Chapter 24, Problem 71QAP

To determine

(a)

The image distance and height and draw the ray diagram

Answer to Problem 71QAP

The image is virtual because $d_1=-4cm$ which is less than 0

The image is tall and upright $h_1=8cm$ which is greater than 0

  

Explanation of Solution

Given:

Power of the diverging lens $=5diopters$

Height $h_o=10cm$

Distance $d_o=+5cm$

Calculation:

Focal length

  $\begin{array}{l}f=\frac{1}{P}\\ =\frac{1}{-5}\\ =-0.20m\\ =-20cm\end{array}$

Image distance:

  $\begin{array}{l}\frac{1}{d_o}+\frac{1}{d_1}=\frac{1}{f}\\ \frac{1}{d_1}=\frac{1}{f}-\frac{1}{d_o}\\ \frac{1}{d_1}=\frac{d_o-f}{f{d}_o}\\ d_1=\frac{f{d}_o}{d_o-f}\end{array}$

  $\begin{array}{l}{d}_1=\frac{-20\times 5}{5-(-20)}\\ =-4cm\end{array}$

Image height:

  $\begin{array}{l}m=\frac{h_1}{h_o}=-\frac{d_1}{d_o}\\ h_1=-\frac{d_1{h}_o}{d_o}\\ =-\frac{-4\times 10}{5}\\ =8cm\end{array}$

Conclusion:

The image is virtual because $d_1=-4cm$ which is less than 0

The image is tall and upright $h_1=8cm$ which is greater than 0

  

To determine

(b)

The image distance and height and to draw the ray diagram

Answer to Problem 71QAP

The image is virtual because $d_1=-6.7cm$ which is less than 0

The image is tall and upright $h_1=6.7cm$ which is greater than 0

  

Explanation of Solution

Given:

Power of the diverging lens $=5diopters$

Distance $d_o=+10cm$

Calculation:

Focal length

  $\begin{array}{l}f=\frac{1}{P}\\ =\frac{1}{-5}\\ =-0.20m\\ =-20cm\end{array}$

Image distance:

  $\begin{array}{l}\frac{1}{d_o}+\frac{1}{d_1}=\frac{1}{f}\\ \frac{1}{d_1}=\frac{1}{f}-\frac{1}{d_o}\\ \frac{1}{d_1}=\frac{d_o-f}{f{d}_o}\\ d_1=\frac{f{d}_o}{d_o-f}\end{array}$

  $\begin{array}{l}{d}_1=\frac{-20\times 10}{10-(-20)}\\ =-6.7cm\end{array}$

Image height:

  $\begin{array}{l}m=\frac{h_1}{h_o}=-\frac{d_1}{d_o}\\ h_1=-\frac{d_1{h}_o}{d_o}\\ =-\frac{-6.7\times 10}{10}\\ =6.7cm\end{array}$

Conclusion:

The image is virtual because $d_1=-6.7cm$ which is less than 0

The image is tall and upright $h_1=6.7cm$ which is greater than 0

  

To determine

(c)

To calculate the image distance and height and to draw the ray diagram

Answer to Problem 71QAP

The image is virtual because $d_1=-10cm$ which is less than 0

The image is tall and upright $h_1=5cm$ which is greater than 0

  

Explanation of Solution

Given:

Power of the diverging lens $=5diopters$

Distance $d_o=+20cm$

Calculation:

Focal length

  $\begin{array}{l}f=\frac{1}{P}\\ =\frac{1}{-5}\\ =-0.20m\\ =-20cm\end{array}$

Image distance:

  $\begin{array}{l}\frac{1}{d_o}+\frac{1}{d_1}=\frac{1}{f}\\ \frac{1}{d_1}=\frac{1}{f}-\frac{1}{d_o}\\ \frac{1}{d_1}=\frac{d_o-f}{f{d}_o}\\ d_1=\frac{f{d}_o}{d_o-f}\end{array}$

  $\begin{array}{l}{d}_1=\frac{20\times 20}{20-(-20)}\\ =-10cm\end{array}$

Image height:

  $\begin{array}{l}m=\frac{h_1}{h_o}=-\frac{d_1}{d_o}\\ h_1=-\frac{d_1{h}_o}{d_o}\\ =-\frac{-10\times 10}{20}\\ =5cm\end{array}$

Conclusion:

The image is virtual because $d_1=-10cm$ which is less than 0

The image is tall and upright $h_1=5cm$ which is greater than 0

  

To determine

(d)

To calculate the image distance and height and to draw the ray diagram

Answer to Problem 71QAP

The image is virtual because $d_1=-14cm$ which is less than 0

The image is tall and upright $h_1=2.9cm$ which is greater than 0

  

Explanation of Solution

Given:

Power of the diverging lens $=5diopters$

Distance $d_o=+50cm$

Calculation:

Focal length

  $\begin{array}{l}f=\frac{1}{P}\\ =\frac{1}{-5}\\ =-0.20m\\ =-20cm\end{array}$

Image distance:

  $\begin{array}{l}\frac{1}{d_o}+\frac{1}{d_1}=\frac{1}{f}\\ \frac{1}{d_1}=\frac{1}{f}-\frac{1}{d_o}\\ \frac{1}{d_1}=\frac{d_o-f}{f{d}_o}\\ d_1=\frac{f{d}_o}{d_o-f}\end{array}$

  $\begin{array}{l}{d}_1=\frac{-20\times 50}{50-(-20)}\\ =-14cm\end{array}$

Image height:

  $\begin{array}{l}m=\frac{h_1}{h_o}=-\frac{d_1}{d_o}\\ h_1=-\frac{d_1{h}_o}{d_o}\\ =-\frac{-14\times 10}{50}\\ =2.9cm\end{array}$

Conclusion:

The image is virtual because $d_1=-14cm$ which is less than 0

The image is tall and upright $h_1=2.9cm$ which is greater than 0