A laser that emits pulses of UV lasting 2.95 ns has a beam diameter of 4.05 mm. If each burst contains an energy of 4.60 J, what is the length in space of each pulse?
What is the average energy per unit volume (the energy density) in one of these pulses?
To find the length in space of each pulse of the laser, we need to convert the beam diameter from millimeters to meters.
Given:
Pulse duration (t) = 2.95 ns = 2.95 × 10^(-9) s
Beam diameter (d) = 4.05 mm = 4.05 × 10^(-3) m
The length in space (l) of each pulse is given by the equation:
l = c × t × (d/2)
Where:
c = speed of light in a vacuum = 3 × 10^8 m/s
Plugging in the given values, we get:
l = 3 × 10^8 × 2.95 × 10^(-9) × (4.05 × 10^(-3)/2)
Simplifying the calculation further:
l = 5.8975 × 10^(-3) meters
Therefore, the length in space of each pulse is approximately 5.90 × 10^(-3) meters.
Moving on to the second question:
To find the average energy per unit volume (energy density) in one of these pulses, we need to calculate the volume of the pulse.
The volume (V) of a cylinder (the shape of the pulse) is given by the equation:
V = π × (d/2)^2 × l
Where:
d/2 = radius of the cylinder
l = length in space of each pulse
Plugging in the given values, we get:
V = π × (4.05 × 10^(-3)/2)^2 × (5.8975 × 10^(-3))
Simplifying the calculation further:
V = 1.2786 × 10^(-11) m^3
Now, to find the energy density (D), we divide the total energy (E) of each pulse by the volume (V):
D = E/V
Given:
Energy (E) = 4.60 J
Plugging in the values:
D = 4.60 / 1.2786 × 10^(-11)
Simplifying the calculation further:
D ≈ 3.593 × 10^11 J/m^3
Therefore, the average energy per unit volume (the energy density) in one of these pulses is approximately 3.59 × 10^11 J/m^3.