Chapter 36, Problem 12SP

A lit candle is a perpendicular distance of 20.0 cm from the front of a flat mirror. (a) Where will its image appear? (b) What kind of image will it be?

To determine

The position of the image formed when the object is placed at $20.0\text{cm}$ in front of the flat mirror.

Answer to Problem 12SP

Solution:

$20.0\text{cm}$, behind the reflecting surface.

Explanation of Solution

Given data:

The object is placed at $20.0\text{cm}$ in front of the flat mirror.

Formula used:

Write the thin mirror equation:

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Here, $f$ is the focal length of the mirror, $s_o$ is the distance of the object from the mirror, and $s_i$ is the distance of the image from the mirror.

Explanation:

Consider the expression forthinmirror:

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Understand that the focal length of a plane mirror is infinity.

Substitute $\infty$ for $f$ and $20.0\text{cm}$ for $s_o$

$\frac{1}{\infty }=\frac{1}{20.0\text{cm}}+\frac{1}{s_i}$

Solve for $s_i$

$s_i=20.0\text{cm}$

Conclusion:

Therefore, the position of the imageof formed by the flat mirroris $20.0\text{cm}$ behind the reflecting surface.

To determine

The nature of the image formed by a flat mirror when the object is placedat $20.0\text{cm}$ in front of the flat mirror.

Answer to Problem 12SP

Solution:

Virtual

Explanation of Solution

Introduction:

Plane mirrors have infinite focal length and hence form the image at the same distance as that of the object from the mirror, but form behind the mirror.

Explanation:

The property of a plane mirror is that it forms an image of the same size as that of the object. Also, the image is erect and behind the mirror. Hence, the nature of the image is virtual.

Conclusion:

Therefore, the image formed by a plane mirror is virtual.