Chapter 24, Problem 92QAP

To determine

(a)

The final image of the flower produced by the lens-mirror.

Answer to Problem 92QAP

The final image is $10.5cm$ tall and inverted.

Explanation of Solution

Given:

A thin conserving lens ( $f_A=25cm$ ) is placed $1.000m$ from a plane mirror on the same principle axis

A flower ( $h_o=8.50cm$ ) is placed in front of the mirror; $d_{O,A}=0.450m=45cm$

Formula used:

  $\frac{1}{f_A}=\frac{1}{d_{O,A}}+\frac{1}{d_{1,A}}$$$

Calculation:

  $\frac{1}{f_A}=\frac{1}{d_{O,A}}+\frac{1}{d_{1,A}}$

  $\begin{array}{l}\frac{1}{d_{1,A}}=\frac{1}{f_A}-\frac{1}{d_{O,A}}\\ \frac{1}{d_{1,A}}=\frac{d_{O,A}-f_A}{f_A{d}_{O,A}}\\ d_{1,A}=\frac{f_A{d}_{O,A}}{d_{O,A}-f_A}\end{array}$

  $\begin{array}{l}{d}_{1,A}=\frac{\left(25\right)\left(45\right)}{\left(45\right)-\left(25\right)}\\ d_{1,A}=\frac{1125}{20}\\ d_{1,A}=56.25cm\end{array}$

Image formed by the mirror:

The image formed by lens is located $56.25cm$ behind the lens,

  $d_{O,B}=100-56.25=43.75cm$

  $\begin{array}{l}\frac{1}{d_{O,B}}+\frac{1}{d_{1,B}}=\frac{1}{f_B}\\ \frac{1}{d_{1,B}}=\frac{1}{f_B}-\frac{1}{d_{O,B}}\\ \frac{1}{d_{1,B}}=\frac{1}{\infty }-\frac{1}{d_{O,B}}\\ d_{1,B}=-d_{O,B}=-43.75cm\end{array}$

The image formed by the mirror is located $43.75cm$ behind the mirror.

Since $d_{1,B}<0$, the image is virtual.

Image height:

  $\begin{array}{l}{m}_{total}=m_A{m}_B\\ m_{total}=\left(-\frac{d_{1,A}}{d_{O,A}}\right)\left(-\frac{d_{1,B}}{d_{O,b}}\right)\\ m_{total}=\left(\frac{56.25}{45}\right)\left(\frac{-43.75}{43.75}\right)\\ m_{total}=1.25\times \left(-1\right)\\ m_{total}=-1.25\\ h_1=m_{total}\times h_o\\ h_1=\left(-1.25\right)\left(8.40\right)\\ h_1=-10.5cm\end{array}$

Conclusion:

The final image is $10.5cm$ tall and inverted.

To determine

(b)

The final image of the flower produced by the lens-mirror when converging lens is replaced by a diverging lens.

Answer to Problem 92QAP

The final image is $3cm$ tall and upright.

Explanation of Solution

Given:

A thin conserving lens ( $f_A=25cm$ ) is placed $1.000m$ from a plane mirror on the same principle axis

A flower ( $h_o=8.50cm$ ) is placed in front of the mirror; $d_{O,A}=0.450m=45cm$

Formula used:

  $\frac{1}{f_A}=\frac{1}{d_{O,A}}+\frac{1}{d_{1,A}}$$$

Calculation:

Image formed by the lens:

  $\frac{1}{f_A}=\frac{1}{d_{O,A}}+\frac{1}{d_{1,A}}$

  $\begin{array}{l}\frac{1}{d_{1,A}}=\frac{1}{f_A}-\frac{1}{d_{O,A}}\\ \frac{1}{d_{1,A}}=\frac{d_{O,A}-f_A}{f_A{d}_{O,A}}\\ d_{1,A}=\frac{f_A{d}_{O,A}}{d_{O,A}-f_A}\end{array}$

  $\begin{array}{l}{d}_{1,A}=\frac{\left(-25\right)\left(45\right)}{\left(45\right)-\left(-25\right)}\\ d_{1,A}=\frac{-1125}{70}\\ d_{1,A}=16.07cm\end{array}$

Image formed by the mirror:

The image formed by lens is located $16.07cm$ behind the lens,

  $d_{O,B}=100+16.07=116.07cm$

  $\begin{array}{l}\frac{1}{d_{O,B}}+\frac{1}{d_{1,B}}=\frac{1}{f_B}\\ \frac{1}{d_{1,B}}=\frac{1}{f_B}-\frac{1}{d_{O,B}}\\ \frac{1}{d_{1,B}}=\frac{1}{\infty }-\frac{1}{d_{O,B}}\\ d_{1,B}=-d_{O,B}=-116.07cm\end{array}$

The image formed by the mirror is located $116.07cm$ behind the mirror.

Since $d_{1,B}<0$, the image is virtual.

Image height:

  $\begin{array}{l}{m}_{total}=m_A{m}_B\\ m_{total}=\left(-\frac{d_{1,A}}{d_{O,A}}\right)\left(-\frac{d_{1,B}}{d_{O,b}}\right)\\ m_{total}=\left(\frac{-16.07}{45}\right)\left(\frac{-116.07}{116.07}\right)\\ m_{total}=0.357\times \left(1\right)\\ m_{total}=0.357\\ h_1=m_{total}\times h_o\\ h_1=\left(0.357\right)\left(8.40\right)\\ h_1=3cm\end{array}$

Conclusion:

The final image is $3cm$ tall and upright.