Chapter 17, Problem 95A

(a)

To determine

To Find: The object position and image position.

Explanation of Solution

Given:

The focal length of concave primary mirror, $f_p=1.0\text{m}$

The focal length of convexsecondary mirror, $f_s=-0.5\text{m}$

Location of the secondary mirror with respect to primary mirror is $0.25\text{m}$ .

Formula Used:

As per Mirror formula:

  $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Here, f is the focal length, u represents distance of the object and v represents the distance of the image.

Calculation:

The position of the object is infinity. The image position for the primary mirror can be calculated as follows:

  $\begin{array}{l}\frac{1}{f_p}=\frac{1}{\infty }+\frac{1}{v}\\ \frac{1}{1.0}=0+\frac{1}{v}\\ v=1.0\text{m}\end{array}$

This becomes the object distance for the secondary mirror.

  $\begin{array}{l}\frac{1}{f_s}=\frac{1}{u}+\frac{1}{v}\\ \frac{1}{-0.50}=\frac{1}{1.0}+\frac{1}{v}\\ v=-0.33\text{m}\end{array}$

Conclusion:

Thus, position of object is infinity and position of the image is $-0.33\text{m}$ from the secondary mirror.

(b)

To determine

To Find: The magnification of the convex mirror and whether it inverts the image or not.

Explanation of Solution

Given:

From part (a), position of object and image are $u=1.0\text{m}$ and $v=-0.33\text{m}$ respectively.

Formula Used:

Magnification produced by a spherical mirror can be obtained by:

  $m=-\frac{v}{u}$

Here, u represents distance of the object and v represents the distance of the image.

Calculation:

Magnification produced by secondary convex mirror is:

  $\begin{array}{c}m=-\frac{-0.33}{1.0}\\ =0.33\end{array}$

As this is less than 1, it means that a diminished image would form. As the magnification is positive, it means that a real and erect image would be obtained.

Conclusion:

Thus, the magnification of the convex mirror is 0.33 and it would not invert the image.