Answer
a) less
b) $E_{rms}=193.79V/m$
c) $E_{rms}$ increases by a factor of $2$ and the intensity increases by a factor of $4$.
Work Step by Step
(a) The $E_{rms}$ of the incident beam is less than where the beam intersects the retina because it is not focused on a small area.
(b) We know that
$E_{rms}=\sqrt{\frac{I_{avg}}{c\epsilon_{\circ}}}$
We plug in the known values to obtain:
$E_{rms}=\sqrt{\frac{1\times 10^{-2}W/cm^2}{(3\times 10^8m/s)(8.85\times 10^{-12}C^2/N.m^2)}}$
This simplifies to:
$E_{rms}=193.79V/m$
(c) As $E_{rms_1}=\sqrt{\frac{I_{avg}}{c\epsilon_{\circ}}}$
$E_{rms_1}=\sqrt{\frac{\frac{p_{avg}}{A_1}}{c\epsilon_{\circ}}}$
$\implies E_{rms_1}=\sqrt{\frac{\frac{p_{avg}}{\frac{\pi d_1^2}{4}}}{c\epsilon_{\circ}}}$.....eq(1)
Similarly
$ E_{rms_2}=\sqrt{\frac{\frac{p_{avg}}{\frac{\pi d_1^2}{4}}}{c\epsilon_{\circ}}}$....eq(2)
Dividing eq(1) by eq(2), we obtain:
$\frac{E_{rms1}}{E_{rms2}}=\frac{d_2}{d_1}$
We plug in the known values to obtain:
$\frac{E_{rms1}}{E_{rms2}}=\frac{\frac{5}{2}\mu m}{5\mu m}$
$\frac{E_{rms1}}{E_{rms2}}=\frac{1}{2}$
Thus, $E_{rms}$ is increased by a factor of 2, that is $E_{rms2}=2E_{rms1}$
Intensity increases as the square of the $E_{rms}$ -- a factor of 4.
