Question

You are standing in front of a large flat mirror. As you step back to get farther from the mirror, the size of your image
gets smaller

gets larger

remains the same

Answer 1

The focal length of a mirror is given by:\[f=\frac{ R }{ 2 }\]where R is the mirror's radius of curvature. A plane mirror has an infinite radius of curvature, there's for its focal length is:\[f=\frac{ \infty }{ 2 }=\infty \]Now let's plug the focal length into the imaging equation:\[\frac{ 1 }{ s _{i} }=\frac{ 1 }{ s _{o} }+\frac{ 1 }{ f }=\frac{ 1 }{ s _{o} }+\frac{ 1 }{ \infty }=\frac{ 1 }{ s _{o} }+0\]Therefore:\[\frac{ 1 }{ s _{i} }=\frac{ 1 }{ s _{o} }\]and so,\[s _{i}=s _{o}\]

Answer 2

"So, what?" you say. Well, magnification is defined as:\[m=\frac{ h' }{ h }=\frac{ s _{i} }{ s _{o} }\]We already found that si=so, therefore m=1. Now if the magnification of a plane mirror is m=1, does the image change size as you step backward? Note that in my equation for magnification since h'--the image height--divided by h--the object height--is equal to one, that would mean that if the image height changes, the object height would have to change the same amount.

So, when you step away from the mirror, do you change height?

Answer 3

Ohhh, that makes sense. So then the answer would be "remains the same".

Answer 4

Can you help with this question?

The diagram illustrates a concave mirror that would be part of sphere. F and F´ are at a distance of one focal length f from the mirror, and with C as the center of the sphere that describes the shape of the mirror. Through which labeled point is the ray of light reflected?
from P through F

from P through C

parallel to the principal axis

in a direction that extends back from P to F´

Answer 5

@PsiSquared

Answer 6

From the imaging equation I gave, we'll let so--the object distance--equal infinity. The equation becomes:\[\frac{ 1 }{ s _{i}}=\frac{ -1 }{ \infty }+\frac{ 1 }{ f }\]so si=f, meaning the ray passes from P through F.

Answer 7

Thank you!