91Ó°ÊÓ

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

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

$f_2$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$\frac{r_o^{\mathrm{′}}}{f_1−d}$.

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

$\begin{array}{r}\frac{r_0}{f_1}=\frac{r_0^{\mathrm{′}}}{f_1−d}\\ r_0^{\mathrm{′}}=\left(\frac{f_1−d}{f_1}\right)r_0\end{array}$

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

$\begin{array}{r}\frac{1}{s_2}+\frac{1}{s_2^{\mathrm{′}}}=\frac{1}{f_2}\\ s_2^{\mathrm{′}}=\frac{s_2{f}_2}{s_2−f_2}\end{array}$

Substitute$s_2=d-f_1$ in above equation

$s_2^{\mathrm{′}}=\frac{\left(d−f_1\right)f_2}{d−f_1−f_2}$

Use$\left|f_2\right|=−f_2$

$\begin{array}{r}{s}_2^{\mathrm{′}}=\frac{−\left(f_1−d\right)f_2}{d−f_1+\left(−f_2\right)}\\ =\frac{\left(f_1−d\right)\left|f_2\right|}{d−f_1+\left|f_2\right|}\end{array}$

Hence, proved that the radius of the ray bundle decreases $r_0^{\mathrm{′}}=r_0\left(f_1−d\right)/f_1$at the$s_2^{\mathrm{′}}=\left|f_2\right|\left(f_1−d\right)/\left(\left|f_2\right|−f_1+d\right)$ 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:

$\begin{array}{r}\frac{r_o^{\mathrm{′}}}{r_o}=\left(\frac{f_1}{f_1−d}\right)\\ \frac{r_o^{\mathrm{′}}}{r_0}=\frac{f}{s_2^{\mathrm{′}}}\end{array}$

And

By above two relation focal length of lens written as:

$\begin{array}{r}\frac{f}{s_2^{\mathrm{′}}}=\frac{f_1}{f_1−d}\\ f=\left(\frac{f_1}{f_1−d}\right)s_2^{\mathrm{′}}\end{array}$

Substitute $s_2^{\mathrm{′}}=\left|f_2\right|\left({\mathrm{f}}_1−\mathrm{d}\right)/\left(\left|{\mathrm{f}}_2\right|−{\mathrm{f}}_1+\mathrm{d}\right)$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.