91Ó°ÊÓ

Object stands on the common central axis of two thin symmetric lenses.

Distance between object and lens 1,$p_1=+12\text{cm}$

Distance between lens 1 and 2,$d=32\text{cm}$

Lens 1 converging, focal length$f_1=8\text{cm}$

Lens 2 converging, focal length$f_2=6\text{cm}$

Using the relation between focal length, image distance and object distance find the image distance ${\mathbf{i}}_{\mathbf{2}}$. Then using the formula for overall magnification find its value.

From the solution of part a and b answer part c, d and e.

Formula are as follows:

Formula for focal length,$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$

Overall magnification,$M=m_1{m}_2$

Magnification,$m=-\frac{i}{p}$

Where,$m$is the magnification, $p$ is the pole, $f$is the focal length, and $i$ is the image distance.

(a)

For lens 1, focal length $f_1$, object distance$p_1$

Using expression for focal length,

$\begin{array}{rcl}\frac{1}{f_1}& =& \frac{1}{p_1}+\frac{1}{i_1}\\ \frac{1}{i_1}& =& \frac{1}{f_1}-\frac{1}{p_1}\\ \frac{1}{i_1}& =& \frac{p_1-f_1}{f_1{p}_1}\\ i_1& =& \frac{f_1{p}_1}{p_1-f_1}\\ & & \end{array}$

$i_1=\frac{f_1{p}_1}{p_1-f_1}…………………….\left(1\right)$

$\begin{array}{rcl}{i}_1& =& \frac{8×12}{12-8}\\ i_1& =& \frac{96}{4}\end{array}$

This serves as an object for lens 2,$p_2=d-i_1=32-24=8cm$ , and it is given that$f_2=6\text{cm}$

Modifying equation 1 for lens 2,

$i_2=\frac{f_2{p}_2}{p_2-f_2}$

$$\begin{gathered} i_2=\frac{6×8}{8-6} \\ {i}_2=\frac{48}{2} \\ {i}_2=24\text{cm} \end{gathered}$$

The image produced by lens 2 is at $24cm$.

Therefore, the image distance for the image produced by lens 2,$i_2=+24\text{cm}$.

(b)

To find overall magnification use the formula,

$M=m_1{m}_2$

Magnification $m=-\frac{i}{p}$

$$\begin{gathered} M=\left(-\frac{i_1}{p_1}\right)×\left(-\frac{i_2}{p_2}\right) \\ M=-\frac{24}{12}×-\frac{24}{8} \\ M=+6.0 \end{gathered}$$

Overall magnification for the given lens system is $+6.0$.

Therefore, the overall lateral magnification, including sign,$M=+6.0$.

(c)

Since lens 1 and 2 are converging, the object for lens 2 is outside the focal point. The final image distance is positive. Hence, the image formed by this lens system is real.

Therefore, the final image is real $\left(R\right)$.

(d)

Overall magnification for this lens system is positive, which shows that the image and the object have the same orientation.

Hence, the image is not inverted.

(e)

The final image distance is positive which shows that it is on the positive side of the lens 2, that is on the opposite side of the object.

The focal length and overall magnification of two-lens system can be found using corresponding formula. The nature of the image can be predicted from the characteristics of image formed due to given two-lens system.