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Q. A girl of height $150 \,cm$ with her eye level at $140 \,cm$ stands in front of plane mirror of height $75 \,cm$ fixed to a wall. The lower edge of the mirror is at a height of $85 \,cm$ above her feet level. The height of her image the girl can see in the mirror is

AP EAMCETAP EAMCET 2019

Solution:

Ray diagram of a girl standing in front of a plane mirror is given below,
image
Applying the property of triangle and from similar triangles $O P M_{1}$ and $O I_{1} I_{2}$
$\tan\, \theta=\frac{P M_{1}}{O P}=\frac{I_{1} I_{2}}{O I_{1}}$
Since,
$P M_{1}=140-85=55\, cm$
Let $O P=a$ and $ O I_{1}=2 a $
$\tan\, \theta=\frac{55}{a} =\frac{x}{2 a}$
$ \Rightarrow x=55 \times 2=110\, cm$
So, the maximum height seen by the girl,
$H=110+10=120 \,cm$
Since, the height above her eyes, has not effected the mirror height,
because it is just in front of the mirror.