Chapter 17, Problem 88A

a.

To determine

The kind of mirror required by a production-line inspector to have a magnification of $7.5$ and an upright image.

Answer to Problem 88A

Concave mirror

Explanation of Solution

Given:

An upright image is formed whose magnification is 7.5. Image is located at a distance of 14 mm from machine part.

A concave mirror always forms a real and inverted image of the object except for the case when the object lies within the focal length of the mirror. For this case, the image formed is magnified and is a virtual and upright one. Whereas the convex mirror always forms a upright, virtual but diminished image of the object.

Since the image formed needed by the production-line inspector is upright and should have a magnification of $7.5$ . Thus, the kind of mirror needed by him is concave.

Conclusion:

Therefore, a concave mirror is needed by the production-line inspector for the given case.

b.

To determine

The radius of the curvature of the mirror.

Answer to Problem 88A

  $32.3\text{mm}$

Explanation of Solution

Given:

The required magnification, $m=+7.5$ (since the image is required to be upright) The distance of object from the mirror, $d_o=14.0\text{mm}$

Formula used:

Mirror equation,

  $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

Where,

  $\begin{array}{l}f=\text{focal length}\\ d_i=\text{ image distance}\\ d_o=\text{object distance}\end{array}$

And,

Magnification, $m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

Where,

  $\begin{array}{l}{h}_i=\text{image height}\\ h_o=\text{object height}\end{array}$

The relation between the radius of curvature of mirror $\left(R\right)$ and its focal length $\left(f\right)$ is,

  $f=\frac{R}{2}$

Calculation:

Using the magnification formula, the image distance will be,

  $\begin{array}{l}m=-\frac{d_i}{d_o}\\ 7.5=-\frac{d_i}{d_o}\\ d_i=-7.5d_o\end{array}$

Using the above and given values in Mirror equation, the focal length will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{f}=\frac{1}{\left(-7.5d_o\right)}+\frac{1}{d_o}\\ =\left(\frac{1}{d_o}\right)\left(-\frac{1}{7.5}+1\right)\\ f=\frac{7.5\times d_o}{6.5}\\ =\frac{7.5\times 14}{6.5}\\ =16.15\text{mm}\end{array}$

Then, the radius of curvature of the concave mirror will be,

  $\begin{array}{c}R=2f\\ =2\left(16.15\text{mm}\right)\\ =32.3\text{mm}\end{array}$

Conclusion:

Therefore, the radius of curvature of the concave mirror is $32.3\text{mm}$ .