirror diameter of 2.4 m. Suppose the telescope is used to photograph stars near the center of our galaxy, 30,000 light years away, using red light with a wavelength of 650 nm. For comparison, what is this distance as a multiple of the distance of Jupiter from the sun?

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