Carbon disulfide (n = 1.63) is poured into a container made of crown glass (n = 1.52). What is the critical angle for internal reflection of a ray in the liquid when it is incident on the liquid-to-glass surface?
Refractive index of glass with respect to liquid:
n = 1.52/1.63 = 0.9325
thus, critical angle, i = sin-1(n) = 68.8 degrees
What is the relation between critical angle and refractive index?
In Optics, the angle of incidence to which the angle of refraction is 90° is called the critical angle. The ratio of velocities of a light ray in the air to the given medium is a refractive index. Thus, the relation between the critical angle and refractive index can be established as the Critical angle is inversely proportional to the refractive index.
Critical Angle and Refractive Index
The relationship between critical angle and refractive index can be mathematically written as –
C is the critical angle.
μ is the refractive index of the medium.
a and b represent two mediums in which light ray travels.
Critical angle is
The Critical Angle
a light ray is in the more dense medium and approaching the less dense medium.
The angle of incidence for the light ray is greater than the so-called critical angle.
When the angle of incidence in water reaches a certain critical value, the refracted ray lies along the boundary, having an angle of refraction of 90-degrees. This angle of incidence is known as the critical angle; it is the largest angle of incidence for which refraction can still occur. For any angle of incidence greater than the critical angle, light will undergo total internal reflection.
Which of the following phenomena takes place inside an optical fiber ?
In fibre optics ,an optical phenomenon known as total internal reflection is used to transmit light rays. In case of the simplest form of optical fiber, light entering one end of the fiber strikes the boundary of the fiber and is reflected inward.
Light is confined within the core of a simple optical fiber by:
Light remains confined within the core of simple optical fibre because of Total internal reflection from core cladding boundary.
Light is confined within the core of a simple optical fiber by. If light hits a boundary of a material of lower refractive index at a steep enough angle, it cannot get out and it's reflected back into the high index medium, as in the figure below.
An optical fibre is a thin rod of high-quality glass. Very little light is absorbed by the glass.Optical fibres can carry more information than an ordinary cable of the same thickness. The signals in optical fibres do not weaken as much over long distances as the signals in ordinary cables.
Total internal reflection. When light traveling in an optically dense medium hits a boundary at a steep angle (larger than the critical angle for the boundary), the light is completely reflected. This effect is used in optical fibers to confine light in the core.
The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.
The critical angle for the material of a prism is 45° and its refracting angle is 30°. A monochromatic ray goes out perpendicular to the surface of emergence from the prism. What is the angle of incidence on the prism?
Using the equation for the critical angle
45o= arcsin(1/n1) à n1=1/sin45o= √2
From the figure,
sin β/sin γ à sin β=n1sin γ/n2 à
β=arcsin(n1.sin γ/n2)=arcsin(√2sin30o/1)=arcsin(√2/2)à β=45o
Total internal reflection occurs when
When the angle of incidence in water reaches a certain critical value, the refracted ray lies along the boundary, having an angle of refraction of 90-degrees.For any angle of incidence greater than the critical angle, light will undergo total internal reflection.
Mirage is caused due to
Mirage is formed by total internal reflection in deserts where due to heating of earth, refraction index of air near the surface of earth becomes lesser than above it.
We have a right angled isosceles prism, Its refractive index is 1.5. If we incident a ray normally on one of the two perpendicular surfaces, which of the following phenomenon will take place?
Critical angle will be, sin−1(1/1.5) = 41.8. The angle of incidence inside the prism will be 45. Since it is more than critical angle, the light ray will be internally reflected.
Total internal reflecting mirrors are preferred over plane mirrors because
Whenever there is reflection on any surface there is refraction also beneath the surface.
The energy of the incident beam is partially reflected and partially absorbed.
Even a good mirror reflects nearly 80%of the incident light only.
At normal incidence (Incident perpendicular to the surface) there is no reflection and all the light rays pass through the surface.
When a ray passes through a denser medium and tries to enter a rarer medium and if the angle of incidence is more than the critical angle, we say there is total internal reflection.
The total here implies that there is no refraction at all. The entire ray is refracted.
Therefore when we use totally reflecting prisms, there is no energy lost due to normal incidence from air to glass and there is TOTAL (no partial refraction) reflection inside the prism.
Thus the light is reflected almost 100% in totally reflecting prisms and hence they are preferred even though it costs high.