Answer
$ 91 \mu m$
Work Step by Step
For this problem, we can use the fact that we have similar triangles in Fig 36-42a to set up a proportion to find the diameter of the retina, D.
$\frac{D^{'}}{D}=\frac{L^{'}}{L}$
Solving for D, we obtain:
$D^{'}=\frac{L^{'}}{L}D$
We are given that $D=2.0mm$, $L=45.0cm$, and $L^{'}=2.0cm$, so then:
$D^{'}=\frac{L^{'}}{L}D$
$D^{'}=\frac{2.0*10^{-2}m}{45.0*10^{-2}m}2.0*10^{-3}m$
$D^{'}=0.0889*10^{-3}m$
Based on Fig. 36-42(b), we find the angle to the x-axis, given that $x=6.0mm$
$\theta = \arctan (D^{'}/(2x))$
$\theta = \arctan (0.0889*10^{-3}m/(2*6.0*10^{-3}m))$
$\theta = 0.424^{\circ}$
Now we have enough information to use equation (36-12) to solve for the diameter of the defect, d.
$\sin \theta= 1.22 \frac{\lambda}{d}$
$d= 1.22 \frac{\lambda}{\sin \theta}$
$d= 1.22 \frac{550*10^{-9}m}{\sin 0.424^{\circ}}$
$d= 9.06*10^{-5}m$
$d= 91 \mu m$
