Why Does the Critical Angle Depend on Refractive Index?
Relation between critical angle and refractive index can be formed because both of them are inversely proportional. But, before going into this detail, you must understand these topics separately.
What is Critical Angle?
In optics as a topic of Physics, the critical angle makes reference to a particular angle of incidence, which gives an angle of refraction of 90 degrees. Additionally, a water–air boundary has a critical value of 48.6 degrees. while the critical angle for crown water and glass boundary is 61.8 degrees.
However, the value of critical angle is always dependent on the mediums situated on both sides of a boundary.
Additionally, the equation of critical angle is:
\[\theta _c\] = sin-1 nr / ni
Here, \[\theta _c\] is the critical angle of refraction, and nr and ni are refraction index and incident index, respectively.
What is Refractive Index?
The extent at which rays of light bend when it enters from one medium source to another is known as its refractive index. The refractive index is represented using the alphabet 'n.' Moreover, it can be written as n = c/v, where ‘c’ is light speed or velocity of a specific wavelength in the medium air. On the other hand, v is light’s velocity or speed in other media.
Furthermore, three factors determine this refractive index - medium nature, light colour, and physical conditions.
Fact: An optically rarer medium is one where light travels faster through it. Whereas, an optically denser medium is one where light travels slower through it.
Refractive Index and the Critical Angle Relationship
The mathematical representation of their relationship is:
sin C= 1/ µab
Here, C = critical angle, µ = refractive index, and a and b are two mediums within which light passes.
Furthermore, take a look at this derivation below!
Snell's Law (also known as the Second Law of refraction) is applied to derive the relation between critical angle and refractive index.
Hence, take a light ray having an incident angle i, refractive angle r = 90 degrees, critical angle = C, and refractive index of rarer and denser medium be µa and µb, respectively.
So, by applying the second Law of refraction or Snell's Law:
sin i / sin r = µa / µb
Therefore, µb sin C = µa sin90o
Therefore, µb / µa = 1 / sin C
Thus, with the help of this equation, critical angle and refractive index relation can be stated as:
µab = 1/ sinC
Solved Numericals
(i) Find out the ratio between sine of incident angle and the sine of reflected angle where their refractive indices are provided. In medium 1 it is 2.33, while in medium 2 it is 1.66.
Solution: Snell's Law gives \[n_1sin\theta _i=n_2sin\theta _r\]
In order to get \[\frac{sin\theta _i}{sin\theta _r}\] , you must note that this ratio is \[\frac{n_2}{n_1}\]
After substituting for n1= 2.33 along with n2 = 1.66
=> 1.66/2.33 = 0.71
(ii) Find the ratio between refractive index of two mediums, 1 and 2. Here, the reflected angle of medium 1 is 300, while that of medium 2 is 450.
Solution: Snell's Law gives \[n_1sin\theta _i=n_2sin\theta _r\]
So, for getting \[\frac{n_2}{n_1}\], this ratio is \[sin\theta _i=sin\theta _r\]
On putting values for θi= 30 and θr= 45
Therefore, \[\frac{sin45}{sin30}\] = 2
Applications of Critical Angle
The principle of critical angle is used in several practical ways in our day-to-day life. The most widely used is fibre optic cables. The concept is the base for the construction of fibre optic cables and how they work.
Applications of total internal reflection are multi-touch screens, spatial filtering of lights, prismatic binoculars, automotive rain sensors, fluorescence microscopes, and ubiquitous fibre optics communications.
Do It Yourself
1. Angle of incidence is equal to the angle of reflection for perfect reflection. Answer true or false.
(a) False
(b) True
Ans: (b) True
2. The higher the value of the refractive index of a given medium, the bending of light will be
(a) zero
(b) smaller
(c) higher
(d) negative
Ans: (c) Higher
3. The refractive index of a medium is the relation between light’s speed in vacuum or air, and
(a) Light’s speed in a medium
(b) Can be a or c
(c) Speed of sound in a medium
(d) none
Ans: (a) Light’s speed in a medium
FAQs on Understanding the Relation Between Critical Angle and Refractive Index
1. What is the critical angle, and what are the two essential conditions required for it to occur?
The critical angle is the specific angle of incidence in a denser medium for which the angle of refraction in the rarer medium is exactly 90 degrees. For the phenomenon of critical angle and subsequent total internal reflection to occur, two conditions must be met:
- The light ray must be travelling from a denser medium to a rarer medium (e.g., from water to air).
- The angle of incidence in the denser medium must be equal to (for critical angle) or greater than (for total internal reflection) the critical angle for that pair of media.
2. What is the fundamental relationship between a medium's refractive index and its critical angle?
The relationship is an inverse one. The critical angle (C) is the arcsin of the ratio of the refractive index of the rarer medium (n₂) to that of the denser medium (n₁). The formula is: sin(C) = n₂ / n₁. For light travelling from a medium with refractive index 'μ' into a vacuum or air (where n ≈ 1), the formula simplifies to μ = 1 / sin(C). This shows that a higher refractive index results in a smaller critical angle.
3. What is Total Internal Reflection (TIR), and how is it a direct consequence of the critical angle?
Total Internal Reflection (TIR) is the phenomenon where a light ray travelling from a denser to a rarer medium is completely reflected back into the denser medium, with no light being refracted. This occurs when the angle of incidence is greater than the critical angle. The critical angle acts as the threshold; once the incident angle surpasses this value, refraction becomes impossible, and all the light energy is reflected internally.
4. How does the brilliance of a cut diamond relate to its critical angle?
The exceptional sparkle of a diamond is a direct result of its very high refractive index (about 2.42), which gives it a very small critical angle of approximately 24.4°. Diamond cutters shape the gem with specific facets so that most light entering it strikes the internal surfaces at an angle greater than this critical angle. This causes the light to undergo Total Internal Reflection multiple times before exiting from the top, creating the intense brilliance and fire for which diamonds are known.
5. Why does a critical angle only exist when light travels from a denser medium to a rarer medium?
This is a direct consequence of Snell's Law, n₁sin(i) = n₂sin(r). For the critical angle (C), the angle of refraction (r) is 90°, so sin(r) = 1. The equation becomes sin(C) = n₂/n₁. Since the maximum value of the sine function is 1, a real solution for C can only exist if n₂/n₁ ≤ 1, which means n₂ ≤ n₁. This condition is only met when light travels from a medium with a higher refractive index (denser, n₁) to one with a lower refractive index (rarer, n₂).
6. How do optical fibres use the principle of critical angle for high-speed data transmission?
Optical fibres transmit data as pulses of light. The fibre consists of a central core (denser medium) surrounded by cladding (rarer medium). Light is launched into the core at an angle of incidence greater than the critical angle for the core-cladding interface. This causes the light signals to undergo continuous Total Internal Reflection, bouncing off the walls of the core and propagating down the fibre with minimal loss of signal strength or speed, enabling efficient long-distance communication.
7. How does the critical angle of a medium change with the colour (or wavelength) of light used?
The critical angle is dependent on the wavelength of light because a medium's refractive index (μ) itself varies with wavelength (a phenomenon called dispersion). Typically, the refractive index is higher for shorter wavelengths (like violet light) and lower for longer wavelengths (like red light). Since the critical angle formula is C = sin⁻¹(1/μ), a higher refractive index results in a smaller critical angle. Therefore, in a given medium like glass, the critical angle for violet light is smaller than the critical angle for red light.
