Question

A mirror at an amusement park shows an upright image of any person who stands 1.2 m in front ofit. If the image is three times the person's height, what is the radius of curvature?

Answer 1

Concepts and reason

The concepts used to solve this problem are mirror equation, magnification and radius of curvature. Find the image distance using the magnification expression. Then, using thin mirror equation find the focal length. Finally, find the radius of curvature by equating with twice of focal length.

Fundamentals

The image formed by any person when he stands in front of the mirror at an amusement park are upright and it means that the magnification is positive. The expression for the magnification isgiven below:

\(m=-\frac{d_{i}}{d_{o}}\)

Here, \(m\) is the magnification, \(d_{o}\) is the object distance, and \(d_{i}\) is the image distance. The person's height and his image formation are related by mirror equation. The distance between the middle of a mirror and the rays converging at a point is called focal length. The expression for mirror equation is given below:

\(\frac{1}{f}=\frac{1}{d_{i}}+\frac{1}{d_{o}}\)

Here, \(f\) is the focal length. The radius of curvature is the distance from the point at which the axis meets a curve to the middle of the curvature. The radius of curvature is equal and twice the focal length. The expression for the radius of curvature is given below:

\(R=2 f\)

Here, \(R\) is the radius of curvature.

The expression for the magnification is given below:

$$ m=-\frac{d_{i}}{d_{o}} $$

The formed image is threetimes the person's height.

Substitute 3 for \(\mathrm{m}\) and \(1.2 \mathrm{~m}\) for \(d_{o}\).

$$ \begin{aligned} 3 &=-\frac{d_{i}}{1.2 \mathrm{~m}} \\ d_{i} &=-(3)(1.2 \mathrm{~m}) \\ &=-3.6 \mathrm{~m} \end{aligned} $$

The increase or decrease in size of the image created in the mirror compared to the person's height is called the magnification. Negative sign represents that the image is formed behind the mirror.

The expression for the thin mirror is given below:

\(\frac{1}{f}=\frac{1}{d_{i}}+\frac{1}{d_{o}}\)

Substitute \(-3.6 \mathrm{~m}\) for \(d_{i}\) and \(1.2 \mathrm{~m}\) for \(d_{o}\)

\(\frac{1}{f}=-\frac{1}{3.6 \mathrm{~m}}+\frac{1}{1.2 \mathrm{~m}}\)

\(=0.56 \mathrm{~m}\)

\(f=\frac{1}{0.56 \mathrm{~m}}\)

\(=1.8 \mathrm{~m}\)

The image distance, object distance, and focal length are related by the basic equation called the mirror equation and the positive sian of the focal length indicates that the mirror is concave.

The expression for the radius of curvature is given below:

\(R=2 f\)

Substitute \(1.8 \mathrm{~m}\) for \(f\).

\(R=2(1.8 \mathrm{~m})\)

\(=3.6 \mathrm{~m}\)

In this problem, the radius of curvature is found to be \(3.6 \mathrm{~m}\).

The radius of curvature depends upon the focal length. Due to the positive value of focal length, the mirror used here is concave.