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Test: Snell Law & Critical Angle


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12 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Snell Law & Critical Angle

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

The Snell’s law can be derived from which type of incidence?

Detailed Solution for Test: Snell Law & Critical Angle - Question 1

Answer: d
Explanation: The oblique incidence refers to the interface between dielectric media. Consider a planar interface between two dielectric media. A plane wave is incident at an angle from medium 1 and reflected from medium 2. The interface plane defines the boundary between the media. This is the oblique medium.

Test: Snell Law & Critical Angle - Question 2

The Snell’s law is given by

Detailed Solution for Test: Snell Law & Critical Angle - Question 2

Answer: a
Explanation: The Snell law states that in an oblique medium, the product of the refractive index and sine of incidence angle in medium 1 is same as that of medium 2. Thus it is given by N1 sin θi = N2 sin θt.

Test: Snell Law & Critical Angle - Question 3

Calculate the ratio of sine of incident angle to the sine of reflected angle when the refractive indices of medium 1 and 2 are given as 2.33 and 1.66 respectively.

Detailed Solution for Test: Snell Law & Critical Angle - Question 3

Answer: a
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. To get sin θi/sin θt, the ratio is N2/N1. On substituting for N1 = 2.33 and N2 = 1.66, we get 1.66/2.33 = 0.71

Test: Snell Law & Critical Angle - Question 4

Find the ratio of the refractive index of medium 1 to that of medium 2, when the incident and reflected angles are given by 300 and 450 respectively.

Detailed Solution for Test: Snell Law & Critical Angle - Question 4

Answer: c
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. For getting N1/N2, the ratio is sin θt/sin θi. On substituting for θi = 30 and θt = 45, we get sin 45/sin 30 = 2.

Test: Snell Law & Critical Angle - Question 5

The refractive index of a medium with permittivity of 2 and permeability of 3 is given by

Detailed Solution for Test: Snell Law & Critical Angle - Question 5

Answer: b
Explanation: The refractive index is given by n = c √(με), where c is the speed of light. Given that relative permittivity and relative permeability are 2 and 3 respectively. Thus n = 3 x 108√(2 x 4π x 10-7 x 3 x 8.854 x 10-12) = 2.45.

Test: Snell Law & Critical Angle - Question 6

The critical angle is defined as the angle of incidence at which the total internal reflection starts to occur. State True/False.

Detailed Solution for Test: Snell Law & Critical Angle - Question 6

Answer: a
Explanation: The critical angle is the minimum angle of incidence which is required for the total internal reflection to occur. This is the angle that relates the refractive index with the angle of reflection in an oblique incidence medium.

Test: Snell Law & Critical Angle - Question 7

The critical angle for two media of refractive indices of medium 1 and 2 given by 2 and 1 respectively is

Detailed Solution for Test: Snell Law & Critical Angle - Question 7

Answer: b
Explanation: The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. To get θc, put n1 = 2 and n2 = 1. Thus we get θc = sin-1(n2/n1) = sin-1(1/2) = 30 degree.

Test: Snell Law & Critical Angle - Question 8

The critical angle for two media with permittivities of 16 and 9 respectively is

Detailed Solution for Test: Snell Law & Critical Angle - Question 8

Answer: a
Explanation: The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. Also n = √ε, thus sin θc = √ε2/√ε1. Put ε1 = 16 and ε2 = 9, we get θc = sin-1(3/4) = 48.59 degree.

Test: Snell Law & Critical Angle - Question 9

The angle of incidence is equal to the angle of reflection for perfect reflection. State True/False.

Detailed Solution for Test: Snell Law & Critical Angle - Question 9

Answer: a
Explanation: For complete wave reflection, the angle of incidence should be same as the angle of the reflection. In such cases, the reflection coefficient is unity and the transmission coefficient is zero.

Test: Snell Law & Critical Angle - Question 10

The angle of incidence of a wave of a wave with angle of transmission 45 degree and the refractive indices of the two media given by 2 and 1.3 is

Detailed Solution for Test: Snell Law & Critical Angle - Question 10

Answer: a
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. To get θi, put N1 = 2, N2 = 1.3, θt = 45 degree. Thus we get θi = sin-1(1.3 sin 45)/2 = 41.68 degree.

Test: Snell Law & Critical Angle - Question 11

The angle at which the wave must be transmitted in air media if the angle of reflection is 45 degree is

Detailed Solution for Test: Snell Law & Critical Angle - Question 11

Answer: a
Explanation: In air media, n1 = n2 = 1. Thus, sin θi=sin θt and the angle of incidence and the angle of reflection are same. Given that the reflection angle is 45, thus the angle of incidence is also 45 degree.

Test: Snell Law & Critical Angle - Question 12

For a critical angle of 60 degree and the refractive index of the first medium is 1.732, the refractive index of the second medium is

Detailed Solution for Test: Snell Law & Critical Angle - Question 12

Answer: b
Explanation: From the definition of Snell law, sin θc = n2/n1. To get n2, put n1 = 1.732 and θc = 60. Thus we get sin 60 = n2/1.732 and n2 = 1.5.

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