Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
Why is an Optical Fibre Generally Called a Light Pipe? Why is the Sun Visible Before the Sunrise and After the Sunset?
Transmission of Light through Optical Fibre
Optical fibre: A beam of light can be sent from one place to another through an optical fibre made of glass, quartz or optical grade plastic, by following successive total internal reflections. As water can be sent from one place to another through a hollow pipe, a fibre can allow light to flow through it from one place to another. Hence, an optical fibre is often loosely called light pipe.
Construction and principle of action: An optical fibre is a long and very thin pipe. Its diameter is about 10 × 10-6 metre. The internal section of the pipe is called core. It is the core through which light travels from one point to another. Above the core there is a coating of a substance having refractive index less than that of the core. This coating is called cladding.
A ray of light entering the fibre through one face undergoes successive total internal reflections at the surface of separation of core and cladding and emerges through the other face [Fig.].
As total internal reflection of light takes place inside a fibre, the intensity of the light remains almost the same. The image of a large object cannot be sent through a single fibre. In that case bundles of fibres or cables of fibres are used. A cable contains about a thousand fibres [Fig.(a)]. The image of an object is focussed on one end of the bundle. If the order of the fibres is properly main-tained, the image obtained at the other end will be an exact reproduction.
In Fig.(b), the letter ‘T’ has been focussed at one end of the bundle and an exact image of ‘T’ has been obtained. Light rays from the different portions of ‘T’ travel through the different fibres and form the image at the other end.
Application of optical fibre: Optical fibres are extensively used in medical science and in the field of communication.
i) These are used to study the interior parts of body which are inaccessible to bare eye, e.g., lungs, tissues, intestines etc.
ii) It can be used to transmit high intensity laser light inside the body for medical purposes.
iii) These are used for sending signals form one place to another. This signal is mainly digital in nature. It is information which the signal carries. This information is used in telephone, television, fax, computer etc. It is to be noted that, different digital signals may be sent through the same fibre at the same time, without any chance of overlapping.
iv) So many times it is needed to collect sample inside from human body to identify disease. In this purpose optical fibre is used. Besides it, optical fibre is used for operation inside human body. Thus, in most cases no major excising is needed outer part of the body.
Advantages of optical fibre over copper wire:
- Comparatively less power is required to send a signal.
- The loss in energy is much less.
- The capacity of carrying information is approximately 1000 times.
- There exists no influence of any external electromagnetic wave signal.
- Electrical resistance is much more.
- It is very light.
- Velocity of the signal is very fast (approximately equal to that of light in vacuum).
- Possibilities of the illegal usage of signal are very low.
Numerical Examples
Example 1.
A point source of light is placed at a depth of h below the calm surface of water. From the source, light rays can only be transmitted to air through a definite circu-lar section,
(i) Draw the circular section of the surface of water by ray diagram and mark its radius r.
(ii) Determine the angle of incidence of a ray of light incident at any point on the circumference of the cir-cular plane. [Given: refractive index of water, µ = \(\frac{4}{3}\); 48°36′ = sin-10.7501]
(iii) Show that r = \(\frac{3}{\sqrt{7}} h\).
Solution:
i) Let MN be the open surface of water [Fig.]. O is the source of light at a depth h below the surface of water. Light rays incident on the surface of water from O at angles less than critical angle transmit in air after refraction. At the points A and B the light rays are incident at angles equal to the critical angle (θ). So the refracted rays at these two
points graze along the surface of separation. So the light rays will transmit outside water only through the circular section of radius r = AP = PB. If the rays are incident on the surface of water excluding this circular section, they will be totally reflected from the surface of water and will return to water.
ii) Let the angle of incidence be θ.
iii) From the triangle AOP,
Example 2.
The water in a pond has a refractive index \(\frac{5}{3}\). A source of light is placed 4 m below the surface of water. Cal-culate the minimum radius of an opaque disc needed to be floated on water so that light does not come out. [CBSE ‘ 01]
Solution:
Minimum radius of the opaque disc,
r = \(\frac{h}{\sqrt{\mu^2-1}}\) = \(\frac{4}{\sqrt{\left(\frac{5}{3}\right)^2-1}}\) = \(\frac{4}{\frac{4}{3}}\) = 3 m
Example 3.
Fig. shows a longitudinal cross section of an opti-cal fibre made of glass of refractive index 1.68. The pipe is coated with a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflection inside the fibre can take place?
Solution:
The refractive index of the outer coating with respect to the glass pipe,
gµc = \(\frac{a^\mu c}{a^\mu}\) = \(\frac{1.44}{1.68}\)
If the critical angle for the total reflection is 6, then
Thus total reflection takes place when i’ > 59° or when r < rmax where rmax = 90° – 59° = 31°
So if the maximum angle of incidence on the fibre is imax then,
sin imax = μgsinrmax = 1.68sin31° = 0.865
∴ max = 60°
So, the range of the angles of incidence for total internal reflec-tion inside the fibre is from 0° to 60° .
Example 4.
In which direction will the sun appear to set if the observer is inside the water of a pond? Refractive index of water, μ = 1.33.
Solution:
For a setting sun the incident rays graze along the sur-face of water, i.e., angle of incidence = 90° .
∴ According to Snell’s law, or, 1.33
µw = \(\frac{\sin i}{\sin r}\) or, 1.33 = \(\frac{\sin 90^{\circ}}{\sin r}\)
or, sinr = \(\frac{1}{1.33}\) = 0.7518 = sin 48.75°
∴ r = 48.75°
Therefore, to see the setting sun the observer in water should look at an angle of 48.75° with the normal.
Atmospheric Refraction
Apparent position of a Star: The whole atmosphere surrounding the earth may be supposed to be divided into dif-ferent horizontal layers. As the height above the earth’s surface increases, the density of the air decreases. Due to this, the refrac-tive index of air also decreases with the increase in altitudes.
For this reason, the ray from a star S (say) proceeding towards the earth’s surface cannot travel straight but continually bends towards the normal at the surface of separation due to refraction as it penetrates from rarer to denser layers [Fig.]. This ray after several refractions reaches the observer at O.
But our vision cannot follow the curved path OS. A tangent OS’ is drawn on OS at O. So, the observer sees the star at S’. This phenomenon is called atmospheric refraction.
So due to atmospheric refraction the apparent angular altitude of a star or any other heavenly bodies increases. If the star is sit-uated at A just above the observer, its angular altitude does not change. The observer sees the star in its actual position. Atmo-spheric refraction is measured by the angle a between its actual position and apparent position. When a heavenly body is at the horizon, the value of a becomes the greatest (nearly 0.5°). But as the body transcends above the horizon, a decreases. Finally when the body is situated just above the observer, the value of a becomes zero.
Visibility of sun before sunrise and after sunset:
The diameter of the sun subtends an angle of 0.5° at the eye of an observer on earth. This value is equal to the deviation of sun-light due to atmospheric refraction, when at the horizon. So, the sun appears to just touch the horizon during sunset and sunrise when it is actually below it. What we see therefore is the raised image of the sun, formed due to atmospheric refraction. As a result, we see the sun a few minutes after sunset or before sunrise. Now, the sun covers a distance equal to its diameter in 2 min. So, the sun becomes visible another 2 min earlier at sunrise and also remains visible for another 2 min after actual sunset. Consequently, 4 min are added to the length of a day. This value is valid for observations from the equatorial region. At higher latitude this time increases.
Oval shape of the sun when it is near the horizon: During sunrise or sunset, the lower edge of the sun remains nearer to the horizon than its upper edge. So, the rays coming from the lower edge of the sun are incident on an atmo-spheric layer at larger angle than that for the rays coming from its upper edge.
As refracting angle increases with the increase of incident angle, the rays coming from the lower edge bend more than the others due to multiple refractions at different atmo-spheric layers. As a result the vertical diameter of the sun appears to be reduced whereas the horizontal diameter remains unaffected.
Twinkling of stars: Due to atmospheric refraction we see the stars twinkle. Light rays from the stars situated far and far away from us come to our eyes passing through various layers of air. The temperature of the layers does not remain constant and changes continuously. So the density of the various layers also changes.
Again the refractive index of the layers changes with the change of density. So, when the rays of light from a star come to our eyes, the direction of the path of the rays changes continuously. As a result, the amount of light reaching our eyes also changes continuously. It seems as if the brightness of the stars is changing. So the stars appear twinkling.
As the planets are nearer to us than the stars, more amount of light comes to us. Therefore, change of brightness of the planets due to change of refractive index of various layers of air is negli-gible. We cannot detect it with our eyes. So it appears that the planets are emitting light steadily.