Problem

A number of factors play a role in determining th…

02:32

Problem 107 Easy Difficulty

A number of factors play a role in determining the focal length of
a 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 is
thinner in the middle will diverge light. Another important factor is the index of refraction of the lens
material, $n_{\text { lens. }}$ For example, imagine comparing two lenses with
identical shapes but made of different materials. The lens with the larger index of refraction bends light more, bringing it to a focus in a
shorter distance. As a result, a larger index of refraction implies a focal
length with a smaller magnitude. In fact, the focal length of a lens
surrounded 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 of
refraction 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 air
or 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 the
magnitude 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.
A converging lens with a focal length in air of $f=+5.25 \mathrm{cm}$ is made from ice. What is the focal length of this lens if it is
immersed in benzene? (Refer to Table 26 -2.)
$\begin{array}{ll}{\text { A. }-20.7 \mathrm{cm}} & {\text { B. }-18.1 \mathrm{cm}} \\ {\text { C. }-12.8 \mathrm{cm}} & {\text { D. }-11.2 \mathrm{cm}}\end{array}$

Answer

$-12.8 \mathrm{cm}$

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Video Transcript

this question. It is about finding the focal line off the lands, uh, off certain reflective index. And that is being, uh, so that's being placed in a fluid, uh, which has, uh, certain effective index as well. Okay, so in this particular question, we are given a converging lens. There is made of ice. Okay. And then the focal hang in air is given to be 5.25 year. Okay, so ah, and it is emerging benzene. Hey. Okay, so you want to find a focal line off this lens? Okay, Shane, we'll be using a formula. The focal line through it is equal to on lens minus one times the refractive index off to it. Divide by the difference and lands minus and to it. Yeah, and then multiplied by the focal length in air. Okay, so the effective index off eyes is 1.31 Is that the sister lens and then the refractive index on fanzine? Yes. You go to 1.50 and then this is a to win. And then the focal lining air is given to be 5.25 p. M. So we can calculate that local and into it. Yes. 1.31 minus one times 1.50 If I buy 1.31 minus 1.50 a multiplied by 5.25 yen. You calculate days you get minus shop. Point it. Yeah, OK. And the answer is C. Right, that's all.

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