Question

A concave shaving mirror has a radius of curvature of 35 cm. It is positioned so that the (upright) image of the man’s face is 2.5 times the size of the face. How far is the mirror from the face?
A. 5.23 cm
B. 21.0 cm
C. 10.5 cm
D. 42.0 cm

Answer 1

Magnification of a device is the ability to produce an enlarged image of the object placed in front of it. Hence, mathematically it is the ratio of size of image to the size of the object. A concave mirror can be used to produce both enlarged and diminished images of any object. Also it could be used to produce both real and virtual images of the object.

Formula used:
$f = \dfrac R2, m = \dfrac{h_i}{h_o} = \dfrac{-v}{u},\ \dfrac 1f = \dfrac 1v + \dfrac 1u$

Complete answer:
A concave mirror can be used to produce enlarged and diminished images of any object depending upon the location of the object in front of it. It will produce an enlarged image of the object (magnification > 1) only if the object is placed between the pole and focus of the mirror. We are given the radius of curvature of the mirror as 35 cm.
Hence focal length = $\dfrac {-35}{2} = -17.5\ cm$
[As focal length of concave mirror is negative]
Now, given magnification = 2.5
Thus, $2.5 = \dfrac{-v}{u}$
Thus, $v = -2.5u$
Now, using mirror formula, we get
$\dfrac 1f = \dfrac 1u + \dfrac 1v$
Putting the values:
$\dfrac {1}{-17.5} = \dfrac {1}{u}+\dfrac {1}{-2.5u} = \dfrac3{5u}$
Hence, $u = \dfrac{-3\times17.5}{5} = -10.5 cm$

Hence option C. is correct.

Note:
Here, a negative sign shows that the object is placed to the left of the mirror. Students should practice image formation by different mirrors in several cases. In this question, we can see that the object is placed in between the focus and pole of the mirror so that magnification is possible. Students who remember the condition for magnification in case of a concave mirror can directly omit options B and D.