Chapter 24, Problem 48QAP

To determine

(a)

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

Answer to Problem 48QAP

Image distance $=10.0cm$

Image height $=20.0cm$

Type of image = Virtual
Orientation of the image = Upright

Explanation of Solution

Given info:

Distance to the object from the mirror = $5.0cm$

Radius of the mirror = $20.0cm$

Height of the object = $10.0cm$

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:

  $f=\frac{R}{2}=\frac{20.0cm}{2}=10.0cm$

  $\begin{array}{l}\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10.0cm}-\frac{1}{5.0cm}\\ q=-10.0cm\end{array}$

  $\begin{array}{l}\frac{h_i}{h_o}=-\frac{q}{p}\\ \frac{h_i}{10.0cm}=-\frac{-10.0cm}{5.0cm}\\ h_i=20.0cm\end{array}$

Conclusion:

Image distance $=10.0cm$

Image height $=20.0cm$

Type of image = Virtual
Orientation of the image = Upright

To determine

(b)

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

Answer to Problem 48QAP

Image distance $=20.0cm$

Image height $=10.0cm$

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 = $20.0cm$

Height of the object = $10.0cm$

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:

  $f=\frac{R}{2}=\frac{20.0cm}{2}=10.0cm$

  $\begin{array}{l}\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10.0cm}-\frac{1}{20.0cm}\\ q=20.0cm\end{array}$

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

Conclusion:

Image distance $=20.0cm$

Image height $=10.0cm$

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 48QAP

Image distance $=12.5cm$

Image height $=0.25cm$

Type of image = Real
Orientation of the image = Inverted

Explanation of Solution

Given info:

Distance to the object from the mirror = $50.0cm$

Radius of the mirror = $20.0cm$

Height of the object = $10.0cm$

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:

  $f=\frac{R}{2}=\frac{20.0cm}{2}=10.0cm$

  $\begin{array}{l}\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10.0cm}-\frac{1}{50.0cm}\\ q=12.5cm\end{array}$

  $\begin{array}{l}\frac{h_i}{h_o}=-\frac{q}{p}\\ \frac{h_i}{10.0cm}=-\frac{12.5cm}{50.0cm}\\ h_i=-0.25cm\end{array}$

Conclusion:

Image distance $=12.5cm$

Image height $=0.25cm$

Type of image = Real
Orientation of the image = Inverted

To determine

(d)

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

Answer to Problem 48QAP

Image distance $=11.1cm$

Image height $=1.11cm$

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 = $20.0cm$

Height of the object = $10.0cm$

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:

  $f=\frac{R}{2}=\frac{20.0cm}{2}=10.0cm$

  $\begin{array}{l}\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10.0cm}-\frac{1}{100.0cm}\\ q=11.1cm\end{array}$

  $\begin{array}{l}\frac{h_i}{h_o}=-\frac{q}{p}\\ \frac{h_i}{10.0cm}=-\frac{11.1cm}{100.0cm}\\ h_i=-1.11cm\end{array}$

Conclusion:

Image distance $=11.1cm$

Image height $=1.11cm$

Type of image = Real
Orientation of the image = Inverted