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Q101P

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Found in: Page 1157

University Physics with Modern Physics

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 ${{\mathbf{f}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{-}}|{f}_{2}|$ 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 $\mathbf{|}{\mathbf{f}}_{\mathbf{2}}\mathbf{|}{\mathbf{>}}\left({f}_{1}-d\right)$. To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius ${{\mathbit{r}}}_{{\mathbf{0}}}$ 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 ${{\mathbf{s}}}_{{\mathbf{2}}}^{{\mathbf{\prime }}}{\mathbf{=}}|{f}_{2}|\left({f}_{1}-d\right){\mathbf{/}}\left(\left|{f}_{2}\right|-{f}_{1}+d\right)$ 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 ${\mathbit{f}}{=}{{\mathbit{f}}}_{{1}}\left|{\mathbit{f}}_{2}\right|{I}\left(\left|{\mathbit{f}}_{2}\right|-{\mathbf{f}}_{1}+\mathbf{d}\right)$ . (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 ?

1. 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.
2. 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.
3. The effective focal length is $f={f}_{1}\left|{f}_{2}\right|/\left(\left|{f}_{2}\right|-{f}_{1}+d\right)$ .
4. 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 .
See the step by step solution

Step 1: Given Data

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

Step 2: Define the focal length.

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

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

Step 3: Find the radius of the ray and final image.

${f}_{2}$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.

Step 4: Find the effective focal length.

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 $f={f}_{1}\left|{f}_{2}\right|/\left(\left|{f}_{2}\right|-{f}_{1}+d\right)$.

Step 5: Find the maximum and minimum focal lengths and distance.

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:

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

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.