Contents
By learning Physics Topics, we can gain a deeper appreciation for the natural world and our place in it.
Define the Terms Critical Angle and Total Internal Reflection? What are the Examples of Total Internal reflection with Explanation?
We know, in refraction from denser to a rarer medium light bents away from the normal. As a result angle of refraction becomes larger than angle of incidence.
In Fig., line AB represents the surface of separation of water and air.
i) Ray P1O travelling through water is incident at O on the surface of separation [Fig.(a)]. A part of the ray is reflected back into water along ORl and another part is refracted into air along OQ1 .The angle of refraction ∠Q1ON is greater than the angle of incidence ∠P1ON1. Greater the angle of incidence, greater is the angle of refraction and in each case both reflection and refraction will take place.
ii) For a particular value of the angle of incidence, the angle of refraction becomes 90°, so that the refracted ray grazes the surface of separation [Fig.(b)]. This limiting angle of incidence in the denser medium is called the critical angle for the two given media. Thus, ∠P2ON1 = critical angle (θc). In this case, the angle of refraction ∠NOQ2 = 90° . Here also a part of the incident ray is reflected back to water along OR2.
iii) If the angle of incidence exceeds the critical angle i.e., if i > θc [as in the case of the incident ray P3O in Fig.(c)] no part of the incident ray is refracted in the second medium. The ray is completely reflected along OR3 into the first medium. This phenomenon is called total internal reflection. In this case the surface of separation of the two media behaves as a mirror.
Critical angle: It is that particular angle of incidence of a ray of light for a given pair of media, passing from denser one to rarer one, for which the corresponding angle of refraction is equal to 90° and the refracted ray grazes along the surface of the interface separating the two media. The critical angle of a pair of media depends on the colour of the incident light and the nature of the two media.
For example, in case of two particular media, the critical angle for red light is greater than that for violet light. Again, critical angle of water to air is 49°, while that of glass to air is 42° .
The statement, ‘critical angle of glass to air is 42° ‘ means that a ray of light from glass being incident on the surface of separation of glass and water at an angle 42°, should go along the surface of separation after refraction i.e., the refracted angle will be 90° .
Total internal reflection: When a ray of light travelling from a denser medium to a rarer medium is incident at the surface of separation of the two media at an angle greater than the critical angle for the media, there is no refraction; rather the whole of the incident ray is totally reflected. This phenomenon is known as total internal reflection.
Condition of total internal reflection: The conditions to be satisfied for total internal reflection are as follows—
- Light must travel from a denser to a rarer medium.
- The angle of incidence must be greater than the critical angle for the two media.
Reason for using the term ‘total’: In ordinary reflection, a part of the incident light is reflected from the surface of separation and the rest is refracted. But in case of internal reflec-tion no part of the incident light is refracted, rather the entire portion of the incident light is reflected back to the first medium from the surface of separation of the two media. So this reflec-tion is called total reflection.
Relation between critical angle and refractive index of the denser medium: Let ∠P2ON1 = θc = critical angle between the two media, water and air [Fig.(b)], which implies that the angle of refraction is 90°. If the refractive index of air with respect to water is wµa, then
wµa = \(\frac{\sin \theta_c}{\sin 90^{\circ}}\) or, sin θc = \(\frac{1}{a^{\mu_w}}\)
So, the value of the critical angle depends on the refractive index of one medium with respect to another.
If the medium is a and b and µa > µb then,
Numerical Examples
Example 1.
If the absolute refractive index of a medium is \(\sqrt{2}\), calculate the critical angle of glass to the medium. Given that the critical angle of glass to air is 30°.
Solution:
If the refractive index of glass with respect to air be aµg, then
aµg = \(\frac{1}{\sin \theta_c}\) = \(\frac{1}{\sin 30^{\circ}}\) = 2
If the refractive index of glass with respect to the medium be mµg then
mµg = \(\frac{a^\mu g}{a^\mu \mu_m}\) = \(\frac{2}{\sqrt{2}}\) = \(\sqrt{2}\)
If the critical angle of glass to the medium be θc, then
sin θc = \(\frac{1}{m^\mu}\) = \(\frac{1}{\sqrt{2}}\) = sin45° or, θc = 45°
Example 2.
Refractive index of carbon disulphide for red light is 1.634 and the difference in the values of critical angle for red and blue light at the surface of separation of carbon disulphide and air is 0°56′. What is the value of refractive index of carbon disulphide for blue light?
Solution:
Let the refractive index of carbon disulphide for red light = µr and critical angle = θr.
Now, sin θr = \(\frac{1}{\mu_r}\) = \(\frac{1}{1.634}\) = 0.6119 = sin37.73°
∴ θr = 37.73°
Let the critical angle for blue light be θb Refractive index increases as wavelength of light decreases. So critical angle decreases.
According to the question,
θb = 37.73° – 0°56′ = 37.73° – 0.93° = 36.8°
So refractive index of carbon disuiphide for blue light,
µb = \(\frac{1}{\sin \theta_b}\) = \(\frac{1}{\sin 36^{\circ} 48^{\prime}}\) = \(\frac{1}{\sin 36.8^{\circ}}\) = \(\frac{1}{0.599}\)
= 1.669
Example 3.
If the refractive index of diamond is 2.42, prove that all the beam of rays having angle of incidence more than 25° will be totally reflected. [sin 24.41° = 0.4132]
Solution:
If the critical angle be θb then,
sin θc = \(\frac{1}{\mu}\) = \(\frac{1}{2.42}\) = 0.4132 = sin24.41°
∴ θc = 24.41°
We know that if the angle of incidence of a ray of light is greater than the critical angle, the ray will be totally reflected. Here the critical angle is 24.41° . So rays of light having angle of incidence greater than 25° will be totally reflected.
Example 4.
A ray of light will go from diamond to glass. What should be the minimum angle of incidence at the sur-face of separation of the two media, diamond and glass, so that the ray of light cannot be refracted in glass? µ of glass = 1.51 and µ of diamond = 2.47; sin37.69° = 0.61134
Solution:
If the light ray is incident at the surface of separation of the two media diamond and glass at critical angle, the ray is grazingly refracted in glass. If the critical angle is θc then
sin θc = \(\frac{1}{g^\mu d}\) = \(\frac{\frac{1}{\mu_d}}{\mu_g}\) = \(\frac{\mu_g}{\mu_d}\) = \(\frac{1.51}{2.47}\)
= 0.61134 = sin 37.69°
∴ θc = 37.69°
So if the angle of incidence of a light ray is greater than 37.69° it cannot be refracted in glass.
∴ Required minimum angle of incidence = 37.69°.
Example 5.
A cube has refractive index µ1. There is a plate of refractive index µ2(µ2 < µ1) upon it. A ray travelling
through air is incident on a side face of the cube. The refracted ray is incident on the upper face of the cube at the minimum angle for total internal reflection to occur. Finally the reflected ray emerges from the opposite face. Show that if the angle of emergence is ϕ then sinϕ = \(\sqrt{\mu_1^2-\mu_2^2}\)
Solution:
In Fig., PQ = incident ray on the side face of the cube, QR = refracted ray inside the cube, RS = totally reflected ray from the upper face of the cube, ST = emergent ray from the opposite face. Let critical angle for total reflection be θc.
According to the question, \(\theta_c^{\prime}\) ≈ θc
Example 6.
A ray of light travelling through a denser medium is incident at an angle i in a rarer medium. If the angle between the reflected ray and the refracted ray is 90°, show that the critical angle of the two media, θc = sin-1(tan i).
Solution:
Suppose the angle of refraction in the rarer medium = r.
Example 7.
A nail is fixed up perpendicularly at the centre of a circular wooden plate. Keeping the nail at bottom, the circular plate is made to float in water. What should be the maximum ratio of the radius of the plate and length of the nail so that the nail will be out of vision? Refractive Index of water = \(\frac{4}{3}\)
Solution:
AB is the circular wooden plate and CD is the nail. Suppose, radius of the plate = r and the length of the nail = h [Fig.]. Since the nail is not seen from air, the angle of incidence of the ray DA will be greater than θc and the ray will be totally reflected.
This is the required ratio.
Example 8.
The critical angle of glass relative to a liquid is 57°20’. Calculate the velocity of light In the liquid. Given, μ of glass = 1.58, velocity of light in vacuum = 3 × 108 m ᐧ s-1 and sin57°20’ = 0.8418.
Solution:
Critical angle of glass relative to the liquid, θc = 57°20’.
If the refractive index of glass with respect to the liquid is lµg then,
Example 9.
A transparent solid cylindrical rod has a refractive index of \(\frac{2}{\sqrt{3}}\). It is surrounded by air. A light ray is incident at the midpoint of one end of the rod as shown in Fig. Determine the incident angle θ for which the light ray grazes along the wall of the rod. [AIEEE ‘09]
Solution:
For refraction of light at point B, we can write by applying Snell’s law,
1 × sinθ = µsinr
[where µ is the refractive index of the solid material]
or, sinθ = \(\frac{2}{\sqrt{3}}\)sinr …. (1)
The light ray BC is incident on point C making critical angle θc and propagates along CD [Fig.].
Examples of Total Internal Reflection
A test tube dipped in water: A glass test tube half filled with water is held obliquely in a beaker containing water [Fig.]. The empty portion of the immersed test tube appears shining if it is seen from above. This happens due to total internal reflection of light.
For the empty portion of the tube, light goes from denser to rarer medium. Rays which are incident at angles greater than the critical angle of glass and air (48.5°) are totally reflected. So this portion of the glass appears shining.
Portion of the tube filled with water does not glow because here light enters water in test tube from water in beaker. Thus, total internal reflection does not occurred here. In this discussion we do not take account of the existence of glass wall of the tube due to its negligible thickness.
A metal ball coated with lampblack immersed in water: If a metal ball coated will lampblack is immersed in water, the ball appears shining. Due to the coating of the lampblack, a thin layer of air surrounds the surface of the ball. Rays incident at an angle greater than the critical angle of water and air, are totally reflected. The ball appears shining when the totally reflected rays reach the eyes of the observer.
Glass tumbler full of water: A glass tumbler full of water is held above the eye level. If now the upper surface of water in the tumbler is seen from any one side, the surface appears shining [Fig.].
Rays coming from the side of the tumbler are incident on the surface of separation of water and air. Hence total internal reflection takes place at particular angles of slantness and the surface of water appears shining.
Air bubbles: The air bubbles rising through water look shiny. Rays travelling through water are incident on the surface of the air bubbles. Those rays which are incident at angles greater than the critical angle are totally reflected. When these reflected rays reach the eyes of the observer, the bubbles appear shining. For the same reason air bubbles existing in paper weights appear to be shining.
Natural Examples of Total Internal Reflection
Mirage: It is an optical illusion brought about by total internal reflection. There are two types of mirage, one observed in hot regions and the other observed in extremely cold regions.
Inferior mirage or mirage in desert: People travelling through desert sometimes see water at a distant place which is actually an optical illusion, called inferior mirage, or simply, mirage.
During day time, the lower regions of the atmosphere become hotter than the upper regions. So density of air in the lower regions is less than that in the higher regions. Let us consider the atmosphere to be made up of layers of air, one above the other. A ray of light starting from a distant tree (P) and travelling downward happens to be going from denser to rarer medium [Fig.].
So its angle of incidence at consecutive layers goes on increasing gradually till it exceeds the critical value when it is reflected back due to total internal reflection. The traveller sees an inverted virtual image (P’) of the tree.
Secondly, due to continuous change in temperature, there exists a temperature gradient in the layers which undergo continuous change of density and hence in the refractive index as well. So the path of the rays coming through the layers of air are also continuously changing. Hence to the traveller, the image of the tree appears to be swaying. This completes the illusion of a pond lined with trees.
2. Superior mirage or mirage in cold countries: in cold countries, the temperature of air in the lower regions is lower than that of the upper region. So the density of air in lower region is greater Mn that of upper region. A ray of light starting from an object (P) travelling upwards, finds itself going from denser to rarer medium [Fig.]. So its angle of incidence at consecutive layers of air gradually goes on increasing till it reaches the critical value. Then it is reflected back due to total internal reflection. To an observer, the ray appears to come from a point above, thus giving an impression that an inverted object (P’) is floating in air which is an optical illusion. This phenomenon is called superior mirage.
View of an observer inside water: To the eye of an observer or a fish inside water, all objects above water appear to exist in a cone of semivertical angle 49° which is the critical angle of water and air. This happens due to total internal reflection of light.
If a ray of light travelling from a denser medium is incident at the critical angle, the refracted ray grazes the surface of separation. Conversely, if a ray of light travelling from a rarer medium is incident at an angle of 90°, the angle of refraction in the denser medium becomes equal to the critical angle. Critical angle of water and air is 49° . So if a ray of light S1A coming from the rising sun S: along the surface of water reaches the eye E along the direction AE, then the angle of refraction in water becomes 49° [Fig.]. As the eye cannot follow ray AS1, so an observer inside water will see the rising sun along the line EA C and this line will make an angle 49° with the line OE. Similarly, the. setting sun S2 will be seen along the line EBD and this line also will make an angle 49° with the line OE. So all the objects above water appear to exist in a cone of angle 98° to the eye of a fish or observer under water.
It is to be noted that the sun describes an arc of 180° to earth- bound observers but to the eyes of a fish it describes an arc of 98° .
Surface of water to the eye of an observer inside water: The diameter of the circular base of the cone AEB is AB. If an observer keeping his eye at E looks at the circular section of water, he can see any object lying above water. But if the observer looks at the rest of the portion of water other than the circular portion, then
1. he cannot see any object above water, rather
2. he can see the images of the objects inside water. Reason explaining 1st incident: Any ray of light coming from outside water can reach the point E only through the circular section but cannot reach the point E if it comes through the remaining portion.
Reason explaining 2nd incident: Suppose the ray of light emerging from the object situated in water, reaches the point E after reflection from the surface of water [Fig.]. This reflection will take place from the surface of water excluding the circular portion. Obviously, this reflection will be total reflection. For example if a ray of light from the object P situated inside water, is incident on the surface of water, the angle of incidence exceeds 49°. So, the ray after total reflection from the surface of water reaches the eye of the observer and he observes the object at P’.
So to the observer situated inside water, the surface of water appears as a mirror with a circular hole in it, because he sees the objects situated outside water through the circular section and sees the images of the objects inside water in the remaining por-tion of the surface of water. Obviously, the radius of the circular hole is OA or OB.
Determination of the radius of the hole: Let the radius of the hole =OA = OB = r and OE = h [Fig.]. If the critical angle is θc, then ∠OEA = θc.
Sparkling of diamond: Diamond is notable for its spar-kle and shine. This characteristic of diamond is based on total internal reflection. Refractive index of diamond is 2.42 and its critical angle relative to air is only 24.4° . This value is quite small as compared to other pairs of media. Therefore, there is a high probability of total internal reflection in case of diamond. If the diamond is cut properly, it will have a large number of faces. Ray of light entering through one face undergoes total internal reflection at a number of faces. As the rays of light enter through many faces and being confined inside they emerge together through only a few faces, these faces appear to sparkle and shine.