Question

A sky wave with a frequency $55MHz$ is incident on $D-$ region of earth's atmosphere at $45^{\circ }.$ The angle of refraction is (electron density for $D-$ region is $400$ electron/ $c{m}^3$ )

• A. $60^{\circ }$
• B. $45^{\circ }$
• C. $30^{\circ }$
• D. $15^{\circ }$

Answer 1

Answer: (B)

$\mu ={\mu }_0\sqrt{1-\frac{81.45N}{v^2}}$
Here $N=400c{m}^{-3}=400\times 10^6{m}^{-3}$
$v=55MHz=55\times 10^{6}Hz\Rightarrow \mu =1\sqrt{1-\frac{81.45\left(400\times 10^6\right)}{{\left(55\times 10^6\right)}^2}}$
$\Rightarrow \mu =1$
$\frac{\sin i}{\sin r}=\mu =1\Rightarrow \frac{\sin 45^{\circ }}{\sin r}=1\text{ or }r=45^{\circ }$