Plane Mirrors — Question 4

Section 34-2 · Physics pages 5578–5579

Question 4 A full-length view needs a mirror only half as tall as the person A person of height H stands distance d from a vertical plane mirror. The eye is at height h above the floor. Treat top of head and feet as points vertically below or above the eye at the same distance from the mirror. H eye mirror segment Schematic; labels, not drawing lengths, specify the data. (a) Find the lowest and highest mirror heights needed to see the feet and top of head. (b) Derive the minimum mirror height and evaluate its endpoints for H=1.80 m and h=1.68 m.
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Question 4 - Solution A full-length view needs a mirror only half as tall as the person H eye mirror segment (a) Method and calculation ybottom = (0 + h)/2 = h/2 . ytop = (H + h)/2 ⇒LM = ytop −ybottom = H/2 . (b) Interpretation and further calculation ybottom = 0.84 m, ytop = 1.74 m . LM = 0.90 m, ∂LM/∂d = 0 . Physical check The halfway heights follow because the eye and virtual image are equally distant from the mirror plane; moving farther away does not reduce the required mirror length.

Question 4

A full-length view needs a mirror only half as tall as the person

A person of height H stands distance d from a vertical plane mirror. The eye is at height h above the floor. Treat top of head and feet as points vertically below or above the eye at the same distance from the mirror.

Diagram shown in the original SVG worksheet.

  1. Find the lowest and highest mirror heights needed to see the feet and top of head.

  2. Derive the minimum mirror height and evaluate its endpoints for H=1.80 m and h=1.68 m.

Question 4 - Solution

A full-length view needs a mirror only half as tall as the person

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ybottom=(0+h)/2=h/2.y_{\mathrm{bottom}}=(0+h)/2=\boxed{h/2}.ytop=(H+h)/2⇒LM=ytop−ybottom=H/2.y_{\mathrm{top}}=(H+h)/2\Rightarrow\boxed{L_M=y_{\mathrm{top}}-y_{\mathrm{bottom}}=H/2}.

(b) Interpretation and further calculation

ybottom=0.84m,ytop=1.74m.\boxed{y_{\mathrm{bottom}}=0.84\,\mathrm m,\quad y_{\mathrm{top}}=1.74\,\mathrm m}.LM=0.90m,∂LM/∂d=0.\boxed{L_M=0.90\,\mathrm m,\quad\partial L_M/\partial d=0}.

Physical check

The halfway heights follow because the eye and virtual image are equally distant from the mirror plane; moving farther away does not reduce the required mirror length.