Chapter 36, Problem 36.78AP

Two converging lenses having focal lengths of f₁ = 10.0 cm and f₂ = 20.0 cm are placed a distance d = 50.0 cm apart as shown in Figure P35.48. The image due to light passing through both lenses is to be located between the lenses at the position x = 31.0 cm indicated. (a) At what value of p should the object be positioned to the left of the first lens? (b) What is the magnification of the final image? (c) Is the final image upright or inverted? (d) Is the final image real or virtual?

To determine

The value of $p$ at which the object should be placed to the left of the first lens.

Answer to Problem 36.78AP

The value of $p$ at which the object should be placed to the left of the first lens is $13.3\text{cm}$ distance in front of the first lens.

Explanation of Solution

Given info: The focal length of first lens $f_1$ is $10.0\text{cm}$, the focal length of second lens $f_2$ is $20.0\text{cm}$, the distance between lenses $d$ is $50.0\text{cm}$ and the image due to the light passing through both the lenses is to be located between the lenses at a distance $31.0\text{cm}$.

Write the expression for the final image distance $q_2$.

$q_2=-\left(d-x\right)$

Here,

$q_2$ is the final image distance.

$d$ is the distance between the lenses.

$x$ is the distance between the image position and first lens.

Substitute $50.0\text{cm}$ for $d$ and $31.0\text{cm}$ for $x$ in the above equation to get the final image distance.

$\begin{array}{c}{q}_2=-\left(50.0\text{cm}-31.0\text{cm}\right)\\ =-19.0\text{cm}\end{array}$

Write the expression for the thin lens equation.

$\begin{array}{c}\frac{1}{p_2}+\frac{1}{q_2}=\frac{1}{f_2}\\ p_2=\frac{q_2{f}_2}{q_2-f_2}\end{array}$

Here,

$p_2$ is the object distance to the second lens.

$f_2$ is the focal length of the second lens.

Substitute $-19.0\text{cm}$ for $q_2$ and $20.0\text{cm}$ for $f_2$ in the above equation to get the object distance from the second lens.

$\begin{array}{c}{p}_2=\frac{\left(-19.0\text{cm}\right)\left(20.0\text{cm}\right)}{\left(-19.0\text{cm}\right)-\left(20.0\text{cm}\right)}\\ =\frac{-380{\text{cm}}^2}{-39.0\text{cm}}\\ =9.74\text{cm}\end{array}$

The image distance to the first lens $q_1$ is,

$q_1=d-p_2$

Here,

$q_1$ is the image distance to the first lens.

Substitute $50.0\text{cm}$ for $d$ and $9.74\text{cm}$ for $p_2$ in the above equation to get the image distance to the first lens.

$\begin{array}{c}{q}_1=50.0\text{cm}-9.74\text{cm}\\ =40.26\text{cm}\end{array}$

Formula to calculate the thin lens equation is,

$\begin{array}{c}\frac{1}{p}+\frac{1}{q_1}=\frac{1}{f_1}\\ p=\frac{q_1{f}_1}{q_1-f_1}\end{array}$

Here,

$p$ is the object distance to the first lens.

$f_1$  is the focal length of the first lens.

Substitute $40.26\text{cm}$ for $q_1$ and $10.0\text{cm}$ for $f_1$ in the above equation to get the object distance to the first lens.

$\begin{array}{c}p=\frac{\left(40.26\text{cm}\right)\left(10.0\text{cm}\right)}{\left(40.26\text{cm}\right)-\left(10.0\text{cm}\right)}\\ =13.3\text{cm}\end{array}$

Conclusion:

Therefore, the value of $p$ at which the object should be placed to the left of the first lens is $13.3\text{cm}$ distance in front of the first lens.

To determine

The magnification of the final image.

Answer to Problem 36.78AP

The magnification of the final image is $-5.91$.

Explanation of Solution

Given info: The focal length of first lens $f_1$ is $10.0\text{cm}$, the focal length of second lens $f_2$ is $20.0\text{cm}$, the distance between lenses $d$ is $50.0\text{cm}$ and the image due to the light passing through both the lenses is to be located between the lenses at a distance $31.0\text{cm}$.

Write the equation for magnification of the first lens.

$M_1=-\frac{q_{\text{l}}}{p}$

Write the equation for magnification of the second lens.

$M_2=-\frac{q_2}{p_2}$

Write the equation for magnification of the final image.

$M=M_1{M}_2$

Substitute $\left(-\frac{q_1}{p}\right)$ for $M_1$ and $\left(-\frac{q_2}{p_2}\right)$ for $M_2$ in the above equation.

$\begin{array}{c}M=\left(-\frac{q_1}{p}\right)\left(-\frac{q_2}{p_2}\right)\\ =\frac{q_1{q}_2}{p{p}_2}\end{array}$

Substitute $40.26\text{cm}$ for $q_1$, $-19.0\text{cm}$ for $q_2$, $13.3\text{cm}$ for $p$ and $9.74\text{cm}$ for $p_2$ in the above equation to get the magnification of the final image.

$\begin{array}{c}M=\frac{\left(40.26\text{cm}\right)\left(-19.0\text{cm}\right)}{\left(13.3\text{cm}\right)\left(9.74\text{cm}\right)}\\ =-5.91\end{array}$

Conclusion:

Therefore, the magnification of the final image is $-5.91$.

To determine

Whether the final image is upright or inverted.

Answer to Problem 36.78AP

The final image is inverted.

Explanation of Solution

Given info: The focal length of first lens is $f_1=10.0\text{cm}$, the focal length of second lens is $f_2=20.0\text{cm}$, the distance between lenses is $d=50.0\text{cm}$ and the image due to the light passing through both the lenses is to be located between the lenses at a distance $x=31.0\text{cm}$.

Here, the value of magnification of the final image is $-5.91$.

If the magnification is found to be negative $M<0$, then an inverted image is formed.

Thus, the image formed is an inverted image.

Conclusion:

Therefore, the final image is inverted.

To determine

Whether the final image is real or virtual.

Answer to Problem 36.78AP

The final image is virtual.

Explanation of Solution

Given info: The focal length of first lens is $f_1=10.0\text{cm}$, the focal length of second lens is $f_2=20.0\text{cm}$, the distance between lenses is $d=50.0\text{cm}$ and the image due to the light passing through both the lenses is to be located between the lenses at a distance $x=31.0\text{cm}$.

Here, the value of final image distance is $-19.0\text{cm}$.

If the value of final image is found to be negative $q_2<0$, then a virtual image is formed.

Thus, the image formed is a virtual image.

Conclusion:

Therefore, the final image is virtual.