Waveguides and Guided Waves MCQ

Answer.1. Total internal reflection

Explanation

• The working of an optical fiber is based on total internal reflection. Hence, option 2 is correct. 
• Optical fibers consist of many long high-quality composite glass/quartz fibers. Each fiber consists of a core and cladding.
• The refractive index of the core (μ1) material is higher than that of the cladding (μ2).
• When the light is incident on one end of the fiber at a small angle, the light passes inside, undergoes repeated total internal reflections along with the fiber, and finally comes out.
• The angle of incidence is always larger than the critical angle of the core material concerning its cladding.
• Even if the fiber is bent, the light can easily travel through along the fiber.

Answer.3. Less than that of core

Explanation

The refractive index of the cladding is always less than the refractive index of the core. This is to ensure the confinement of optical rays within the core. It is this property of core and cladding which makes light propagate inside the fiber. Thus, In a step-index optical fiber refractive index of the core is higher than the cladding.

Answer.2. Reduce dispersion and thereby increase the data rate

Explanation

• Graded-index (GI) fiber reduces multimode dispersion by grading the refractive index of the core so that it smoothly tapers between the core center and the cladding.
• Graded index fiber is used to minimize dispersion. This is because, in the graded-index fiber, the profile of the refractive index is parabolic and due to this refocusing of the signal within the core is increased, which eventually increases the data rate.

Answer.1. 15 mm

Explanation

The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.

The cut-off frequency for a rectangular waveguide with dimension ‘a (length)’ and ‘b (width)’ is given as:

${f_{c\left( {mn} \right)}} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

‘m’ and ‘n’ represents the possible modes.

c = speed of light = 3 × 1010 cm/s

Calculation:

TE₁₀ mode means m = 1, n = 0

The cut-off frequency of the dominant mode TE10 of the rectangular waveguide is:

f_c = c/2a

Where a is the dimension of the inner broad wall

a = (3 × 10¹⁰)/(2 × 10 × 10⁹)

a = 1.5 cm

a = 15mm

Answer.4. Horn

Explanation

A Horn antenna is best excited by a waveguide. The signal is fed from and received through a waveguide connected to a horn antenna.

Horn antennas are quite often used as feed systems for launching and detecting THz waves, particularly at longer wavelengths. A horn antenna can be regarded as a flared section of waveguide that acts as a smooth transition to free space. If the flare is gentle, then the field structure at the horn aperture.

Answer.3. Greater than

Explanation

The refractive index of the core is always greater than the refractive index of the cladding. This is to ensure the confinement of optical rays within the core. It is this property of core and cladding which makes light propagate inside the fiber. Thus, In a step-index optical fiber refractive index of the core is higher than the cladding.

Answer.4. 8.791 GHz

Explanation

For the circular waveguide dominant mode is TE₁₁

For the dominant mode, TE₁₁ cutoff frequency is given by

fc = 1.841c/2πa Hz

Where,

c = velocity of light = 3×10⁸ m/s

a = radius of the circular waveguide in meter

Calculations:

Given

a = 1 cm = 0.01 m

The cutoff frequency is given by

fc = 1.841c/2πa

fc = 1.841 × 3×10⁸ /(2π × 0.01)

fc = 8.791 GHz

Answer.2. 1.73

Explanation

The cutoff frequency of TE_mn mode is given by:

${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

for TE₁₀ mode

${f_{{c_{10}}}} = \frac{c}{2}\left( {\frac{1}{a}} \right)$ ——- (1)

for TE11 mode

${f_{{c_{11}}}} = \frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}}$ —— (2)

Dividing equation 1 by equation 2 we get

$\begin{array}{l} \frac{1}{2} = \frac{{\frac{c}{2}\left( {\frac{1}{a}} \right)}}{{\frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}\\ \\ = \frac{1}{2} = \frac{{\frac{c}{{2a}}}}{{\frac{c}{2}\sqrt {\frac{{{a^2} + {b^2}}}{{ab}}} }}\\ \\ = \frac{b}{{\sqrt {{b^2} + {a^2}} }}\\ \\ \frac{1}{2} = \frac{1}{{\sqrt {1 + {{\left( {\frac{a}{b}} \right)}^2}} }}\\ \\ {\left( {\frac{a}{b}} \right)^2} = {\left( {\frac{w}{h}} \right)^2} \end{array}$

= 4 − 1 = 3

ω/h = √3 = 1.732

Answer.2. 1.5 cm

Explanation

The cut off frequency is mathematically calculated as:

${f_{c\left( {min} \right)}} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

Where a and b are the dimensions of the waveguide

m and n are mode numbers TE_mn.

Analysis:

The dominant mode of a standard rectangular waveguide with a > b is TE₁₀, i.e. with m = 1, and n = 0, the cut-off frequency will be:

fc_(min) = c/2a

fc_(min) = 10 GHz

10 × 10⁹ = (3 × 10¹⁰)/2a

a = (3 × 10¹⁰)/(2 × 10 × 10⁹)

a = 1.5 cm

Answer.4. 4.5 × 10⁸ m/s

Explanation

Group velocity × Phase velocity = c²

2 × 10⁸ × phase velocity = (3 × 10⁸)²

V_P = (3 × 10¹⁶)/(2 × 10⁸ )

V_P = 4.5 × 10⁸ m/s