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A number of factors play a role in determining the focal length ofa lens. First and foremost is the shape of the lens. As a general rule,a lens that is thicker in the middle will converge light, a lens that isthinner in the middle will diverge light. Another important factor is the index of refraction of the lensmaterial, $n_{\text { lens. }}$ For example, imagine comparing two lenses withidentical shapes but made of different materials. The lens with the larger index of refraction bends light more, bringing it to a focus in ashorter distance. As a result, a larger index of refraction implies a focallength with a smaller magnitude. In fact, the focal length of a lenssurrounded by air $(n=1)$ is given by the lens maker's formula:$$\frac{1}{f_{\text { in air }}}=\left(n_{\text { lens }}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$A lens is not always surrounded by air, however. More generally, the fluid in which the lens is immersed may have an index ofrefraction given by $n_{\text { finid }} .$ In this case, the focal length is given by$$\frac{1}{f_{\text { in fluid }}}=\left(\frac{n_{\text { lens }}-n_{\text { fluid }}}{n_{\text { fluid }}}\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$It follows, then, that the focal lengths of a lens surrounded by airor by a general fluid are related by$$f_{\text { in fluid }}=\left[\frac{\left(n_{\text { lens }}-1\right) n_{\text { fluid }}}{n_{\text { lens }}-n_{\text { fluid }}}\right] f_{\text { in air }}$$This relation shows that the surrounding fluid can change themagnitude of the focal length, or even cause it to become infinite.The fluid can also change the sign of the focal length, which determines whether the lens is diverging or converging.Suppose a lens is made from fused quartz (glass), and that its focal length in air is $-7.75 \mathrm{cm} .$ What is the focal length of thislens if it is immersed in benzene? (Refer to Table $26-2 . )$$\begin{array}{ll}{\text { A. }-130 \mathrm{cm}} & {\text { B. } 134 \mathrm{cm}} \\ {\text { C. } 141 \mathrm{cm}} & {\text { D. }-145 \mathrm{cm}}\end{array}$

134 $\mathrm{cm}$

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

Cornell University

Simon Fraser University

Hope College

University of Winnipeg

Lectures

02:51

In physics, wave optics is…

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$\textbf{Figure P34.101}$ …

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this question is about finding the focal land off the lens when it is place inside obsolete. Okay, so, given that, uh, this is a lens made off used courts que and other focal line in air is equal to negative 7.75 PM and then it is Imus. Well, benzene. Okay. Okay. So to calculate, to find this focal lane, uh, you'll be using a formula, uh, f into it is equal to and length minus one times effective index of fluid. Divide by the difference between the refractive index off the lens and fluid. Okay. And then multiplied by the focal lane in air. Okay, so, um, of effective index off the lens, it's equal to 1.46 So this is a reflective index. Off views, quotes. Okay, the refractive index off Louis eyes 1.50 Kate is this magazine. And then the focal line is given us negative 7.75 centimeters. Okay, so the focal line into it, we just substitute their numbers 1.46 minus one times 1.5 right by 1.46 minus 1.5. But to provide negative 7.75 Yeah. Okay, so you get 134th year, and the answer is B. Okay, so this is the answer, and that's all.

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