91Ó°ÊÓ

1. Object distance,$p=10cm$
2. Distance of observer,${d_e}_y=20\mathrm{cm}$
3. Diameter of the pupil,$D=5\mathrm{mm}$

The area of the mirror can be determined by using the object's distance.

Formula:

The area of the circle,$A=\frac{{\mathrm{Ï€»å}}^2}{4},d=diameterofthecircle$ (i)

Consider the eye of an observer and the object to be in the same line perpendicular to the mirror surface. Consider the two light rays, $r$and localid="1663053204988" $r^r$, which are closest to and on either sides of the normal ray. Each of these rays has an angle of incidence equal to $\mathrm{θ}$when they reach the mirror. Consider that these two rays strike the top and bottom edge of the pupil after they have reflected. If ray $r$ strikes the mirror at point $A$and ray $r\text{'}$strikes the mirror at, the distance between $A$and $B$, say $X$is given as follows:

$x=2d_0\tan \mathrm{θ}............\left(a\right)$

Where,$d_0=p$is the distance from the mirror to the object.

We can construct a right triangle starting with the image point of the object$d_0$. One side of the triangle follows the extended normal axis, and the hypotenuse is along the extension of raylocalid="1663052236991" $r$. The distance from pupil to Iis,localid="1663052334127" $d_{eye}+d_0$and the small angle in this triangle is again$\mathrm{θ}$. Thus,

$tan\mathrm{θ}=\frac{R}{d_{ey}+d_0}………………\left(b\right)$

Where, $R$is the pupil radius$2.5mm$.

Substituting equation (b) in equation (a) with given data, we get the distance between A and B as follows:

$$\begin{gathered} x=2d_0\frac{R}{d{}_{ey}+d_0} \\ =2×100mm×\frac{2.5mm}{200mm+100mm} \\ =1.67mm \end{gathered}$$

Now, x serves as the diameter of a circular area A on the mirror, in which all rays that reflect will reach the eye. Thus, the area of the circular area is given using the data in equation (i) as follows:

localid="1663053097431" $\begin{array}{l}A=\frac{1}{4}×\mathrm{π}×{(1.67mm)}^2\\ =2.2{\mathrm{mm}}^2\end{array}$

Hence, the value of the area is$2.2{\mathrm{mm}}^2$.