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Test: Refractive Index & Numerical Aperture


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12 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Refractive Index & Numerical Aperture

Test: Refractive Index & Numerical Aperture for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Refractive Index & Numerical Aperture questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Refractive Index & Numerical Aperture MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Refractive Index & Numerical Aperture below.
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Test: Refractive Index & Numerical Aperture - Question 1

The expression for refractive index is given by

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 1

Answer: b
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.

Test: Refractive Index & Numerical Aperture - Question 2

Numerical aperture is expressed as the

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 2

Answer: a
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.

Test: Refractive Index & Numerical Aperture - Question 3

For total internal reflection to occur, which condition must be satisfied?

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 3

Answer: b
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.

Test: Refractive Index & Numerical Aperture - Question 4

Find the refractive index of a medium having a velocity of 1.5 x 108

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 4

Answer: d
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.

Test: Refractive Index & Numerical Aperture - Question 5

The refractive index of water will be

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 5

Answer: d
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).

Test: Refractive Index & Numerical Aperture - Question 6

The refractive index of air is unity. State True/False.

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 6

Answer: a
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.

Test: Refractive Index & Numerical Aperture - Question 7

The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 7

Answer: d
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.

Test: Refractive Index & Numerical Aperture - Question 8

Find the acceptance angle of a material which has a numerical aperture of 0.707 in air.

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 8

Answer: c
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.

Test: Refractive Index & Numerical Aperture - Question 9

The numerical aperture of a material with acceptance angle of 60 degree in water will be

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 9

Answer: a
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.

Test: Refractive Index & Numerical Aperture - Question 10

The core refractive index should be lesser than the cladding refractive index for a coaxial cable. State True/False

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 10

Answer: b
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.

Test: Refractive Index & Numerical Aperture - Question 11

The refractive index is 2.33 and the critical angle is 350. Find the numerical aperture.

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 11

Answer: b
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).

Test: Refractive Index & Numerical Aperture - Question 12

Choose the optical fibre material from the given materials.

Detailed Solution for Test: Refractive Index & Numerical Aperture - Question 12

Answer: c
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.

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