Chapter 36, Problem 36.77AP

The lens and mirror in Figure P36.77 are separated by d = 1.00 m and have focal lengths of +80.0 cm and

-50.0 cm, respectively. An object is placed p = 1.00 m to the left of the lens as shown, (a) Locate the final image, formed by light that has gone through the lens twice. (b) Determine the overall magnification of the image and (c) state whether the image is upright or inverted.

To determine

The exact position of the final image formed by light.

Answer to Problem 36.77AP

The exact position of the final image formed by light is $160\text{cm}$ to the left of the lens.

Explanation of Solution

Given info: The object distance from the lens is $1.00\text{m}$, the focal length of the lens is $+80.0\text{cm}$ and the focal length of the mirror is $-50.0\text{cm}$.

Write the expression for the image distance from the lens,

$\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{f_1}$

Here,

$p_1$ is the object distance from the lens.

$f_1$ is focal length of the lens.

Substitute $1.00\text{m}$ for $p_1$ and $+80.0\text{cm}$ for $f_1$ in the above equation.

$\begin{array}{c}\frac{1}{1.00\text{m}\left(\frac{100\text{cm}}{1.00\text{m}}\right)}+\frac{1}{q_1}=\frac{1}{+80.0\text{cm}}\\ q_1=400\text{cm}\end{array}$

Thus, the image distance from the lens is $400\text{cm}$.

This image distance act as an object for the mirror with an object distance of,

$p_2=100\text{cm}-q_1$

Substitute $400\text{cm}$ for $q_1$ in the above equation.

$\begin{array}{c}{p}_2=100\text{cm}-400\text{cm}\\ =-300\text{cm}\end{array}$

Write the expression for the image distance from the mirror,

$\frac{1}{p_2}+\frac{1}{q_2}=\frac{1}{f_2}$

Here,

$p_2$ is the object distance from the mirror.

$f_2$ is focal length of the mirror.

Substitute $-300\text{cm}$ for $p_1$ and $-50.0\text{cm}$ for $f_1$ in the above equation.

$\begin{array}{c}\frac{1}{-300\text{cm}}+\frac{1}{q_2}=\frac{1}{-50.0\text{cm}}\\ q_2=-60\text{cm}\end{array}$

Thus, the image distance from the mirror is $-60\text{cm}$.

This image distance act as an object for the lens when the light passes through right to left,

The object distance for the lens is,

$p_3=100\text{cm}-q_2$

Substitute $-60\text{cm}$ for $q_2$ in the above equation.

$\begin{array}{c}{p}_3=100\text{cm}-\left(-60\text{cm}\right)\\ =160\text{cm}\end{array}$

Write the expression for the image distance from the thin lens,

$\frac{1}{p_3}+\frac{1}{q_3}=\frac{1}{f_1}$

Here,

$p_3$ is the object distance from the thin lens.

$f_1$ is focal length of the thin lens.

Substitute $160\text{cm}$ for $p_3$ and $+80.0\text{cm}$ for $f_1$ in the above equation.

$\begin{array}{c}\frac{1}{160\text{cm}}+\frac{1}{q_3}=\frac{1}{+80.0\text{cm}}\\ q_3=160\text{cm}\end{array}$

Conclusion:

Therefore, the exact position of the final image formed by light is $160\text{cm}$ to the left of the lens.

To determine

The overall magnification of the image.

Answer to Problem 36.77AP

The overall magnification of the image is $-0.800$.

Explanation of Solution

Given info: : The object distance from the lens is $1.00\text{m}$, the focal length of the lens is $+80.0\text{cm}$ and the focal length of the mirror is $-50.0\text{cm}$.

From the part (a) the image distance from the lens is $400\text{cm}$, image distance from the mirror is $-60\text{cm}$, exact position of the final image formed by light is $160\text{cm}$, the object distance for mirror is $-300\text{cm}$ and the object distance for lens is $160\text{cm}$ when the light passes through right to left.

Formula to calculate the overall magnification is,

$M=M_1\cdot M_2\cdot M_3$ (1)

Formula to calculate the magnification of lens is,

$M_1=\left(-\frac{q_1}{p_1}\right)$

Formula to calculate the magnification of mirror is,

$M_2=\left(-\frac{q_2}{p_2}\right)$

Formula to calculation the magnification of lens when the light passes through right to left is,

$M_3=\left(-\frac{q_3}{p_3}\right)$

Substitute $\left(-\frac{q_1}{p_1}\right)$ for $M_1$, $\left(-\frac{q_2}{p_2}\right)$ for $M_2$ and $\left(-\frac{q_3}{p_3}\right)$ for $M_3$ in the equation (1).

$M=\left(-\frac{q_1}{p_1}\right)\cdot \left(-\frac{q_2}{p_2}\right)\cdot \left(-\frac{q_3}{p_3}\right)$

Substitute $400\text{cm}$ for $q_1$, $1.00\text{m}$ for $p_1$, $-60\text{cm}$ for $q_2$, $-300\text{cm}$ for $p_2$, $160\text{cm}$ for $q_3$ and $160\text{cm}$ for $p_3$ in the above equation.

$\begin{array}{c}M=\left(-\frac{400\text{cm}}{1.0\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)}\right)\cdot \left[-\frac{\left(-60\text{cm}\right)}{\left(-300\text{cm}\right)}\right]\cdot \left(-\frac{160\text{cm}}{160\text{cm}}\right)\\ =-\left(4\right)\left(0.2\right)\\ =-0.800\end{array}$

Conclusion:

Therefore, the overall magnification of the image is $-0.800$.

To determine

Whether the image is upright or inverted.

Answer to Problem 36.77AP

The image is inverted.

Explanation of Solution

From part (b) the overall magnification of the image is $-0.800$.

Since, the overall magnification of the image is less than zero $\left(M_{\text{overall}}<0\right)$ and it gives the inverted final image.

Conclusion:

Therefore, the image is inverted.