Chapter 24, Problem 56QAP

To determine

(a)

Construct ray diagrams to locate the images and estimate the image height in each of the following cases.

A 10-cm-tan object located 5 cm in front of a spherical convex mirror with a radius of curvature of 20 cm.

Answer to Problem 56QAP

    

$\text{height of the mirror image}$ = $6.66\text{ cm}$

Explanation of Solution

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:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\frac{20.00\text{ cm}}{2}\\ f=10.00\text{ cm}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=10.00\text{ cm}\\ u=-5.00\text{ cm}\\ \frac{1}{10}=\frac{1}{v}+(-\frac{1}{5})\\ v=3.33\text{ cm}\end{array}$

  $\begin{array}{l}\frac{v}{u}=\frac{H_i}{H_o}\\ \frac{3.33\text{ cm}}{5.00\text{ cm}}=\frac{H_i}{10.0\text{ cm}}\\ H_i=6.66\text{ cm}\end{array}$

Conclusion:

$\text{height of the mirror image}$ = $6.66\text{ cm}$

To determine

(b)

Construct ray diagrams to locate the images and estimate the image height in each of the following cases.

A 10-cm-tall object located 10 cm in front of a spherical convex mirror with a radius of curvature of 20 cm.

Answer to Problem 56QAP

    $\text{height of the mirror image}$ = $5.00\text{ cm}$

Explanation of Solution

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:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\frac{20.00\text{ cm}}{2}\\ f=10.00\text{ cm}\end{array}$

  $\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ f=10.00\text{ cm}\\ u=-10.00\text{ cm}\\ \frac{1}{10}=\frac{1}{v}+(-\frac{1}{10})\\ v=5.00\text{ cm}\end{array}$

  $\begin{array}{l}\frac{v}{u}=\frac{H_i}{H_o}\\ \frac{\text{5}\text{.00 cm}}{10.00\text{ cm}}=\frac{H_i}{10.0\text{ cm}}\\ H_i=5.00\text{ cm}\end{array}$

Conclusion:

$\text{height of the mirror image}$ = $5.00\text{ cm}$

To determine

(c)

Construct ray diagrams to locate the images and estimate the image height in each of the following cases.

A 10-cm-tall object located 20 cm in front of a spherical convex minor with a radius of curvature of 20 cm.

Answer to Problem 56QAP

    $\text{height of the mirror image}$ = $3.33\text{ cm}$

Explanation of Solution

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:

  $\begin{array}{l}f=\frac{R}{2}\\ f=\frac{20.00\text{ cm}}{2}\\ f=10.00\text{ cm}\end{array}$

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

  $\begin{array}{l}\frac{v}{u}=\frac{H_i}{H_o}\\ \frac{\text{6}\text{.66 cm}}{20.00\text{ cm}}=\frac{H_i}{10.0\text{ cm}}\\ H_i=3.33\text{ cm}\end{array}$

Conclusion:

$\text{height of the mirror image}$ = $3.33\text{ cm}$