Chapter 32, Problem 58P

(a)

To determine

The ray diagram for the two locations.

Explanation of Solution

Introduction:

The ray diagram is related to show the focal length $f$ , the separation between the two lens position $D$ and the distance between the object and screen $L$ .

Draw a ray diagram to show the position of object, image and screen.

  

Conclusion:

Thus, the ray diagram is given above.

(b)

To determine

The focal length of the lens using Bessel’s equation.

Explanation of Solution

Given:

The object to image distance is $1.70\text{m}$ .

The distance between the two position of the lens is $72\text{cm}$ .

Formula used:

Write the expression for focal length using Bessel’s method.

  $f=\frac{L^2-D^2}{4L}$ …… (1)

Here, $f$ is the focal length, $L$ is object to image distance and $D$ is the distance between two position of the lens.

Calculation:

Substitute $1.70\text{m}$ for $L$ and $72\text{cm}$ for $D$ in equation (1).

  $\begin{array}{c}f=\frac{{\left(1.70\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\right)}^2-{\left(72\text{cm}\right)}^2}{4\left(1.70\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\right)}\\ f=35\text{cm}\end{array}$

Conclusion:

Thus, the focal length of the lens is $35\text{cm}$ .

(c)

To determine

The two locations of the lens with respect to the object

Explanation of Solution

Given:

The object to image distance is $1.70\text{m}$ .

The distance between the two position of the lens is $72\text{cm}$ .

Formula used:

Write the expression for the first lens equation.

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

Here, $u_1$ is the distance of the first object and $v_1$ is the distance of the first image

Write the expression for the second lens equation.

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

Here, $u_2$ is the distance of the second object and $v_2$ is the distance of the second image

Write the expression for the distance between object and screen for first lens.

  $L=u_1+v_1$ …… (2)

Here, $L$ is the separation between the object and screen.

Write the expression for the distance between object and screen for first lens.

  $L=u_2+v_2$

Write the distance between the two lenses in terms of first image.

  $D=u_2-u_1$ …… (3)

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

Write the distance between the two lenses in terms of second image.

  $D=v_2-v_1$

Calculation:

Subtract equation (3) from equation (2).

  $\begin{array}{c}L-D=u_1+v_1-u_2+u_1\\ L-D=2u_1+v_1-u_2\end{array}$

Substitute $u_2$ for $v_1$ in the above equation.

  $L-D=2u_1$

Rearrange the above equation.

  $u_1=\frac{L-D}{2}$

Substitute $1.70\text{m}$ for $L$ and $72\text{cm}$ for $D$ in the above equation.

  $\begin{array}{c}{u}_1=\frac{\left(1.70\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\right)-72\text{cm}}{2}\\ u_1=49\text{cm}\end{array}$

Add equation (3) to equation (2).

  $\begin{array}{c}L+D=u_1+v_1+u_2-u_1\\ L-D=v_1+u_2\end{array}$

Substitute $u_2$ for $v_1$ in the above equation.

  $L+D=2u_2$

Rearrange the above equation.

  $u_2=\frac{L+D}{2}$

Substitute $1.70\text{m}$ for $L$ and $72\text{cm}$ for $D$ in the above equation.

  $\begin{array}{c}{u}_2=\frac{\left(1.70\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\right)+72\text{cm}}{2}\\ u_2=121\text{cm}\end{array}$

Conclusion:

Thus, the two locations of the lens with respect to object are $49\text{cm}$ and $121\text{cm}$ .

(d)

To determine

The magnification of the images for two different positions of the lens.

Explanation of Solution

Given:

The object to image distance is $1.70\text{m}$ .

The distance between the two position of the lens is $72\text{cm}$ .

The distance of the first object is $49\text{cm}$ .

The distance of the second object is $121\text{cm}$

Formula used:

Write the expression for the distance between object and screen for first lens.

  $L=u_1+v_1$

Here, $L$ is the separation between the object and screen.

Rearrange the above equation.

  $v_1=L-u_1$

Write the expression for the distance between object and screen for first lens.

  $L=u_2+v_2$

Rearrange the above equation.

  $v_2=L-u_2$

Write the expression for the lateral magnification of the first image.

  $m_1=-\frac{v_1}{u_1}$ …… (4)

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

Write the expression for the lateral magnification of the second image.

  $m_2=-\frac{v_2}{u_2}$ …… (5)

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

Calculation:

Substitute $\left(L-u_1\right)$ for $v_1$ in equation (4).

  $m_1=-\frac{\left(L-u_1\right)}{u_1}$

Substitute $49\text{cm}$ for $u_1$ and $1.70\text{m}$ for $L$ in the above equation.

  $\begin{array}{c}{m}_1=-\frac{\left(1.70\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)-49\text{cm}\right)}{49\text{cm}}\\ m_1=-2.5\end{array}$

Substitute $\left(L-u_2\right)$ for $v_2$ in equation (5).

  $m_2=-\frac{\left(L-u_2\right)}{u_2}$

Substitute $121\text{cm}$ for $u_2$ and $1.70\text{m}$ for $L$ in the above equation.

  $\begin{array}{c}{m}_1=-\frac{\left(1.70\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)-121\text{cm}\right)}{121\text{cm}}\\ m_1=-0.41\end{array}$

Conclusion:

Thus, the magnification of the two images for two lens positions are $-2.5$ and $-0.41$ .