The lens shown in FIGURE CP24.52 is called an achromatic...

The lens shown in FIGURE CP24.52 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 $R$. a. For parallel light rays coming from the left, show that the effective focal length of this two-lens system is $f=R /\left(2 n_{2}-n_{1}-1\right),$ where $n_{1}$ and $n_{2}$ are, respectively, the indices of refraction of the diverging and the converging lenses. Don't forget to make the thin-lens approximation. b. Because of dispersion, either lens alone would focus red rays and blue rays at different points. Define $\Delta n_{1}$ and $\Delta n_{2}$ as $n_{\mathrm{blue}}-n_{\mathrm{red}}$ for the two lenses. Find an expression for $\Delta n_{2}$ in terms of $\Delta n_{1}$ that makes $f_{\text {blue }}=f_{\text {red }}$ for the two-lens system. That is, the two-lens system does not exhibit chromatic aberration. c. Indices of refraction for two types of glass are given in the table. To make an achromatic doublet, which glass should you use for the converging lens and which for the diverging lens? Explain. (TABLE CAN'T COPY) d. What value of $R$ gives a focal length of $10.0 \mathrm{cm} ?$

The lens shown in FIGURE CP24.52 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 $R$.
a. For parallel light rays coming from the left, show that the effective focal length of this two-lens system is $f=R /\left(2 n_{2}-n_{1}-1\right),$ where $n_{1}$ and $n_{2}$ are, respectively, the indices of refraction of the diverging and the converging lenses. Don't forget to make the thin-lens approximation.
b. Because of dispersion, either lens alone would focus red rays and blue rays at different points. Define $\Delta n_{1}$ and $\Delta n_{2}$ as $n_{\mathrm{blue}}-n_{\mathrm{red}}$ for the two lenses. Find an expression for $\Delta n_{2}$ in terms of $\Delta n_{1}$ that makes $f_{\text {blue }}=f_{\text {red }}$ for the two-lens system. That is, the two-lens system does not exhibit chromatic aberration.
c. Indices of refraction for two types of glass are given in the table. To make an achromatic doublet, which glass should you use for the converging lens and which for the diverging lens? Explain. (TABLE CAN'T COPY)
d. What value of $R$ gives a focal length of $10.0 \mathrm{cm} ?$

Key Concepts

Achromatic Doublet

An achromatic doublet is an optical component made by combining a converging lens and a diverging lens in such a manner that they correct each other’s chromatic dispersion. The design minimizes chromatic aberration by ensuring that two different wavelengths (typically red and blue) are brought to the same focus. This concept is fundamental in lens design for achieving sharper and clearer images.

Chromatic Aberration

Chromatic aberration is the optical phenomenon where light rays of different wavelengths fail to converge at the same point after refracting through a lens. This is primarily due to the dispersion of the lens material, where the refractive index varies with wavelength. Correcting chromatic aberration is important to prevent color fringing and ensure accurate image reproduction in optical systems.

Thin Lens Approximation

The thin lens approximation assumes that the thickness of the lens is negligible compared to its focal length, allowing the lens to be treated as a single plane where refraction occurs. This simplification makes it easier to analyze and combine the optical powers of multiple lenses, as it permits the direct summation of their individual effects.

Effective Focal Length

The effective focal length of a system of lenses is the distance at which parallel light rays converge after passing through the entire system. In the thin lens approximation, the system’s overall power is the sum of the individual lens powers, and the effective focal length is calculated as the reciprocal of this total power. This concept is crucial when designing compound lens systems.

Dispersion

Dispersion refers to the dependence of a material's refractive index on the wavelength of light. In optical materials, shorter wavelengths (blue light) are typically refracted more than longer wavelengths (red light), leading to the separation of colors. Understanding dispersion is essential in lens design, especially when correcting chromatic aberration.

Lens Maker's Formula

The Lens Maker’s Formula relates the focal length of a lens to its radii of curvature and the refractive index of the material. It is a fundamental equation in geometrical optics that is used to design lenses with desired focal lengths and to understand how changes in curvature or refractive index affect the focusing ability of a lens.

Optical System Design for Aberration Correction

Designing optical systems that correct for aberrations involves choosing materials and geometries that counteract undesirable effects like chromatic aberration. By carefully selecting the refractive indices and dispersion properties of different lens elements, designers can balance the variation in focal lengths for different wavelengths, thereby achieving a system where all colors come to a common focus.

Transcript

00:01 Remember, we're going from the first lens, which as in his reflection of n1, to the second lens, n2, and then to the second lens through open air and air.

00:12 So we start the equation n1 over s plus n2 over s prime is equal to n2 minus n1 over r.

00:26 We're dealing with incoming parallel light rays, so this is going to go to 0.

00:30 Because s will be infinity and so we'll get that s prime is equal to r n2 over n2 minus n1 now the image position of the first lens becomes the object of the second lens so we can go to i guess the same equation but for our second process over here and so we'll have n2 over s plus n air over s prime is equal to nair minus n2 over r.

01:08 Now s prime will become this s and s prime down here will become our focal length.

01:16 So we'll have n2 over r and 2 minus n1 plus 1 over focal length, which is actually negative based on our lenses.

01:32 And that is equal to 1 minus n2 over r.

01:36 Now we can do a little algebra.

01:38 These n2s here will cancel.

01:40 We'll have n2 minus n1 over r plus 1 over negative f is equal to 1 minus n2 over r.

01:51 Whoops.

01:53 And finally we'll get n2 minus n1 minus 1 plus n2 over r is equal to 1 over f.

02:04 So f is then equal to 2 and 2 is uh excuse me now r over 2 n2 minus n1 minus 1.

02:16 Now i need to find an expression using delta n1 and delta n2 where the two focal lengths for blue and red are equal.

02:24 So let's start with f blue is equal to f red.

02:28 F blue is equal to r over 2n2 blue minus n1 blue minus 1 is equal to r over 2n1 red minus 2n2 red minus 1.

02:52 The r is will cancel and we just need to deal with the denominator.

02:56 So you'll get 2n2 blue minus n2 red is equal to n1 blue minus n1 red.

03:07 We know that this is equal to delta n2 and this is equal to delta n1...

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