Chapter 32, Problem 59P

(a)

To determine

The distance between the object and the final image.

Explanation of Solution

Given:

The distance of an object is $17.5\text{cm}$ to the left of a lens.

The focal length of the lens is $+8.50\text{cm}$ .

The focal length of a second lens is $-30.0\text{cm}$ .

The distance of the second lens is $+5.00\text{cm}$ to the right of the first lens.

Formula used:

Draw a ray diagram to show the image distance and its properties.

  

Write the expression for the thin lens equation for the first lens.

  $\frac{1}{u_1}+\frac{1}{v_1}=\frac{1}{f_1}$

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

Rearrange the above equation to calculate the image distance for first lens.

  $v_1=\frac{f_1{u}_1}{u_1-f_1}$ …… (1)

Write the expression for the thin lens equation for the second lens.

  $\frac{1}{u_2}+\frac{1}{v_2}=\frac{1}{f_2}$

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

Rearrange the above equation to calculate the image distance for second lens.

  $v_2=\frac{f_2{u}_2}{u_2-f_2}$ …… (2)

Write the expression for the object distance for second lens.

  $u_2=l-v_1$ …… (3)

Here, $l$ is the distance between the two lenses.

Write the expression for object to image distance.

  $d=u_1+l+v_2$ …… (4)

Here, $d$ is the object to image distance.

Calculation:

Substitute $8.50\text{cm}$ for $f_1$ and $17.5\text{cm}$ for $u_1$ in equation (1).

  $\begin{array}{c}{v}_1=\frac{\left(8.50\text{cm}\right)\left(17.5\text{cm}\right)}{17.5\text{cm}-8.50\text{cm}}\\ v_1\approx 16.53\text{cm}\end{array}$

Substitute $5.00\text{cm}$ for $l$ and $16.53\text{cm}$ for $v_1$ in the equation (3).

  $\begin{array}{c}{u}_2=5.00\text{cm}-16.53\text{cm}\\ =-11.53\text{cm}\end{array}$

Substitute $\left(-30.0\text{cm}\right)$ for $f_2$ and $\left(-11.53\text{cm}\right)$ for $u_2$ in the equation (2).

  $\begin{array}{c}{v}_2=\frac{\left(-30.0\text{cm}\right)\left(-11.53\text{cm}\right)}{\left(-11.53\text{cm}\right)-\left(-30.0\text{cm}\right)}\\ v_2\approx 18.73\text{cm}\end{array}$

Substitute $17.5\text{cm}$ for $u_1$ , $5.00\text{cm}$ for $l$ and $18.73\text{cm}$ for $v_2$ in equation (4).

  $\begin{array}{c}d=17.5\text{cm}+5.00\text{cm}+18.73\text{cm}\\ d\approx 41\text{cm}\end{array}$

Conclusion:

Thus, the object to the final image distance is $41\text{cm}$ from object.

(b)

To determine

The overall magnification of the system.

Explanation of Solution

Given:

The object distance for the first lens is $17.5\text{cm}$ .

The image distance for the first lens is $16.53\text{cm}$ .

The object distance for the second lens is $\left(-11.53\text{cm}\right)$ .

The image distance for the second lens is $18.73\text{cm}$ .

Formula used:

Write the expression for the lateral magnification of the image formed by the first lens.

  $m_1=-\frac{v_1}{u_1}$

Here, $m_1$ is the lateral magnification of the image formed by the first lens.

Write the expression for the lateral magnification of the image formed by the second lens.

  $m_2=-\frac{v_2}{u_2}$

Here, $m_2$ is the lateral magnification of the image formed by the second lens.

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

  $m=m_1{m}_2$

Here, $m$ is the overall magnification of a system.

Substitute $\left(-\frac{v_1}{u_1}\right)$ for $m_1$ and $\left(-\frac{v_2}{u_2}\right)$ for $m_2$ in the above equation.

  $\begin{array}{c}m=\left(-\frac{v_1}{u_1}\right)\left(-\frac{v_2}{u_2}\right)\\ m=\frac{v_1{v}_2}{u_1{u}_2}\end{array}$ …… (4)

Calculation:

Substitute $17.5\text{cm}$ for $u_1$ , $16.53\text{cm}$ for $v_1$ , $\left(-11.53\text{cm}\right)$ for $u_2$ and $18.73\text{cm}$ for $v_2$ in the equation (4).

  $\begin{array}{c}m=\frac{v_1{v}_2}{u_1{u}_2}\\ m=\frac{\left(16.53\text{cm}\right)\left(18.73\text{cm}\right)}{\left(17.5\text{cm}\right)\left(-11.53\text{cm}\right)}\\ m\approx -1.53\end{array}$

Conclusion:

Thus, the overall magnification is $-1.53$ .

(c)

To determine

Whether the object is real or virtual and upright or inverted.

Explanation of Solution

Given:

The final image distance from the second lens is $18.73\text{cm}$ .

The overall magnification for the system of two lenses is $-1.53$ .

Introduction:

A real image is formed by an object when all the outgoing parallel rays from the object are appeared to converge to a point. For positive image distance the image is real and for negative image distance the image is virtual.

The image distance, $v_2>0$ , so the image is real but the overall magnification for the system, $m<0$ , so the image is inverted.

Conclusion:

Thus, the image is real and inverted.