Chapter 32, Problem 50P

(a)

To determine

The position of the final image.

Explanation of Solution

Given:

The focal length of a converging lens is $10\text{cm}$ and focal length of a diverging lenses is $\left(-15\text{cm}\right)$ .

The separation between the two lenses is $35\text{cm}$ .

The distance of the object from the converging lens is $20\text{cm}$ .

Formula used:

Draw a ray diagram to show the final image position and size.

  

Write the expression for thin lens equation for first image.

  $\frac{1}{u_1}+\frac{1}{v_1}=\frac{1}{f_1}$ …… (1)

Here, $u_1$ is the distance of the object from converging lens, $v_1$ is the distance of the first image and $f_1$ is the focal length of the converging lens.

Write the expression for thin lens equation for second image.

  $\frac{1}{u_2}+\frac{1}{v_2}=\frac{1}{f_2}$ …… (2)

Here, $u_2$ is the distance of the object from diverging lens, $v_2$ is the distance of the second image and $f_2$ is the focal length of the diverging lens.

Write the expression for the lateral magnification of image due to converging lens.

  $m_1=-\frac{v_1}{u_1}$ …… (3)

Here, $m_1$ is the lateral magnification due to converging lens.

Write the expression for the lateral magnification due to the diverging lens.

  $m_2=-\frac{v_2}{u_2}$ …… (4)

Here, $m_2$ is the lateral magnification due to the diverging lens.

Calculation:

Rewrite equation (1) to calculate the first image distance.

  $v_1=\frac{f_1{u}_1}{u_1-f_1}$

Substitute $10\text{cm}$ for $f_1$ and $20\text{cm}$ for $u_1$ in the above equation.

  $\begin{array}{c}{v}_1=\frac{\left(10\text{cm}\right)\left(20\text{cm}\right)}{\left(20\text{cm}\right)-\left(10\text{cm}\right)}\\ v_1=20\text{cm}\end{array}$

Substitute $20\text{cm}$ for $u_1$ and $20\text{cm}$ for $v_1$ in equation (3).

  $\begin{array}{c}{m}_1=-\frac{20\text{cm}}{20\text{cm}}\\ m_1=-1.0\end{array}$

The distance of the object from diverging lens is calculated below.

  $u_2=d-v_1$

Here, $d$ is the separation between the two lenses.

Substitute $20\text{cm}$ for $v_1$ and $35\text{cm}$ for $d$ in the above equation.

  $\begin{array}{c}{u}_2=35\text{cm}-20\text{cm}\\ u_2=15\text{cm}\end{array}$

Rewrite equation (2) to calculate the second image distance.

  $v_2=\frac{f_2{u}_2}{u_2-f_2}$

Substitute $\left(-15\text{cm}\right)$ for $f_2$ and $15\text{cm}$ for $u_2$ in the above equation.

  $\begin{array}{c}{v}_2=\frac{\left(-15\text{cm}\right)\left(15\text{cm}\right)}{\left(15\text{cm}\right)-\left(-15\text{cm}\right)}\\ v_2=-7.5\text{cm}\end{array}$

Substitute $15\text{cm}$ for $u_2$ and $\left(-7.5\text{cm}\right)$ for $v_2$ in equation (4).

  $\begin{array}{c}{m}_2=-\frac{-7.5\text{cm}}{15\text{cm}}\\ m_2=0.50\end{array}$

The final distance of the image is calculated below.

  $v=v_2+d+u_1$

Here, $v$ is the distance of the final image.

Substitute $\left(-7.5\text{cm}\right)$ for $v_2$ , $20\text{cm}$ for $u_1$ and $35\text{cm}$ for $d$ in the above equation.

  $\begin{array}{c}v=\left(-7.5\text{cm}\right)+35\text{cm}+20\text{cm}\\ v\approx 48\text{cm}\end{array}$

Conclusion:

Thus, the distance of the final image is $48\text{cm}$ to the right of the object.

(b)

To determine

Whether the final image is real or virtual and upright or inverted.

Explanation of Solution

Given:

The distance of the second image is $\left(-7.5\text{cm}\right)$ .

The lateral magnification of the image due to the converging lens is $\left(-1.0\right)$ .

The lateral magnification of the image due to the diverging lens is $0.50$ .

Formula used:

Write the expression for the overall lateral magnification for the system of two lenses.

  $m=m_1{m}_2$ …… (5)

Here, is the overall lateral magnification for the system of two lenses.

Calculation:

Substitute $\left(-1.0\right)$ for $m_1$ and $0.50$ for $m_2$ in equation (5).

  $\begin{array}{c}m=\left(-1.0\right)\left(0.50\right)\\ m=-0.50\end{array}$

Conclusion:

Thus, the image is virtual and inverted. Because the image distance, $u_2<0$ , so the image is virtual and the overall magnification, $m<0$ , so the image is inverted and half of the size of the object.

(c)

To determine

The overall magnification of the system.

Explanation of Solution

Given:

The lateral magnification of the image due to the converging lens is $\left(-1.0\right)$ .

The lateral magnification of the image due to the diverging lens is $0.50$ .

Formula used:

Write the expression for the overall lateral magnification for the system of two lenses.

  $m=m_1{m}_2$

Calculation:

Substitute $\left(-1.0\right)$ for $m_1$ and $0.50$ for $m_2$ in equation (5).

  $\begin{array}{c}m=\left(-1.0\right)\left(0.50\right)\\ m=-0.50\end{array}$

Conclusion:

Thus, the overall magnification for the system of two lenses is $\left(-0.50\right)$ .