A sky wave with a frequency 55 MHz is incident on D-region of earth-s atmosphere at 45- The angle of refraction is electron density for D-region is 400 electron-cm 3 A- 15-B- 60-C- 45-D- 30-

The correct option is C 45°n_eff = n_0 √(1 (80.5 N/v^2)) = 1 √(1 80.5 × (400 ×10^6)/(55 ×10^6)^2)≈ 1 Also n_eff = sin i/sin r⇒ sin r = sin i ⇒ r = i = 45^∘ ...

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Standard XII

Physics

Propagation of Waves

A sky wave wi...

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/ ${c m}^{3}$ )

60°

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45°

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30°

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15°

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Solution

The correct option is B 45°
$n_{e f f} = n_{0} \sqrt{1 - (\frac{80.5 N}{v^{2}})}$ = 1 $\sqrt{1 - \frac{80.5 \times (400 \times 10^{6})}{{(55 \times 106)}^{2}}} \approx 1$
Also $n_{e f f} = \frac{s i n i}{s i n r} \Rightarrow s i n r = s i n i \Rightarrow r = i = 45^{\circ}$

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A sky wave with a frequency 55 MHz is incident on D-region of earth's atmosphere at 45∘. The angle of refraction is (electron density for D-region is 400 electron/cm3)

The correct option is B 45°neff=n0√1−(80.5Nv2) = 1 √1−80.5×(400×106)(55×106)2≈1 Also neff=sinisinr⇒sinr=sini⇒r=i=45∘

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/ ${c m}^{3}$ )

The correct option is B 45°
$n_{e f f} = n_{0} \sqrt{1 - (\frac{80.5 N}{v^{2}})}$ = 1 $\sqrt{1 - \frac{80.5 \times (400 \times 10^{6})}{{(55 \times 106)}^{2}}} \approx 1$
Also $n_{e f f} = \frac{s i n i}{s i n r} \Rightarrow s i n r = s i n i \Rightarrow r = i = 45^{\circ}$