Question

A sky wave with a frequency $55\, MHz$ 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/ $cm ^{3}$ )

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

Answer 1

Answer: (B)

$\mu=\mu_{0} \sqrt{1-\frac{81.45 N}{v^{2}}}$
Here $N=400\, cm ^{-3}=400 \,\times 10^{6} m ^{-3}$
$ v =55 MHz =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} $