Question

a laser beam of wavelength 630 nm coherence width 8x10 -3 m and power 10mw shines on a surface 100 m away. deduce the illumination. compare it with that due to a collimated beam from a torch filament of diameter 0.1 cm , lence of focal length 10 cm and power 10 w

Answer 1

Answer on Question #53724, Physics Solid State Physics

a laser beam of wavelength $630~\mathrm{nm}$ 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 $0.1\mathrm{cm}$ , lence of focal length $10\mathrm{cm}$ and power $10\mathrm{w}$

Solution

Fig.1

Solid angle (see Fig.1)

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

where $\theta = \lambda / a$ is the semi angle of cone of laser beam; $\lambda$ is the wavelength of light; $a$ is the coherence width.

Illumination

$$E = \frac {P}{\Delta S} = \frac {a ^ {2} P}{\pi \lambda^ {2} r ^ {2}} = \frac {\left(8 \cdot 1 0 ^ {- 3}\right) ^ {2} \cdot 1 0 \cdot 1 0 ^ {- 2}}{3 . 1 4 \cdot \left(6 3 0 \cdot 1 0 ^ {- 9}\right) \cdot 1 0 0} = 5 7 W / m ^ {2},$$

where $\Delta S = r^2\Omega = \pi \theta^2 r^2$ is the areal speed

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

$$\theta^ {\prime} = \frac {0 . 1 c m}{1 0 c m} = 1 0 ^ {- 2} r a d$$

So, illumination

$$E = \frac {P ^ {\prime}}{\Delta S ^ {\prime}} = \frac {P ^ {\prime}}{\pi (\theta^ {\prime}) ^ {2} r ^ {2}} = \frac {1 0}{3 . 1 4 \cdot (0 . 0 1) ^ {2} \cdot (1 0 0) ^ {2}} = 3. 2 W / m ^ {2}$$

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