Question

A 9 × 1014 Hz laser emits a 8.8 μs pulse, 5.0 mm in diameter, with a beam energy density of 0.8 J/m3. The number of wavelengths in the length of the beam is closest to?

Answer 1

To solve this problem we will apply the concepts related to Frequency (Reverse to Period) and the description of the wavelength as a function of the speed of light at the rate of frequency.

Our Laser frequency is given as

$f = 9*10^{14}Hz$

Therefore the laser wavelength would be

$\lambda = \frac{c}{f}$

Where,

c = Speed of light

f = Frequency

$\lambda = \frac{3*10^8}{ 9*10^{14}}$

$\lambda = 3.33*10^{-7}$

The laser pulse is emitted at a period (T) of $8.8*10^{-6}s$

Therefore the pulse wavelength would be

$\lambda' = \frac{c}{f}$

$\lambda' = c \frac{1}{f} \rightarrow \frac{1}{f} = T$

$\lambda' = c *T$

$\lambda' = (3*10^8)(8.8*10^{-6})$

$\lambda' = 2640m$

Finally the number of wavelengths is the ratio between the two wavelengths, then

$n = \frac{\lambda'}{\lambda}$

$n = \frac{2640}{3.33*10^{-7} }$

$n = 7.927*10^9$

The number of wavelengths in the beam length is closer to $7.927*10^9$