This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Refractive Index and Numerical Aperture”.
1. The expression for refractive index is given by
a) N = v/c
b) N = c/v
c) N = cv
d) N = 1/cv
View Answer
Explanation: The refractive index is defined as the ratio of the velocity of light in a vacuum to its velocity in a specified medium. It is given by n = c/v. It is constant for a particular material.
2. Numerical aperture is expressed as the
a) NA = sin θa
b) NA = cos θa
c) NA = tan θa
d) NA = sec θa
View Answer
Explanation: The numerical aperture is the measure of how much light the fiber can collect. It is the sine of the acceptance angle, the angle at which the light must be transmitted in order to get maximum reflection. Thus it is given by NA = sin θa.
3. For total internal reflection to occur, which condition must be satisfied?
a) N1 = N2
b) N1 > N2
c) N1 < N2
d) N1 x N2=1
View Answer
Explanation: The refractive of the transmitting medium should be greater than that of the receiving medium. In other words, the light must flow from denser to rarer medium, for total internal reflection to occur.
4. Find the refractive index of a medium having a velocity of 1.5 x 108.
a) 0.5
b) 5
c) 0.2
d) 2
View Answer
Explanation: The refractive index is given by the ratio of the speed of light to the velocity in a particular medium. It is given by n = c/v. On substituting for v = 1.5 x 108 and c = 3 x 108, we get n = 3/1.5 = 2. The quantity has no unit.
5. The refractive index of water will be
a) 1
b) 2.66
c) 5
d) 1.33
View Answer
Explanation: The velocity of light in water as medium will be 2.25 x 108. On substituting for the speed of light, we get refractive index as n = 3/2.25 = 1.33(no unit).
6. The refractive index of air is unity. State True/False.
a) True
b) False
View Answer
Explanation: The velocity of light in the air medium and the speed of light are both the same. Since light travels at maximum velocity in air only. Thus the refractive index n = c/v will be unity.
7. The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is
a) 3.73
b) 0.83
c) 3.46
d) 1.86
View Answer
Explanation: The numerical aperture is given by NA = √(n12 – n22), where n1 and n2 are the refractive indices of core and cladding respectively. On substituting for n1 = 2.33 and n2 = 1.4, we get NA = √(2.332-1.42) = 1.86.
8. Find the acceptance angle of a material which has a numerical aperture of 0.707 in air.
a) 30
b) 60
c) 45
d) 90
View Answer
Explanation: The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is unity in air. Thus NA = sin θa. To get θ= sin-1(NA), put NA = 0.707, thus θa = sin-1(0.707) = 45 degree.
9. The numerical aperture of a material with acceptance angle of 60 degree in water will be
a) 1.15
b) 2.15
c) 5.21
d) 1.52
View Answer
Explanation: The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is 1.33 for water medium. Given that the acceptance angle is 60, we get NA = 1.33 sin 60 = 1.15.
10. The core refractive index should be lesser than the cladding refractive index for a coaxial cable. State True/False
a) True
b) False
View Answer
Explanation: The light should pass through the core region only, for effective transmission. When light passes through cladding, losses will occur, as cladding is meant for protection. Thus core refractive index must be greater than the cladding refractive index.
11. The refractive index is 2.33 and the critical angle is 350. Find the numerical aperture.
a) 2
b) 1.9
c) 2.33
d) 12
View Answer
Explanation: The numerical aperture is given by NA = n cos θc, where θc is the critical angle and n is the refractive index. On substituting for n = 2.33 and θc = 35, we get NA = 2.33 cos 35 = 1.9(no unit).
12. Choose the optical fibre material from the given materials.
a) Glass
b) Plastic
c) Silica
d) Quartz
View Answer
Explanation: Silica is the most dominant optical fibre material. This is because of its hardness, flexibility, melting point. Also it is an easily available material.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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