Chapter 24, Problem 70QAP

To determine

(a)

The image distance and height and to draw the ray diagram

Answer to Problem 70QAP

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

The image is tall and upright $h_1=13cm$ which is greater than 0

  

Explanation of Solution

Given:

Power of the converging lens $=5diopters$

Distance $d_o=+5cm$

Calculation:

Focal length

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

Image distance:

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

  $\begin{array}{l}{d}_1=\frac{20\times 5}{5-20}\\ =-6.7cm\end{array}$

Image height:

  $\begin{array}{l}m=\frac{h_1}{h_o}=-\frac{d_1}{d_o}\\ h_1=-\frac{d_1{h}_o}{d_o}\\ =-\frac{-6.7\times 10}{5}\\ =13cm\end{array}$

Conclusion:

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

The image is tall and upright $h_1=13cm$ which is greater than 0

  

To determine

(b)

The image distance and height and to draw the ray diagram

Answer to Problem 70QAP

The image is virtual because $d_1=-20cm$ which is less than 0

The image is tall and upright $h_1=20cm$ which is greater than 0

  

Explanation of Solution

Given:

Power of the converging 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:

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

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

Image height:

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

Conclusion:

The image is virtual because $d_1=-20cm$ which is less than 0

The image is tall and upright $h_1=20cm$ which is greater than 0

  

To determine

(c)

The image distance and height and to draw the ray diagram

Answer to Problem 70QAP

No image is formed

  

Explanation of Solution

Given:

Power of the converging 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:

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

  $\begin{array}{l}{d}_1=\frac{20\times 20}{20-20}\\ =\infty \end{array}$

Conclusion:

No image is formed

  

To determine

(d)

The image distance and height and to draw the ray diagram

Answer to Problem 70QAP

The image is Real because $d_1=33cm$ which is greater than 0

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

  

Explanation of Solution

Given:

Power of the converging 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:

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

  $\begin{array}{l}{d}_1=\frac{20\times 50}{50-20}\\ =33cm\end{array}$

Image height:

  $\begin{array}{l}m=\frac{h_1}{h_o}=-\frac{d_1}{d_o}\\ h_1=-\frac{d_1{h}_o}{d_o}\\ =-\frac{-33\times 10}{50}\\ =-6.7cm\end{array}$

Conclusion:

The image is Real because $d_1=33cm$ which is greater than 0

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