Chapter 36, Problem 53P

To determine

The position, height and nature of the image.

Answer to Problem 53P

The position, height and nature of the image are $\underset{\_}{7.47\text{cm}}$ in front of the second lens, $\underset{\_}{1.07\text{cm}}$, upright and virtual respectively.

Explanation of Solution

Write the expression for thin lens equation.

    $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$                                                                                                                 (I)

Here, $p$ is the object distance, $q$ is the image distance and $f$ is the focal length.

Rewrite equation (I) to find the image distance of the first lens.

`    $q_1=\frac{p_1{f}_1}{p_1-f_1}$                                                                                                              (II)

Write the expression for the magnification of the first lens.

    $M_1=\frac{-q_1}{p_1}$                                                                                                                (III)

Write the expression to find the object distance of the second lens.

    $p_2=6.00\text{cm}+\left|q_1\right|$                                                                                                  (IV)

Write the expression to find the image distance of the second lens.

`    $q_2=\frac{p_2{f}_2}{p_2-f_2}$                                                                                                            (V)

Write the expression for the magnification of the second lens.

    $M_2=\frac{-q_2}{p_2}$                                                                                                              (VI)

Write the expression for total magnification.

    $M=M_1{M}_2$                                                                                                           (VII)

Write the expression for the height of the final image.

`    $h^{\prime }=Mh$                                                                                                               (VIII)

Here, $h^{\prime }$ is the height of the final image and $h$ is the height of the object.

Conclusion:

Substitute $4.00\text{cm}$ for $p_1$ and $8.00\text{cm}$ for $f_1$ in equation (II).

    $\begin{array}{c}{q}_1=\frac{\left(4.00\text{cm}\right)\left(8.00\text{cm}\right)}{4.00\text{cm}-8.00\text{cm}}\\ =-8.00\text{cm}\end{array}$

Substitute $4.00\text{cm}$ for $p_1$ and $-8.00\text{cm}$ for $q_1$ in equation (III).

    $\begin{array}{c}{M}_1=\frac{-\left(-8.00\text{cm}\right)}{4.00\text{cm}}\\ =+2.00\end{array}$

Substitute $-8.00\text{cm}$ for $q_1$ in equation (IV).

    $\begin{array}{c}{p}_2=6.00\text{cm}+\left|-8.00\text{cm}\right|\\ =+14.00\text{cm}\end{array}$

Substitute $14.00\text{cm}$ for $p_2$ and $-16.0\text{cm}$ for $f_2$ in equation (V).

    $\begin{array}{c}{q}_2=\frac{\left(14.00\text{cm}\right)\left(-16.0\text{cm}\right)}{14.00\text{cm}-\left(-16.0\text{cm}\right)}\\ =-7.47\text{cm}\end{array}$

Substitute $14.00\text{cm}$ for $p_2$ and $-7.47\text{cm}$ for $q_2$ in equation (VI).

    $\begin{array}{c}{M}_1=\frac{-\left(-7.47\text{cm}\right)}{14.00\text{cm}}\\ =+0.533\end{array}$

Substitute $+2.00$ for $M_1$ and $+0.533$ for $M_2$ in equation (VII).

    $\begin{array}{c}M=\left(+2.00\right)\left(+0.533\right)\\ =+1.07\end{array}$

Substitute $+1.07$ for $M$ and $1.00\text{cm}$ for $h$ in equation (VIII).

    $\begin{array}{c}{h}^{\prime }=\left(+1.07\right)\left(1.00\text{cm}\right)\\ =1.07\text{cm}\end{array}$

The final image is upright as $M>0$ and the image is virtual as $q<0$.

Therefore, the position, height and nature of the image are $\underset{\_}{7.47\text{cm}}$ in front of the second lens, $\underset{\_}{1.07\text{cm}}$, upright and virtual respectively.