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Q101P

Expert-verifiedFound in: Page 1157

Book edition
14th edition

Author(s)
Hugh D. Young, Roger A. Freedman

Pages
1596 pages

ISBN
9780321973610

**Focal Length of a Zoom Lens. Figure P34.101 shows a simple version of a zoom lens. The converging lens has focal length ${{\mathbf{f}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{-}}{\left|{f}_{2}\right|}$**** and the diverging lens has focal length ****. The two lenses are separated by a variable distance **** that is always less than **** Also, the magnitude of the focal length of the diverging lens satisfies the inequality $\mathbf{\left|}{\mathbf{f}}_{\mathbf{2}}\mathbf{\right|}{\mathbf{>}}{\left({f}_{1}-d\right)}$****. To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius ${{\mathit{r}}}_{{\mathbf{0}}}$**** entering the converging lens. (a) Show that the radius of the ray bundle decreases ${{\mathbf{r}}}_{{\mathbf{0}}}^{{\mathbf{\prime}}}{\mathbf{=}}{{\mathbf{r}}}_{{\mathbf{0}}}{\left({f}_{1}-d\right)}{\mathbf{/}}{{\mathbf{f}}}_{{\mathbf{1}}}$**** at the point that it enters the diverging lens. (b) Show that the final image is formed a distance ${{\mathbf{s}}}_{{\mathbf{2}}}^{{\mathbf{\prime}}}{\mathbf{=}}{\left|{f}_{2}\right|}{\left({f}_{1}-d\right)}{\mathbf{/}}{\left(\left|{f}_{2}\right|-{f}_{1}+d\right)}$**** to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended ****backward to the left of the diverging lens, they will eventually expand to the original radius **** at some point ****. The distance from the final image I′ to the point **** is the effective focal length of the lens combination; if the combination were replaced by a single lens of focal length f placed at ****, parallel rays would still be brought to a focus at . Show that the effective focal length is given by ${\mathit{f}}{=}{{\mathit{f}}}_{{1}}\left|{\mathit{f}}_{2}\right|{I}\left(\left|{\mathit{f}}_{2}\right|-{\mathbf{f}}_{1}+\mathbf{d}\right)$ ****. (d) If **** and the separation **** is adjustable between **** and **** find the maximum and minimum focal lengths of the combination. What value d of **** gives f = 30.0 cm ****?**

- The radius of the ray bundle decreases ${r}_{0}^{\mathrm{\prime}}={r}_{0}\left({f}_{1}-d\right)/{f}_{1}$ at the point that it enters the diverging lens.
- The final image is formed a distance ${s}_{2}^{\mathrm{\prime}}=\left|{f}_{2}\right|\left({f}_{1}-d\right)/\left(\left|{f}_{2}\right|-{f}_{1}+d\right)$ to the right of the diverging lens.
- The effective focal length is $f={f}_{1}\left|{f}_{2}\right|/\left(\left|{f}_{2}\right|-{f}_{1}+d\right)$ .
- The maximum and minimum focal lengths are 36 cm and 21.6 cm respectively and for the focal length f = 30 cm the d is 1.2 cm .

Focal length of converging lens: 12.0 cm

Focal length of diverging lens: - 18.0 cm

Separation between lens: 0 cm to 4.0 cm

**The focal length is the distance between the convex or concave mirror and the focal point of the mirror.**

The relation between the distance of object , the distance of the image s' and the focal length f is

$\frac{1}{f}=\frac{1}{s}+\frac{1}{{s}^{\mathrm{\prime}}}$

${f}_{2}$From the given figure, smaller and larger triangle are similar and ratio of their sides of larger triangle is $\frac{{r}_{0}}{{f}_{1}}$ while ratio of sides of smaller triangle is $\frac{{r}_{o}^{\mathrm{\prime}}}{{f}_{1}-d}$.

As the ratio of sides of two triangle are same therefore,

$\begin{array}{r}\frac{{r}_{0}}{{f}_{1}}=\frac{{r}_{0}^{\mathrm{\prime}}}{{f}_{1}-d}\\ {r}_{0}^{\mathrm{\prime}}=\left(\frac{{f}_{1}-d}{{f}_{1}}\right){r}_{0}\end{array}$

From the figure, for converging lens the distance of the image is $\left({f}_{1}-d\right)$ and for diverging lens the distance of the image is ${s}_{2}=d-{f}_{1}$ , where ${f}_{1}$ is focal length of lens.

By the relation between the distance of object ${s}_{2}$ , the distance of the image $s{\text{'}}_{2}$ and the focal length , the distance of image $s{\text{'}}_{2}$ written as

$\begin{array}{r}\frac{1}{{s}_{2}}+\frac{1}{{s}_{2}^{\mathrm{\prime}}}=\frac{1}{{f}_{2}}\\ {s}_{2}^{\mathrm{\prime}}=\frac{{s}_{2}{f}_{2}}{{s}_{2}-{f}_{2}}\end{array}$

Substitute ${s}_{2}=d-{f}_{1}$ in above equation

${s}_{2}^{\mathrm{\prime}}=\frac{\left(d-{f}_{1}\right){f}_{2}}{d-{f}_{1}-{f}_{2}}$

Use $\left|{f}_{2}\right|=-{f}_{2}$

$\begin{array}{r}{s}_{2}^{\mathrm{\prime}}=\frac{-\left({f}_{1}-d\right){f}_{2}}{d-{f}_{1}+\left(-{f}_{2}\right)}\\ =\frac{\left({f}_{1}-d\right)\left|{f}_{2}\right|}{d-{f}_{1}+\left|{f}_{2}\right|}\end{array}$

Hence, proved that the radius of the ray bundle decreases ${r}_{0}^{\mathrm{\prime}}={r}_{0}\left({f}_{1}-d\right)/{f}_{1}$ at the ${s}_{2}^{\mathrm{\prime}}=\left|{f}_{2}\right|\left({f}_{1}-d\right)/\left(\left|{f}_{2}\right|-{f}_{1}+d\right)$ point that it enters the diverging lens. And the final image is formed a distance to the right of the diverging lens.

The relation between the ${r}_{0}$ and ${r}_{0}\text{'}$ are:

$\begin{array}{r}\frac{{r}_{o}^{\mathrm{\prime}}}{{r}_{o}}=\left(\frac{{f}_{1}}{{f}_{1}-d}\right)\\ \frac{{r}_{o}^{\mathrm{\prime}}}{{r}_{0}}=\frac{f}{{s}_{2}^{\mathrm{\prime}}}\end{array}$

And

By above two relation focal length of lens written as:

$\begin{array}{r}\frac{f}{{s}_{2}^{\mathrm{\prime}}}=\frac{{f}_{1}}{{f}_{1}-d}\\ f=\left(\frac{{f}_{1}}{{f}_{1}-d}\right){s}_{2}^{\mathrm{\prime}}\end{array}$

Substitute ${s}_{2}^{\mathrm{\prime}}=\left|{f}_{2}\right|\left({\mathrm{f}}_{1}-\mathrm{d}\right)/\left(\left|{\mathrm{f}}_{2}\right|-{\mathrm{f}}_{1}+\mathrm{d}\right)$ in above expression

$\begin{array}{r}f=\left(\frac{{f}_{1}}{{f}_{1}-d}\right)\left[\frac{\left({f}_{1}-d\right)\left|{f}_{2}\right|}{d-{f}_{1}+\left|{f}_{2}\right|}\right]\\ f=\frac{{f}_{1}\left|{f}_{2}\right|}{\left|{f}_{2}\right|-{f}_{1}+d}\end{array}$

Hence, proved that the effective focal length is $f={f}_{1}\left|{f}_{2}\right|/\left(\left|{f}_{2}\right|-{f}_{1}+d\right)$.

For maximum focal length ${f}_{1}$ is 12 cm, d is zero and $\left|{f}_{2}\right|$ is 18 cm .

$\begin{array}{r}{f}_{max}=\frac{{f}_{1}\left|{f}_{2}\right|}{\left|{f}_{2}\right|-{f}_{1}+d}\\ =\frac{\left(12\mathrm{cm}\right)\left(18\mathrm{cm}\right)}{18\mathrm{cm}-12\mathrm{cm}+0\mathrm{cm}}\\ =36\mathrm{cm}\end{array}$

For minimum focal length ${f}_{1}$ is 12 cm , d is 4 and $\left|{f}_{2}\right|$ is 18 cm.

$\begin{array}{r}{f}_{min}=\frac{{f}_{1}\left|{f}_{2}\right|}{\left|{f}_{2}\right|-{f}_{1}+d}\\ =\frac{\left(12\mathrm{cm}\right)\left(18\mathrm{cm}\right)}{18\mathrm{cm}-12\mathrm{cm}+4\mathrm{cm}}\\ =21.6\mathrm{cm}\end{array}$

For the focal length f = 30 cm distance is:

$\begin{array}{r}d=\frac{{f}_{1}\left|{f}_{2}\right|}{f}-\left(\left|{f}_{2}\right|-{f}_{1}\right)\\ =\frac{\left(12\mathrm{cm}\right)\left(18\mathrm{cm}\right)}{30\mathrm{cm}}(18\mathrm{cm}-12\mathrm{cm})\\ =1.2\mathrm{cm}\end{array}$

Hence, the maximum and minimum focal lengths are 36 cm and 21.6 cm respectively and for the focal length f = 30 cm the d is 1.2 cm.

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