Chapter 24, Problem 49QAP

To determine

(a)

The image distance, height, type and orientation of the image

Answer to Problem 49QAP

Image distance $=30.0cm$

Image height $=60.0cm$

Type of image = Real
Orientation of the image = Inverted

Explanation of Solution

Given info:

Distance to the object from the mirror = $10.0cm$

Radius of the mirror = $15.0cm$

Height of the object = $20.0cm$

Formula used:

  $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

  $f=$ Focal length
  $p=$ Object distance
  $q=$ Image distance
  $f=\frac{R}{2}$

  $R=$ Radius
  $M=-\frac{q}{p}$

  $M=$ Magnification
  $M=\frac{h_i}{h_o}$

  $h_i=$ Image height
  $h_o=$ Object height

Calculation:

  $f=\frac{R}{2}=\frac{15.0cm}{2}=7.5cm$

  $\begin{array}{l}\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{7.5cm}-\frac{1}{10.0cm}\\ q=30.0cm\end{array}$

  $\begin{array}{l}\frac{h_i}{h_o}=-\frac{q}{p}\\ \frac{h_i}{20.0cm}=-\frac{30.0cm}{10.0cm}\\ h_i=60.0cm\end{array}$

Conclusion:

Image distance $=30.0cm$

Image height $=60.0cm$

Type of image = Real
Orientation of the image = Inverted

To determine

(b)

The image distance, height, type and orientation of the image

Answer to Problem 49QAP

Image distance $=12.0cm$

Image height $=12.0cm$

Type of image = Real
Orientation of the image = Inverted

Explanation of Solution

Given info:

Distance to the object from the mirror = $20.0cm$

Radius of the mirror = $15.0cm$

Height of the object = $20.0cm$

Formula used:

  $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

  $f=$ Focal length
  $p=$ Object distance
  $q=$ Image distance
  $f=\frac{R}{2}$

  $R=$ Radius
  $M=-\frac{q}{p}$

  $M=$ Magnification
  $M=\frac{h_i}{h_o}$

  $h_i=$ Image height
  $h_o=$ Object height

Calculation:

  $f=\frac{R}{2}=\frac{15.0cm}{2}=7.5cm$

  $\begin{array}{l}\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{7.5cm}-\frac{1}{20.0cm}\\ q=12.0cm\end{array}$

  $\begin{array}{l}\frac{h_i}{h_o}=-\frac{q}{p}\\ \frac{h_i}{20.0cm}=-\frac{12.0cm}{20.0cm}\\ h_i=-12.0cm\end{array}$

Conclusion:

Image distance $=12.0cm$

Image height $=12.0cm$

Type of image = Real
Orientation of the image = Inverted

To determine

(c)

The image distance, height, type and orientation of the image

Answer to Problem 49QAP

Image distance $=8.11cm$

Image height $=1.62cm$

Type of image = Real
Orientation of the image = Inverted

Explanation of Solution

Given info:

Distance to the object from the mirror = $100.0cm$

Radius of the mirror = $15.0cm$

Height of the object = $20.0cm$

Formula used:

  $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

  $f=$ Focal length
  $p=$ Object distance
  $q=$ Image distance
  $f=\frac{R}{2}$

  $R=$ Radius
  $M=-\frac{q}{p}$

  $M=$ Magnification
  $M=\frac{h_i}{h_o}$

  $h_i=$ Image height
  $h_o=$ Object height

Calculation:

  $f=\frac{R}{2}=\frac{15.0cm}{2}=7.5cm$

  $\begin{array}{l}\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{7.5cm}-\frac{1}{100.0cm}\\ q=8.11cm\end{array}$

  $\begin{array}{l}\frac{h_i}{h_o}=-\frac{q}{p}\\ \frac{h_i}{20.0cm}=-\frac{8.11cm}{100.0cm}\\ h_i=-1.62cm\end{array}$

Conclusion:

Image distance $=8.11cm$

Image height $=1.62cm$

Type of image = Real
Orientation of the image = Inverted