Question

Find the length of field of view on $\text{Y}$-axis when the object is at $\text{A}$ $\text{(2, 0)}$ and $\text{MN}$ is a plane mirror as shown in the figure below.

Answer 1

Ray originating from $\text{A}$ which strikes the mirror at $\text{M}$ gets reflected as the ray ${\text{MM}}^{\prime }$ which appears to come from $\text{A'}$.

Now, the ray ${\text{A}}^{\prime }$ $\text{N}$ is extended to cut $\text{Y-axis}$ at ${\text{N}}^{\prime }$ (Ray originating from $\text{A}$ which strikes the mirror at $\text{N}$ gets reflected as the ray $\text{NN}\text{'}$ which appears to come from ${\text{A}}^{\prime }$).

From the diagram, we need to find the length ${\text{M}}^{\prime }$${\text{N}}^{\prime }$.

Since, $\Delta A^{\prime }{M}^{\prime }O$ and $\Delta A^{\prime }MX$ are similar.
$A^{\prime }X=6-4=2\text{units}$
$A^{\prime }O=6\text{units}$
$MX=2\text{units}$


$\therefore \frac{M^{\prime }O}{MX}=\frac{A^{\prime }O}{A^{\prime }X}$

$\Rightarrow M^{\prime }O=\frac{6}{2}\times 2=6\text{units}$

Similarly,

$\Delta A^{\prime }{N}^{\prime }O$ and $\Delta A^{\prime }NX$ are similar.
$A^{\prime }X=2\text{units}$
$A^{\prime }O=6\text{units}$
$NX=3\text{units}$

$\therefore \frac{N^{\prime }O}{NX}=\frac{A^{\prime }O}{A^{\prime }X}$

$\Rightarrow N^{\prime }O=\frac{6}{2}\times 3=9\text{units}$

$\therefore$ Field of view on $Y$ axis is $M^{\prime }{N}^{\prime }=O{N}^{\prime }-O{M}^{\prime }$
$\therefore M^{\prime }{N}^{\prime }=9-6=3\text{units}$

Accepted answer : $3,3.0,3.00$