Chapter 38, Problem 76PQ

Figure P38.76 shows an object placed a distance d_o1 from one of two converging lenses separated by s = 1.00 m. The first lens has focal length f₁ = 22.0 cm, and the second lens has focal length f₂ = 45.0 cm. An image is formed by light passing through both lenses at a distance d_i2 = 15.0 cm to the left of the second lens. a. What is the value of d_o1 that will result in this image position? b. Is the final image formed by the two lenses real or virtual? c. What is the magnification of the final image? d. Is the final image upright or inverted?

Figure P38.76

To determine

The value of $d_{01}$.

Answer to Problem 76PQ

The value of $d_{01}$ is $29.25\text{cm}$.

Explanation of Solution

Write the thin lens equation for first lens.

    $\frac{1}{f_1}=\frac{1}{d_0}+\frac{1}{d_{\text{i1}}}$

Here, $f_1$ is the focal length, $d_{01}$ is the object distance, and $d_{i1}$ is the image distance for the first lens.

Rearrange the above equation to find $d_{o1}$ and $d_{02}$ we get,

    $\begin{array}{c}\frac{1}{d_0}=\frac{1}{f_1}-\frac{1}{d_{\text{i1}}}\\ d_{01}=\frac{f_1{d}_{\text{i1}}}{d_{\text{i}1}-f_1}\end{array}$                                         (I)

Similarly,

    $d_{02}=\frac{f_2{d}_{\text{i2}}}{d_{\text{i2}}-f_2}$                                       (II)

The image of first lens acts as object for second lens. Hence, object distance for second lens is,

    $d_{02}=s-d_{\text{i1}}$                              (III)

Conclusion:

Substitute $-15.0\text{cm}$ for $d_{\text{i2}}$ and $45.0\text{cm}$ for $f_2$. in the equation (II) to find $d_{02}$.

$\begin{array}{c}{d}_{02}=\frac{\left(-15.0\text{cm}\right)\left(45.0\text{cm}\right)}{-15.0\text{cm - 45}\text{.0}\text{cm}}\\ =11.25\text{cm}\end{array}$

From equation (III), we get,

$d_{\text{i1}}=s-d_{02}$

Substitute $1.00\text{m}$ for $s$ and $11.25\text{cm}$ for $d_{02}$ in the above equation to find $d_{\text{i1}}$.

    $\begin{array}{c}{d}_{\text{i1}}=1.00\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)-11.25\text{cm}\\ \text{=88}\text{.75}\text{cm}\end{array}$

Substitute $88.75\text{cm}$ for $d_{\text{i1}}$ and $22.0\text{cm}$ for $f_1$ in the equation (I) to find $d_{01}$.

    $\begin{array}{c}{d}_{01}=\frac{\left(22.0\text{cm}\right)\left(88.75\text{cm}\right)}{88.75\text{cm}\text{-}\text{22}\text{.0}\text{cm}}\\ =29.25\text{cm}\end{array}$

Thus, the value of $d_{01}$ is $29.25\text{cm}$.

To determine

Whether the image formed by the two lenses is real or virtual.

Answer to Problem 76PQ

The image formed by the two lenses is Virtual.

Explanation of Solution

Since, the image is formed on the left side of the second lens, the image is Virtual.

To determine

The magnification of the final image.

Answer to Problem 76PQ

The total magnification of the final image is $-4.04$.

Explanation of Solution

Write the expression for the total magnification of the final image.

    $m=m_1{m}_2$ (IV)

Here, $m_1$ is the magnification of the first lens and is the magnification of the second lens.

Write the expression for the magnification of the first lens.

    $m_1=-\frac{d_{\text{i1}}}{d_{01}}$                                              (V)

Here, $m_1$ is the magnification, $d_{\text{i1}}$ is the image distance and $d_{01}$ is the object distance for first lens.

Write the expression for the magnification of the second lens.

    $m_2=-\frac{d_{\text{i2}}}{d_{02}}$                                            (VI)

Here, $m_2$ is the magnification, $d_{\text{i2}}$ is the image distance and $d_{02}$ is the object distance for first lens.

Conclusion:

Substitute the equations (V) and equation (VI) in the equation (IV) to find $m$.

    $m=\left(-\frac{d_{\text{i1}}}{d_{01}}\right)\left(-\frac{d_{\text{i2}}}{d_{02}}\right)$

Substitute $88.75\text{cm}$ for $d_{\text{i1}}$, $29.25\text{cm}$ for $d_{01}$, $-15.0\text{cm}$ for $d_{\text{i2}}$ and $11.25\text{cm}$ for $d_{02}$ in the above equation to find $m$.

    $\begin{array}{c}m=\left(-\frac{88.75\text{cm}}{29.25\text{cm}}\right)\left(-\frac{-15.0\text{cm}}{11.25\text{cm}}\right)\\ =-4.04\end{array}$

Thus, the total magnification of the final image is $-4.04$.

To determine

Whether the final image is upright or inverted.

Answer to Problem 76PQ

The final image produced will be inverted.

Explanation of Solution

Since, the total magnification has a negative sign, the image formed will be inverted.