Chapter 24, Problem 57QAP

To determine

(a)

The image formed when a 10.0-cm-tall object is positioned 20.0 cm from the mirror. provide the image distance, the image height, the type of image (real or virtual), and the orientation of the image (upright or inverted).

Answer to Problem 57QAP

Mirror image appear $6.66\text{ cm}$ behind the mirror
Height the image= $3.33\text{ cm}$

Virtual, upright image

Explanation of Solution

Given info:

Distance to object from mirror= $20.00\text{ cm}$

Radius of the curvature= $20.00\text{ cm}$

Height of the object= $10.00\text{ cm}$

Formula used:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\text{focal length}\\ R=\text{radius of curvature}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=\text{focal length}\\ v=\text{distance from image to mirror}\\ u=\text{distance from object to mirror}\end{array}$

  $\begin{array}{l}m=\frac{v}{u}=\frac{H_i}{H_o}\\ m=\text{magnification}\\ v=\text{distance from image to mirror}\\ H_i=\text{height of the mirror image}\\ H_o=\text{height of the object}\end{array}$

Calculations:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\frac{20.00\text{ cm}}{2}\\ f=10.0\text{ cm}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=10.00\text{ cm}\\ u=-20.00\text{ cm}\\ \frac{1}{10}=\frac{1}{v}+(-\frac{1}{20})\\ v=6.66\text{ cm}\end{array}$

  $\begin{array}{l}\frac{v}{u}=\frac{H_i}{H_o}\\ \frac{6.66\text{ cm}}{20.00\text{ cm}}=\frac{H_i}{10.0\text{ cm}}\\ H_i=3.33\text{ cm}\end{array}$

Conclusion:

Mirror image appear $6.66\text{ cm}$ behind the mirror
Height the image= $3.33\text{ cm}$

Virtual, upright image

To determine

(b)

The image formed when a 10.0-cm-tall object is positioned 50.0 cm from the mirror, provide the image distance, the image height, the type of image (real or virtual), and the orientation of the image (upright or inverted).

Answer to Problem 57QAP

Mirror image appear $8.33\text{ cm}$ behind the mirror
Height the image= $1.67\text{ cm}$

Virtual, upright image

Explanation of Solution

Given info:

Distance to object from mirror= $50.00\text{ cm}$

Radius of the curvature= $20.00\text{ cm}$

Height of the object= $10.00\text{ cm}$

Formula used:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\text{focal length}\\ R=\text{radius of curvature}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=\text{focal length}\\ v=\text{distance from image to mirror}\\ u=\text{distance from object to mirror}\end{array}$

  $\begin{array}{l}m=\frac{v}{u}=\frac{H_i}{H_o}\\ m=\text{magnification}\\ v=\text{distance from image to mirror}\\ H_i=\text{height of the mirror image}\\ H_o=\text{height of the object}\end{array}$

Calculations:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\frac{20.00\text{ cm}}{2}\\ f=10.0\text{ cm}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=10.00\text{ cm}\\ u=-50.00\text{ cm}\\ \frac{1}{10}=\frac{1}{v}+(-\frac{1}{50})\\ v=8.33\text{ cm}\end{array}$

  $\begin{array}{l}\frac{v}{u}=\frac{H_i}{H_o}\\ \frac{8.33\text{ cm}}{50.00\text{ cm}}=\frac{H_i}{10.0\text{ cm}}\\ H_i=1.67\text{ cm}\end{array}$

Conclusion:

Mirror image appear $8.33\text{ cm}$ behind the mirror
Height the image= $1.67\text{ cm}$

Virtual, upright image

To determine

(c)

The image formed when a 10.0-cm-tall object is positioned 100.0 cm from the mirror. For each case, provide the image distance, the image height, the type of image (real or virtual), and the orientation of the image (upright or inverted).

Answer to Problem 57QAP

Mirror image appear $9.10\text{ cm}$ behind the mirror
Height the image= $0.91\text{ cm}$

Virtual, upright image

Explanation of Solution

Given info:

Distance to object from mirror= $100.00\text{ cm}$

Radius of the curvature= $20.00\text{ cm}$

Height of the object= $10.00\text{ cm}$

Formula used:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\text{focal length}\\ R=\text{radius of curvature}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=\text{focal length}\\ v=\text{distance from image to mirror}\\ u=\text{distance from object to mirror}\end{array}$

  $\begin{array}{l}m=\frac{v}{u}=\frac{H_i}{H_o}\\ m=\text{magnification}\\ v=\text{distance from image to mirror}\\ H_i=\text{height of the mirror image}\\ H_o=\text{height of the object}\end{array}$

Calculations:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\frac{20.00\text{ cm}}{2}\\ f=10.0\text{ cm}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=10.00\text{ cm}\\ u=-100.00\text{ cm}\\ \frac{1}{10}=\frac{1}{v}+(-\frac{1}{100})\\ v=9.10\text{ cm}\end{array}$

  $\begin{array}{l}\frac{v}{u}=\frac{H_i}{H_o}\\ \frac{9.10\text{ cm}}{100.00\text{ cm}}=\frac{H_i}{10.0\text{ cm}}\\ H_i=0.91\text{ cm}\end{array}$

Conclusion:

Mirror image appear $9.10\text{ cm}$ behind the mirror
Height the image= $0.91\text{ cm}$

Virtual, upright image