# Antennas Questions and Answers – Space Wave Propagation – Refractivity

This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Space Wave Propagation – Refractivity”.

1. Which of the following varies with the refractive index in the atmosphere?
b) Optical horizon
c) Line of sight (LOS)
d) Both radio horizon and LOS

Explanation: LOS is the distance covered by a direct space wave from transmitting to receiving antenna. It depends on the height of the transmitting and receiving antennas and effective earth’s radius factor k. Radio horizon varies with the refractive index in the atmosphere. Optical horizon is less than LOS.

2. The expression for refractivity in terms of the Pressure and temperature coefficient is given by ____________
a) $$\frac{77.6}{T}P+4810 \frac{e}{T}$$
b) $$\frac{77.6}{P}T+4810 \frac{e}{T}$$
c) $$\frac{77.6}{P}T+4810 \frac{T}{e}$$
d) $$\frac{77.6}{T}P+4810 \frac{T}{e}$$

Explanation: The expression used to calculate refractivity is given by
$$N=\frac{77.6}{T} P+4810 \frac{e}{T}$$
Where p is the total pressure in millibars, e is the partial pressure of water vapor and T is the absolute temperature.

3. Gradient of refractive index is ____________
a) Change in the refractivity with respect to height
b) Change in the height with respect to refractivity
c) Change in the refractivity with respect to Pressure
d) Change in the refractivity with respect to temperature

Explanation: Change in the refractivity with respect to height is known as the gradient of refractive index. Radio horizon varies with the refractive index in the atmosphere.

4. What is the standard value for the radius of curvature rc to be equal to the radius of earth re?
a) rc=4re
b) rc=$$\frac{4r_e}{3}$$
c) rc=re
d) rc=$$\frac{1}{4}$$ re

Explanation: The radius of curvature is four times the radius of earth. rc=4re
The effective radius factor of earth is 4/3 times actual radius of earth.

5. Expression for effective radius factor k in terms of radius of curvature rc and radius of earth re?
a) $$k=\frac{1}{1-\frac{r_c}{r_e}}$$
b) $$k=\frac{1}{1-\frac{r_e}{r_c}}$$
c) $$k= 1-\frac{r_e}{r_c}$$
d) $$k= 1-\frac{r_c}{r_e}$$

Explanation: Effective radius factor k in terms of radius of curvature rc and radius of earth re is
$$k=\frac{1}{1-\frac{r_e}{r_c}}$$
The radius of curvature is four times the radius of earth. rc=4re
The effective radius factor of earth is 4/3 times actual radius of earth.
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6. What is the effective radius factor when radius of curvature is 4 times the radius of earth?
a) 4
b) 4/3
c) 3
d) 3/4

Explanation: Effective radius factor k in terms of radius of curvature rc and radius of earth re is
$$k=\frac{1}{1-\frac{r_e}{r_c}}$$
The radius of curvature is four times the radius of earth. rc=4re
⇨ K=1/(1-(1/4)) =4/3
The effective radius factor of earth is 4/3 times actual radius of earth.

7. Expression for refractivity is given by ___________
a) N=(n-1)×106
b) N=(n-1)×10-6
c) N=1/(n-1)
d) N=$$\frac{1}{n-1}$$×10-6

Explanation: Refractivity is to observe the change in refractive index due to the smallest change in the relative dielectric constant.
N=(n-1)×106

8. On which of the following the refractivity depends on?
a) Air pressure
b) Water pressure
c) Temperature
d) Air pressure, water pressure and temperature

Explanation: The expression used to calculate refractivity is given by
$$N=\frac{77.6}{T} P+4810 \frac{e}{T}$$
Where p is the total pressure in millibars, e is the partial pressure of water vapor and T is the absolute temperature.

9. Refractive index is directly proportion to ___________
a) $$\sqrt{ϵ_r}$$
b) $$1/\sqrt{ϵ_r}$$
c) ϵr
d) 1/ϵr

Explanation: Refractive index is ratio of velocity of light in air to velocity of light in medium
n=c/v
$$n=\sqrt{ϵ_r μ}$$

10. Relation between gradient of refractive index with height and dielectric constant gradient is ____________
a) $$\frac{dϵ_r}{dH}=2n \frac{dn}{dH}$$
b) $$\frac{dϵ_r}{dH}=n \frac{dn}{dH}$$
c) $$\frac{dϵ_r}{dH}=2n/\frac{dn}{dH}$$
d) $$\frac{dϵ_r}{dH}=-2n \frac{dn}{dH}$$

Explanation: Refractive index is $$n=n=\sqrt{ϵ_r μ} = n=\sqrt{ϵ_r} (μ=1)$$
Differentiating with respective heights treating n, ϵ_r as dependent variables
$$\frac{dn}{dH}=\frac{1}{2} \frac{1}{\sqrt{ϵ_r}} \frac{dϵ_r}{dH}$$
$$\frac{dϵ_r}{dH}=2n \frac{dn}{dH}$$

Sanfoundry Global Education & Learning Series – Antennas.

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