Radio waves coming at $\angle \mathrm{\alpha }$ to vertical are received by radar after reflection from a nearby water surface and directly. The height of the antenna from the water surface so that it records a maximum intensity, should be (wavelength $=\mathrm{\lambda }$)

• Correct answer
  A

  $\frac{\text{λ}}{\text{4cos}\text{α}}$
• B

  $\frac{\text{2λ}}{\text{4cos}\text{α}}$
• C

  $\frac{\text{3λ}}{\text{6cos}\text{α}}$
• D

  $\frac{\text{5λ}}{\text{7cos}\text{α}}$

Solution

Path difference $=\text{BN}+\text{BA}+\frac{\text{λ}}{2}$

(because reflection occurs from denser medium)



In $\Delta \text{ACB}$

$\frac{\mathrm{h}}{\text{AB}}=\text{cos α}$

$\text{AB}=\frac{\mathrm{h}}{\text{cos α}}\text{....(1)}$

from right angle $\Delta \text{ANB}$

$\frac{\text{BN}}{\text{AB}}=\text{cos 2}\text{α}$

$\mathrm{BN}=\mathrm{AB}\cos 2\alpha$                       ....(2)

Path difference $=\left(\text{AB}+\text{AB }\text{cos 2}\text{α}\right)+\frac{\text{λ}}{2}$

$\text{λ}=\text{AB}\left(1+\text{cos 2}\text{α}\right)+\frac{\text{λ}}{2}$

$\frac{\text{λ}}{2}=\frac{\mathrm{h}}{\text{cos α}}\times 2{\text{cos}}^2\text{α}$

$\frac{\text{λ}}{2}=2\text{h cos α}$

$\mathrm{h}=\frac{\text{λ}}{\text{4 cos α}}$

Hence, the correct answer is option: 'A'.