Answer
Greater than
Work Step by Step
We know that
$P_1=\frac{1.49\times 10^{19}phtons/s(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{2(488\times 10^{-9}m)}$
$P_1=3.035W$
and $P_2=\frac{(1.49\times 10^{19}photons/s)(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{2(514.5\times 10^{-9}m)}=2.88W$
$P=3.035W+2.88W=5.92W$
Now $n=\frac{hc}{\lambda}$
We plug in the known values to obtain:
$n=\frac{(1.49\times 10^{19})(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{414\times 10^{-9}m}$
$n=7.16W$
Thus, this is greater than the power calculated in problem 88.