Question

A He-Ne laser emits a beam of diameter 2×10−3m and wavelength 630 nm. It is directed
towards an aeroplane flying at a height of 11 km. Calculate the diameter of the light patch
produced on the surface of the aeroplane

Answer 1

Answer on Question#51648 - Physics - Optics

A He-Ne laser emits a beam of diameter $d = 2 \times 10^{-3} \, \mathrm{m}$ and wavelength $\lambda = 630 \, \mathrm{nm}$. It is directed towards an aeroplane flying at a height of $H = 11 \, \mathrm{km}$. Calculate the diameter of the light patch produced on the surface of the aeroplane.

Solution:

The angle of diffraction (first minimum) of the laser with such diameter of the outlet is given by

$$\sin \theta = 1.22 \frac{\lambda}{d}$$

Therefore the diameter of the light patch on the surface of the plane is

$$D = H \cdot \sin \theta = 1.22 \cdot H \frac{\lambda}{d} = 1.22 \cdot 11000 \, \mathrm{m} \cdot \frac{630 \cdot 10^{-9} \, \mathrm{m}}{2 \cdot 10^{-3} \, \mathrm{m}} = 4.23 \, \mathrm{m}$$

**Answer**: $D = 1.22 \cdot H \frac{\lambda}{d} = 4.23 \, \mathrm{m}$.