# Total Internal Reflection

• Last Updated : 21 Sep, 2021

You’ve probably seen diamonds gleaming. Do you know why diamonds shine so brightly? Alternatively, imagine you place a penny beneath the glass and fill it with water. The coin is not visible when viewed from the side, but it is visible when viewed from the top. What causes this to happen? Internal introspection is the key to this. Let’s take a closer look at the complete interior reflection.

### Total Internal Reflection

Let us do a short task to better grasp the notion of comprehensive internal reflection. Take a coin and a glass and put them together. Place the penny at the bottom of the glass and fill it with water. What happened to the coin? We can see the coin from the top of the glass, but not when we look at it sideways. We can’t see the coin since it’s invisible to us. What caused this to happen?

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This is due to the fact that when there is water in the glass, light from the coin travels through the glass at a certain angle to our sight. Further, when water is added to the glass, the light from the coil strikes the interior of the glass at a higher angle than the critical angle. The glass inside reflects 100% of the light. Internal reflection is what it is.

The full reflection of a light ray at the boundary of two media when the beam is in the medium with a higher refractive index is known as total internal reflection.

When light beams go from a more optically dense material to a less optically dense medium, this phenomenon happens. There are two situations in which total internal reflection occurs:

• When the light is approaching the less dense medium from the more dense medium.
• If the incidence angle is larger than the critical angle.

### Illustration of Total Internal Reflection

Consider the case below. A beam of light travels from a watery medium to one of air. The lightray will be refracted at the point where the two mediums meet. The refracted light beam bends away from the normal as it passes from a medium with a higher refractive index to one with a lower refractive index.

• The incident beam of light is refracted in such a way that it travels over the water’s surface at a certain angle of incidence.
• The critical angle is the angle of incidence at which something happens. The refraction angle is 90° here.
• The incident ray is reflected back to the medium when the angle of incidence is larger than the critical angle. This event is referred to as total internal reflection.

From the above observation, we can conclude two formulas:

• Total Internal Reflection

n1 ⁄ n2 = sin r ⁄ sin i

where,

• i is the angle of incidence,
• r is the angle of refraction,
• n1 is the denser medium, &
• n2 is the rarer medium.
• Critical Angle

θc = sin−1(n2 ⁄ n1)

where, θc is the critical angle.

### Examples of Total Internal Reflection

• Optical fibre

Because the angle generated by the incoming beam is higher than the critical angle, total internal reflection occurs when it strikes the cladding. Optical fibres have revolutionised the speed with which signals are carried across cities, nations, and continents, making telecommunication one of the quickest forms of information transportation. Endoscopy also uses optical fibres.

• Mirage

It’s an optical illusion that causes the water layer to appear at short distances in the desert or on the road. Total internal reflection, which happens as a result of atmospheric refraction, is an example of a mirage.

• Diamond

When light rays penetrate a diamond, they are completely internalised and reflected on all of the diamond’s facets. The crucial angle for a beam of light going from a diamond to air is just 24°. As a result, the majority of incoming photons experience complete internal reflections.

Furthermore, diamonds are often cut in such a way that when a beam of light penetrates them, it is subjected to entire internal reflections on multiple faces. When the angle of incidence at any face is less than 24°, light shines through, making the diamond look brilliant. Optical fibres also employ total internal reflection. Video and audio signals are sent across great distances using optical fibres.

### Sample Questions

Question 1: The glass with the refractive index n1 = 1.33 is made up of an optical fibre and is surrounded by another glass with the refractive index n2. Determine the cladding’s refractive index n2 so that the critical angle between the two claddings is 30°.

Given:

Critical angle, θ = 50°

Refractive index, n1 = 1.33

Critical Angle

θc = sin−1(n2 ⁄ n1)

n2 = n1 sinθc

= 1.33 × sin 30°

= 0.665

Hence, the refractive index of the second glass is 0.665.

Question 2: Determine the refractive index of a medium with a critical angle of 45°.

Critical angle, θc = 45°

Refractive index of the medium, μ = 1 ⁄ sin θc

μ = 1 ⁄ sin 45°

= 1.414

Hence, the refractive index of a medium is 1.414.

Question 3: Assume that a beam of light travels from a media having a refractive index of 1.33 to a medium with a refractive index of 1.50. What is the critical angle if the incident ray is at a 50° angle with respect to the normal?

Light is said to go from one medium to another in response to this query. We’re provided with the refraction indices for both media, as well as the incidence angle relative to the normal, and we’re asked to calculate the critical angle. It’s crucial to keep in mind that as the light goes from one medium to another, the speed at which it travels changes.

The index of refraction of a medium is determined by how fast light travels in that medium compared to how fast light travels in a vacuum. Light bends more strongly towards the normal in denser media, whereas it bends farther away from the normal in a less dense medium. The light must move from a medium with a higher density to a medium with a lower density in order to reach a critical angle. To put it another way, light must move from a medium with a higher refraction index to one with a lower refraction index.

As a result, if a ray of light is moving from a medium with a low index of refraction to a media with a greater index of refraction, there will be no critical angle. This is exactly what is going on in the question stem. As a result, there can’t be a critical angle.

Question 4: When the wavelengths of light in two liquids, x and y are 250 nm and 500 nm respectively, what will be the critical angle of x relative to y?

Given:

Wavelengths, λx = 250 nm and λy = 500 nm

θc = sin−1(ny ⁄ nx) = sin−1x ⁄ λy)

θc = sin−1(250 nm ⁄ 500 nm)

= 30°

Hence, the critical angle of x relative to y is 30°.

Question 5: Why do complete internal reflection pictures appear to be brighter than those created by mirrors or lenses?