Chapter 23, Problem 11P

An object 3.0 mm high is placed 16 cm from a convex mirror of radius of curvature 16 cm. (a) Show by ray tracing that the image is virtual, and estimate the image distance. (b) Show that the (negative) image distance can be computed from Eq. 23-2 using a focal length of-8.0 cm. (c) Compute the image size, using Eq. 23-3.

To determine

(a) To determine:

Show by ray tracing that the image is virtual, and estimate the image distance.

Answer to Problem 11P

Solution:

Ray diagram is given in explanation and image distance is 5 cm.

Explanation of Solution

Given info:

Object height, $h_o =3\text{mm}$

Object distance, $u=16\text{cm}$

Radius of curvature, $r=16\text{cm}$

Formula used:

Radius of curvature, $r=2f$

$1/u+1/v=1/f$

Here, v and f are image and focal length.

Magnification, $m=h_i/h_o=-v/u$

Calculation:

Focal length, $f=16/2=8\text{cm}$

Therefore,

$\begin{array}{l}1/v-1/16=1/8\\ 1/v=1/8+1/16\\ v=5.33\text{cm}\end{array}$

The image distance is 5 cm

From figure, the image is behind the mirror, so it is virtual and image distance is 5 cm

To determine

(b) To determine:

Show that the (negative)image distance can be computed from figure using a focal length – 8 cm.

Answer to Problem 11P

Solution:

The image distance by using a focal length -8 cm is -5.33 cm.

Explanation of Solution

Calculation:

If focal length is – 8 cm, we can locate the image from

$\begin{array}{l}1/v+1/u=1/f\\ 1/16+1/v=-1/8\\ v=-5.33\text{cm}\end{array}$

To determine

(c) To determine:

The image size.

Answer to Problem 11P

Solution:

Image size is 1 mm

Explanation of Solution

Calculation:

From magnification formula,

$\begin{array}{l}m=h_i/h_o=-v/u\\ h_i/(3.0\text{mm)= -5}\text{.33/16}\\ h_i=1\text{mm}\\ \end{array}$