A point object is 10cmaway from a plane mirror, and the eye of an observer (with pupil diameter5.0mm) is 20cmaway. Assuming the eye and the object to be on the same line perpendicular to the mirror surface, find the area of the mirror used in observing the reflection of the point.

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,$A=\frac{{\mathrm{\pi d}}^2}{4},d=diameterofthecircle$ (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 $\theta$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:

$x=2d_0\tan \theta ............\left(a\right)$

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$\theta$. Thus,

$tan\theta =\frac{R}{d_{ey}+d_0}\dots \dots \dots \dots \dots \dots \left(b\right)$

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:

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

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" $\begin{array}{l}A=\frac{1}{4}\times \mathrm{\pi }\times {(1.67mm)}^2\\ =2.2{\mathrm{mm}}^2\end{array}$

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