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

Question

Aicharya · Accepted Answer

Ray diagram of a girl standing in front of a plane mirror is given below,

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, $\quad P M_1=140-85=55 \mathrm{~cm}$ Let $ \begin{gathered} O P=a \text { and } O I_1=2 a \\ \tan \theta=\frac{55}{a}=\frac{x}{2 a} \Rightarrow x=55 \times 2=110 \mathrm{~cm} \end{gathered} $ So, the maximum height seen by the girl, $ H=110+10=120 \mathrm{~cm} $ Since, the height above her eyes, has not effected the mirror height, because it is just in front of the mirror. Hence, the correct option is (c)

A girl of height $150 \mathrm{~cm}$ with her eye level at $140 \mathrm{~cm}$ stands in front of plane mirror…

Solution