Chapter 34: Q1P (page 1038)

A man looks through a camera toward an image of a hummingbird in a plane mirror. The camera is $4.30 m$ in front of the mirror. The bird is at the camera level, $5.00 m$ to the man’s right and $3.30 m$ from the mirror. What is the distance between the camera and the apparent position of the bird’s image in the mirror?

Short Answer

Expert verified

The distance between the camera and the apparent position of the bird’s image in the mirror is $9.1 m$ .

Step by step solution

Step 1: Given data:

Distance between the camera and the mirror, $d_{1} = 4.3 m$ .

Distance between the mirror and the bird, $d_{2} = 3.30 m$ .

Distance between the camera and the bird, $d_{3} = 5.00 m$ .

Determining the concept:

Use Pythagorean Theorem to find the distance between the camera and the apparent position of the bird’s image in the mirror. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

Formulae:

For the right triangle, if $c$ is the hypotenuse and $a$ and $b$ are the other two sides, then

$a^{2} + b^{2} = c^{2}$

(a) Determining the distance between the camera and the apparent position of the bird’s image in the mirror:

The bird is at a distance $d_{2}$ in front of the mirror, and the plane of its image is at the same distance $d_{2}$ behind the mirror.

The lateral distance between the camera and the bird is $d_{3}$ and the distance between the camera, and the bird is $d_{1}$ .

Construct a right triangle out of $d_{3}$ , and the distance between the camera, and the image plane $d_{1} + d_{2}$ .

Consider the distance between the camera and the apparent image of the bird $d$ .

By applying the Pythagorean Theorem, the distance between the camera, and the apparent image is,

role="math" localid="1662977311966" $\begin{array}{rcl} d & = & \sqrt{{(d_{1} + d_{2})}^{2} + d_{3}^{2}} \\ = & \sqrt{{(4.30 + 3.30)}^{2} + {(5.00)}^{2}} \\ = & \sqrt{57.76 + 25} \\ = & 9.1 m \end{array}$