Answer
(a) $NIF$
(b) $NIF$
(c) $NIF$
Work Step by Step
(a) We can see that $3.00MJ$ of energy is greater than $40.0KJ$ of energy, hence NIF laser produces more energy in each pulse.
(b) We know that
$P=\frac{Energy}{time}$
We plug in the known values to obtain:
$P_{NOVA}=\frac{40.0\times 10^3J}{2.50\times 10^{-9}s}$
$P_{NOVA}=16\times 10^{12}W$
and $P_{NIF}=\frac{3.00\times 10^6J}{10.0\times 10^{-9}J}$
$P_{NIF}=3.00\times 10^{14}W$
Thus, the NIF laser produces a greater average power during each pulse than the NOVA laser.
(c) We know that
$\frac{I_{INF}}{I_{NOVA}}=\frac{\frac{P_{INF}}{A_{INF}}}{\frac{P_{NOVA}}{A_{NOVA}}}$
$\frac{I_{INF}}{I_{NOVA}}=(\frac{P_{NIF}}{A_{NIF}})(\frac{A_{NOVA}}{P_{NOVA}})$
As the diameters are the same, so $A_{NOVA}=A_{NIF}$
From part(b), we can see that the NIF laser produces greater average power than the NOVA laser $P_{NIF}\gt P_{NOVA}$
Thus, the NIF laser produces the greater intensity.
