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When a man stands 1.52 m in front of a shaving mirror, the image produced is inverted and has an image distance of 18.0 cm. How close to the mirror must the man place his face if he wants an upright image with a magnification of 2.0?

Sagot :

The nature of shaving mirror is concave

Now using magnification formula

[tex]\begin{gathered} magnification=\frac{v}{u}\begin{cases}v={image\text{ distance}} \\ u={object\text{ distance}}\end{cases} \\ \end{gathered}[/tex]

Again

[tex]\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v}; \\ \frac{1}{f}=\frac{1}{15.2}\text{ +}\frac{1}{18}; \\ \frac{1}{f}=\text{ }\frac{18+15.2}{15.2\times18}=\frac{33.2}{273.6}=\frac{332}{2736}; \\ f=\text{ }\frac{2736}{332}\text{ cm = 8.24cm} \end{gathered}[/tex]

For upright image , object should be placed between principle focus and pole

Using magnification formula we get

[tex]\begin{gathered} magnification=\text{ }\frac{v}{u}; \\ \therefore2=\frac{18}{u}; \\ u=\frac{18}{2}=9\text{ cm} \end{gathered}[/tex]

To satisfy both conditions object must be put at a distance less than 8.24 cm

Answer is:- face distance from mirror is less than 8.24 cm