Question

A laser beam of wavelength 6X10-7 m , coherence width 8X10-3m and power 10mW shines on a surface 100m away. Deduce the illumination , compare it with that due to a collimated beam from a torch filament of diameter 0.1cm , lens of focal length 10cm and power 10W.

Answer 1

Answer on Question #53759, Physics Solid State Physics

A laser beam of wavelength $6 \times 10^{-7} \, \mathrm{m}$, coherence width $8 \times 10^{-3} \, \mathrm{m}$ and power $10 \, \mathrm{mW}$ shines on a surface $100 \, \mathrm{m}$ away. Deduce the illumination, compare it with that due to a collimated beam from a torch filament of diameter $0.1 \, \mathrm{cm}$, lens of focal length $10 \, \mathrm{cm}$ and power $10 \, \mathrm{W}$.

Solution

The semi angle of cone of laser beam

$$\theta = \frac{\lambda}{a} = \frac{6 \cdot 10^{-7}}{8 \cdot 10^{-3}} = 7.5 \cdot 10^{-5} \, \mathrm{rad}$$

Solid angle

$$\Omega = \frac{\Delta S}{r^2} = \frac{\pi (r \theta)^2}{r^2} = \pi \theta^2$$

Areal speed

$$\Delta A = r^2 \Omega = \pi \theta^2 r^2 = 3.14 \cdot (7.5 \cdot 10^{-5})^2 \cdot (100)^2 \, \mathrm{m}^2$$

Illumination

$$E = \frac{P}{\Delta A} = \frac{10 \cdot 10^{-2}}{1.76 \cdot 10^{-4}} = 57 \, \mathrm{W} / \mathrm{m}^2$$

For a torch, the angle subtended by the filament size at the lens,

$$\theta' = \frac{0.1 \, \mathrm{cm}}{10 \, \mathrm{cm}} = 10^{-2} \, \mathrm{rad}$$

Areal speed

$$\Delta A = r ^ {2} \Omega = \pi \left(\theta^ {\prime}\right) ^ {2} r ^ {2} = 3.14 \cdot (0.01) ^ {2} \cdot (100) ^ {2} = 3.14 m ^ {2}$$

Illumination

$$E = \frac {P ^ {\prime}}{\Delta A ^ {\prime}} = \frac {10}{3.14} = 3.2 W / m ^ {2}$$

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