A man 1.80 m tall stands in front of a mirror and sees his full height, no more and no less. If his eyes are 0.14 m from the top of his head, what is the minimum height of the mirror?

Question

GOAL: Apply the properties of a flat mirror.

STRATEGY: Figure 23.4 shows two rays of light, one from the man’s feet and the other from the top of his head, reflecting off the mirror and entering his eye. The ray from his feet just strikes the bottom of the mirror, so if the mirror were longer, it would be too long, and if shorter, the ray would not be reflected. The angle of incidence and the angle of reflection are equal, labeled θ. This means the two triangles, ABD and DBC, are identical because they are right triangles with a common side (DB) and two identical angles θ. Use this key fact and the small isosceles triangle FEC to solve the problem.

Accepted Answer

We need to find BE, which equals d. Relate this length to lengths on the man’s body:

[latex]({{1}})\,\,\,\,\,\,B E=D C+ extstyle{\frac{1}{2}}C F[/latex]

We need the lengths DC and CF. Set the sum of sides opposite the identical angles θ equal to AC:

[latex]({{2}})\ \ \ A D+D C=A C=(1.80-0.14)=1.66\ \mathrm{m}[/latex]

AD = DC, so substitute into Equation (2) and solve for DC:

[latex]A D+D C=2D C=1.66\ m\quad ightarrow\quad D C=0.83\ \mathrm{m}[/latex]

CF is given as 0.14 m. Substitute this value and DC into Equation (1):

[latex]B E=\,d={D C}+ extstyle{\frac{1}{2}}C F=0.83\,{\mathrm{m}}\,+ extstyle{\frac{1}{2}}(0.14\,{\mathrm{m}})\,=\ \ 0.90\,{\mathrm{m}}[/latex]

REMARKS The mirror must be exactly equal to half the height of the man for him to see only his full height and nothing more or less. Notice that the answer doesn’t depend on his distance from the mirror.

Question 23.1: A man 1.80 m tall stands in front of a mirror and sees his f...

Related Answered Questions

Verified Answer:

Verified Answer:

The Case of a Diverging Lens Goal Calculate geometric quantities associated with a diverging lens. Problem Repeat the problem of Example 23.7 for a diverging lens of focal length 10.0 cm. Strategy Once again, substitution into the thin-lens equation and the associated magnification equation, ...

Verified Answer:

Images Formed by a Converging Lens Goal Calculate geometric quantities associated with a converging lens. Problem A converging lens of focal length 10.0 cm forms images of an object situated at various distances. (a) If the object is placed 30.0 cm from the lens, locate the image, state whether ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Images Formed by a Convex Mirror Goal Calculate properties of a convex mirror. Problem An object 3.00 cm high is placed 20.0 cm from a convex mirror with a focal length of 8.00 cm. Find (a) the position of the image, (b) the magnification of the mirror, and (c) the height of the image. Strategy ...

Verified Answer:

Verified Answer:

The Case of a Diverging Lens GOAL Calculate geometric quantities associated with a diverging lens. PROBLEM Repeat the problem of Example 23.7 for a diverging lens of focal length 10.0 cm. STRATEGY Once again, substitution into the thin-lens equation and the associated magnification equation, ...

Verified Answer: