Courses

# Test: Snell Law And Critical Angle

## 12 Questions MCQ Test Electromagnetic Theory | Test: Snell Law And Critical Angle

Description
This mock test of Test: Snell Law And Critical Angle for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 12 Multiple Choice Questions for Electrical Engineering (EE) Test: Snell Law And Critical Angle (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Snell Law And Critical Angle quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Snell Law And Critical Angle exercise for a better result in the exam. You can find other Test: Snell Law And Critical Angle extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

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

Solution:

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.

QUESTION: 2

### The Snell’s law is given by

Solution:

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.

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.

Solution:

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

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.

Solution:

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.

QUESTION: 5

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

Solution:

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.

QUESTION: 6

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

Solution:

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.

QUESTION: 7

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

Solution:

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.

QUESTION: 8

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

Solution:

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.

QUESTION: 9

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

Solution:

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.

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

Solution:

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.

QUESTION: 11

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

Solution: