Plane Mirrors — Question 5

Question 5

A finite mirror selects an interval of visible object heights

A mirror at x=0 extends from height y=0 to y=L. An eye is at (-e,h). Object points lie on the line x=-d, where d and e are positive. Ignore obstructions.

1. Derive the mirror crossing height for an object point of height y.

2. Find the interval of visible y values; evaluate it for e=d, h=L/2.

Question 5 - Solution

A finite mirror selects an interval of visible object heights

(a) Method and calculation

$$O^\prime=(d,y),\quad\lambda_{x=0}=e/(d+e).$$$$\boxed{y_M=h+\frac{e}{d+e}(y-h)=\frac{dh+ey}{d+e}}.$$

(b) Interpretation and further calculation

$$0\le y_M\le L\Rightarrow\boxed{-dh/e\le y\le[(d+e)L-dh]/e}.$$$$e=d,\ h=L/2:\quad\boxed{-L/2\le y\le3L/2}.$$

Physical check

A finite mirror limits which rays reach the eye; it does not change the locations of the virtual image points that are visible.