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?
a) Radio horizon
b) Optical horizon
c) Line of sight (LOS)
d) Both radio horizon and LOS
View Answer

Answer: a
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.
advertisement

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) \(\frac{77.6}{T}P+4810 \frac{T}{e} \)
View Answer

Answer: a
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
View Answer

Answer: a
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.
advertisement
advertisement

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
View Answer

Answer: a
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) \(k= 1-\frac{r_c}{r_e} \)
View Answer

Answer: b
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.
advertisement

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
View Answer

Answer: b
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
View Answer

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

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
View Answer

Answer: d
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
View Answer

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

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) \(\frac{dϵ_r}{dH}=-2n \frac{dn}{dH}\)
View Answer

Answer: a
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.

To practice all areas of Antennas, here is complete set of 1000+ Multiple Choice Questions and Answers .

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement

Leave a Comment

Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter