Question

A boy of height 1m stands in front of a convex mirror. His distance from the mirror is equal to its focal length. The height of his image is: (Assume paraxial ray approximation holds)
A. 0.25m
B. 0.33m
C. 0.5m
D. 0.67m

Answer 1

Hint: The height of image is obtained by dividing the image distance by object distance and multiplying the answer by height of object. The distance of the image is calculated from the mirror formula.

Complete step by step answer:A convex mirror is made by silvering the inner surface of the hollow sphere such that reflection takes place from the outer surface. A boy whose height is 1m stands in front of a convex mirror. His distance from the mirror (u) is equal to its focal length (f) i.e., u = f. The focal length is the distance of the focus from the pole of the mirror and let the distance of the image from the mirror of the boy be v, height of object be h2 and height of image be h1.
From the mirror formula,
$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$ [ In a convex mirror, focal length is positive]
$\dfrac{1}{{ - f}} + \dfrac{1}{v} = \dfrac{1}{f}$ [ The distance of the object from the mirror is taken as negative ]
$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{f}$
$\dfrac{1}{v} = \dfrac{2}{f}$
$v = \dfrac{f}{2}$ [ The value of v is positive as the image is formed behind the mirror]
Magnification = $\dfrac{{{h_1}}}{{{h_2}}} = - \dfrac{v}{u}$
\[\dfrac{{{h_1}}}{{{h_2}}} = - \left( {\dfrac{{\dfrac{f}{2}}}{{ - f}}} \right)\]
$\dfrac{{{h_1}}}{1} = \dfrac{1}{2}$
${h_1} = \dfrac{1}{2} \times 1 = \dfrac{1}{2} = 0.5$
Thus, the height of the image is 0.5 m and a positive sign indicates the image is upright, same as the object.
Therefore, option C is correct.

Note:The distance of the object from the mirror is always taken as negative as all the distances measured in front of a mirror is taken as negative. The focal length is positive for convex mirrors. The magnification is equal to the negative of image distance divided by object distance.