Unit 2.6)Refraction of light

Question 1

What is the approximate refractive index of air?

Answer 1

1.

Question 2

When light enters a more optically dense medium does it bend towards or away from the normal?

Answer 2

Towards the normal.

Question 3

When does total internal reflection occur?

Answer 3

When light is at a boundary to a less optically dense medium and the angle incidence is greater than the critical angle.

Question 4

What is the purpose of the cladding in a step index optical fibre?

Answer 4

-Protects core from scratches which would allow light to escape and degrade the signal.
-Allows TIR as it has a lower refractive index than the core.

Question 5

How does signal degradation by absorption in an optical fibre affect the received signal?

Answer 5

Part of the signals energy is absorbed by the fibre so its amplitude is reduced.

Question 6

State the advantages of optical fibres over traditional copper wires?

Answer 6

-Signal can carry more information as light has a high frequency.
-No energy lost as heat.
-No electrical interference.
-Cheaper.
-Very fast.

Question 7

What path does a light ray take when the angle of incidence is equal to the critical angle?

Answer 7

It goes along the boundary ie. the angle of refraction is 90degrees.

Question 8

What formula can be used to find the critical angle for 2 materials whose refractive indices are known?

Answer 8

SinC=n2/n1 where n1 > n2.
C = critical angle.
n1 = refractive index of material 1.
n2= refractive index of material 2.

Question 9

Find the critical angle of a water to air boundary if water has a refractive index of 1.33?

Answer 9

sinC = n2/n1 n2= air = 1 n1 = water= 1.33.
C = sin^-1 (1/1.33).
C = 48.8 degrees.

Question 10

Using snell’s law of refraction, find the angle of refraction in a material with RI = 1.53 when the angle of incidence is 32 degrees from a material with RI = 1.23?

Answer 10

n1sin i=n2sin r.
1.23sin32= 1.53sin r.
sin r=1.23 sin 32/1.53.
sin r = 0.426 r = 25.2 degrees.

Question 11

Glass has a refractive index of 1.5, water has a refractive index of 1.33, which is more optically dense?

Answer 11

Glass.

Question 12

What formula is used to determine the refractive index of a material?

Answer 12

n=c/v.
n=refractive index.
c= speed of light in vacuum, 3x 10^8 m/s.
v=speed of light in material.

Question 13

What are multimode fibres?

Answer 13

Multimode fibres are just fibres which have multiple paths of light travelling within them. Since different paths are taken, the modes(the individual paths) have varying propagation velocities. This means that the signal does not arrive at the same time and appears distorted which is called multimode dispersion. Shorter distances should be used with multimode fibres such that the effect of this is smaller. Also, you cannot have too many nodes in the fibre otherwise the multimode dispersion effect will become exaggerated and so the rate of data transfer is limited.

Question 14

What are monomode fibres?

Answer 14

These are optical fibres which have only a single mode which is parallel to the wire (through the centre). These have higher data transfer rates and can transfer data over much longer distances because there is little multimode dispersion(resulting form the single light ray and smaller core).

Question 15

The refractive index, n, of a medium being?

Answer 15

c/v in which v is the speed of light in the medium and c is the speed of light in a vacuum.

Question 16

How snells law relates to the wave model of light propagation and for diagrams of plane waves approaching a plane boundary obliquely, and being refracted

Answer 16

Certainly! Let’s explore how Snell’s law relates to the wave model of light propagation and discuss diagrams of plane waves approaching a plane boundary obliquely and being refracted.

1. Snell’s Law:
   
   • Snell’s law describes how light refracts at a boundary between two different media with varying indices of refraction.
   • It states that the angle of incidence (the angle between the incident ray and the normal to the boundary) and the angle of refraction (the angle between the refracted ray and the normal) are related by:
     n1 sin theta1 = n2 sin theta2
     where:
     n1 and n2 are the indices of refraction of the two media.
     -theta1 is the angle of incidence.
     
     • theta2 is the angle of refraction.
2. Wave Model of Light Propagation:
   
   • In the wave model, light is described as an electromagnetic wave.
   • When light encounters a boundary between two media, it can be thought of as a plane wave (a continuous wavefront) approaching the boundary.
   • The interaction between the wave and the boundary leads to reflection and refraction.
3. Plane Waves Approaching a Boundary:
   
   • Imagine a plane wave (such as light) incident on a boundary (e.g., air-to-glass interface).
   • The wavefronts are parallel and perpendicular to the direction of propagation.
   • The incident angle determines how the wave interacts with the boundary.
4. Refraction:
   
   • When the plane wave enters a medium with a different index of refraction, it changes direction.
   • The refracted wavefronts continue to propagate, but at a different angle.
   • Snell’s law governs this change in direction.

Remember that Snell’s law plays a crucial role in understanding how light bends when transitioning between different media.

Question 17

Time and distance calculations on incident and refracted wave fronts?

Answer 17

Certainly! When dealing with incident and refracted wavefronts, understanding time and distance calculations is essential. Let’s break it down:

1. Time Calculations:
   
   • Time can be calculated using the relationship:
     Time = Distance/speed
   • For example, if you know the distance traveled by a wavefront and the speed of the wave, you can find the time it takes to cover that distance.
2. Distance Calculations:
   
   • Distance can be calculated using the relationship:
     Distance = speed/time
   • If you have the speed of the wave and the time it takes to travel, you can determine the distance covered by the wavefront.

Question 18

How to apply the concept of total internal reflection to multimode optical fibres?

Answer 18

Certainly! Total internal reflection is a critical concept in the operation of multimode optical fibers. Let’s explore how it applies:

1. Optical Fiber Basics:
   
   • Optical fibers consist of a glass core surrounded by a cladding layer.
   • The core has a higher refractive index than the cladding.
2. Total Internal Reflection (TIR):
   
   • When light travels within the core and encounters the core-cladding boundary, it can be trapped within the core.
   • TIR occurs when the angle of incidence is below a certain critical angle.
   • The core-cladding refractive index difference ensures that light reflects back into the core, allowing efficient signal transmission.

In summary, multimode optical fibers utilize total internal reflection to guide light along the fiber, enabling data transmission over long distances.

Question 19

Problem of multimode dispersion with optical fibres? Calculations for the transit times of pulses via the straight path and zig zag path?

Answer 19

Certainly! Let’s delve into the concept of multimode dispersion in optical fibers and its impact on data transfer rates and transmission distances. Additionally, we’ll explore how transit times of pulses can be calculated for both straight and zigzag paths.

1. Multimode Dispersion:
   
   • Multimode fibers (MMFs) allow multiple modes (light paths) to propagate simultaneously.
   • However, due to different path lengths and refractive indices, these modes arrive at the end with varying timing.
   • This modal dispersion results in a narrower bandwidth and shorter transmission distance for MMFs compared to single-mode fibers (SMFs).
2. Data Transfer Rate and Transmission Distance:
   
   • MMFs suffer from limitations in data transfer rates and reach due to dispersion effects.
   • As the data rate increases, the impact of dispersion becomes more pronounced.
   • Longer transmission distances exacerbate the problem.
3. Calculations for Transit Times:
   
   • To address this, candidates must understand transit times for pulses:
     
     • Straight Path: Calculate the time it takes for a pulse to travel directly through the MMF.
     • Zigzag Path: Consider the additional time due to mode mixing and scattering as the pulse follows a zigzag route.

In summary, managing multimode dispersion is crucial for optimizing data transfer rates and extending transmission distances.

Question 20

How has the introduction of monomode optical fibres allowed for much greater transmission rates and distances?

Answer 20

Single-mode fibers, also known as monomode fibers, are optical fibers designed to support only a single propagation mode (LP 01) per polarization direction for a given wavelength. Here’s how they allow for greater transmission rates and distances:

1. Propagation Mode: Single-mode fibers have a small core diameter (typically a few micrometers) and a small refractive index difference between the core and cladding. They support only the fundamental mode (LP 01) and do not allow higher-order modes (e.g., LP 11, LP 20). Cladding modes, which are not localized around the fiber core, exist instead.
2. Transverse Intensity Profile: The transverse intensity profile at the fiber output has a fixed shape, independent of launch conditions and injected light properties. This property ensures efficient coupling into the guided mode.
3. Advantages:
   
   • Intermodal Dispersion: Single-mode fibers eliminate intermodal dispersion, making them suitable for high data rates (multiple Gbit/s) over long distances.
   • Long-Haul Data Transmission: Due to their low losses and dispersion characteristics, single-mode fibers are exclusively used for long-haul data transmission.
   • Distance: Single-mode fibers can transmit up to 100 km without signal regeneration, far exceeding multimode fibers.
4. Applications: Single-mode fibers are commonly used in outdoor long-distance applications, such as optical fiber communications, cable TV broadcasting, and long-distance telephony.

For short-distance indoor use, multimode fibers are more common because they allow the use of cheaper multimode data transmitters based on light-emitting diodes (LEDs) instead of laser diodes.