Q142 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Physics
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
- A130 cm
- B140 cm
- Correct answerC120 cm
- D150 cm
Solution
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).
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).
Hence, the correct answer is option: 'C'.