Introduction
Ray optics is one of the most interesting branches of physics that deals with the behavior of moving light. In ray optics, we have come across many interesting concepts such as reflection, refraction, transmission, etc. One more important concept among those is the total internal reflection. The total internal reflection is a natural process that involves the reflection of all the incident light off the boundary.
The total internal reflection can take place only when both of the following two conditions are met. The first condition says that a light ray is in the more dense medium and approaching the less dense medium and the second condition is the angle of incidence for the light ray is greater than the critical angle. Now, what is the critical angle? The critical angle can be defined as the angle of incidence that offers an angle of refraction of about 90 degrees and the corresponding formula is known as the critical angle formula or formula of critical angle.
In this article, we will discuss the critical angle formula, critical angle of water, critical angle of glass, how to find critical angle along with a few solved examples.
Critical Angle of Water
The critical angle of the matter when passing a light ray from that matter to vacuum is most important as we must use it to determine whether total internal reflection will occur. Generally, we consider the denser medium as water and the rarer medium like air, thus knowing the critical angle of water will be helpful in determining when a light ray passes through the rarer medium to water.
The critical angle in geometric optics refers to a specific angle of incidence. Beyond the critical angle, the total internal reflection of light will occur. The path of a ray of light will deviate from the normal path when it strikes a medium that has a lower refractive index. As a result, the critical angle of the emerging light ray will always be greater than the angle of incidence. This physical phenomenon is known as the total internal reflection.
Hence, the critical angle can be defined as the angle of incidence of a light ray that provides a 90-degree angle of refraction. One of the important points that must be noted here is that the critical angle is nothing but just an angle of incidence value.
Let us have a look at the critical angle formula with a small derivation. Consider two different mediums of different refractive indices. Let the first medium is of refractive index \[n_{1}\]and another one of refractive index \[n_{2}\], let us assume these two mediums, one as incident medium and another refracted medium.
Now, according to the definition of critical angle, we know that the critical angle is the \[\theta_{i}\] that incident angle which gives a \[\theta_{r}\]the value of 90-degrees. Using Snell's law and substituting all these values into Snell's Law equation, a mathematical expression for the critical angle can be derived. Thus, we write:
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We know that snell’s law is given by:
\[\Rightarrow n_{1} \sin \theta_{i} = n_{2} \sin \theta_{r}\].....(1)
Here, we know that angle of incidence is the critical angle and thus the angle of refraction is 900,thus we get:
\[\Rightarrow n_{1} \sin \theta_{crit} = n_{2} \sin 90 = n_{2}\]
\[\Rightarrow \sin \theta_{crit} = \frac{n_{2}}{n_{1}}\]
\[\Rightarrow \theta_{crit} = \sin^{-1} (\frac{n_{2}}{n_{1}})\]…..(2)
Equation (2) is known as the critical angle formula, considering relevant data we can derive the critical angle of water, the critical angle of glass, etc...
Let us have a look at the critical angle of different materials.
The critical angle of water= \[49^{0}\]
The critical angle of glass= \[42^{0}\]
The critical angle of diamond= \[24.4^{0}\]
The critical angle of water to air= \[48.6^{0}\]
Critical Angle of Glass
Let us calculate the critical angle of glass. We will consider air to glass medium. In order to estimate the critical angle from the critical angle of water, we must note down certain parameters such as the refractive indices of the involved mediums. Here, the incident medium is glass and the refractive medium is air.
The refractive index of air= \[n_{2} = 1\]
The refractive index of glass= \[n_{2} = 1.52\]
So, now from the critical angle formula we have:
\[\Rightarrow \theta_{crit} = \sin^{-1} (\frac{n_{2}}{n_{1}})\]
Where,
\[n_{1}\]-The refractive index of the incident medium
\[n_{2}\]-The refractive index of the refracted medium
Substituting, the given data in the above equation and simplify we get:
\[\Rightarrow \theta_{crit} = \sin^{-1} (\frac{1}{{1.52}}) = \sin^{-1} (0.6578)\]
\[\Rightarrow \theta_{crit} = 41.139^{0}\]
Therefore, the critical angle of glass to air medium is around $42^{0}$ theoretically, practically it is around $41.2^{0}$
Solved Example
Now let us understand how to calculate critical angle with a few solved examples.
1. How to find critical angle for the diamond to air medium.
Sol:
Given,
We are considering the light ray moving from diamond to air boundary, then we have:
The refractive index of diamond= \[n_{1} = 2.42\]
The refractive index of air= \[n_{2} = 1\]
So, now from the critical angle formula we have:
\[\Rightarrow \theta_{crit} = \sin^{-1} (\frac{n_{2}}{n_{1}})\]
Where,
\[n_{1}\]-The refractive index of the incident medium
\[n_{2}\]-The refractive index of the refracted medium
Substituting, the given data in the above equation and simplify we get:
\[\Rightarrow \theta_{crit} = \sin^{-1} (\frac{1}{{2.42}}) = \sin^{-1} (0.4132)\]
\[\Rightarrow \theta_{crit} = 24.4^{0}\]
Therefore, the critical angle of the diamond to the air boundary is around \[24.4^{0}\].
2. Estimate the angle of incidence for a light ray passing through water to glass such that it results in the total internal reflection. Where the refractive index of water is \[n_{1}\] = 1.3and the refractive index of glass is \[n_{2}\] = 1.52?
Sol:
Given,
The refractive index of glass=\[n_{2}\] = 1.52
The refractive index of water=\[n_{1}\] = 1.3
We are asked to calculate the angle of incidence such that the incident light ray undergoes TIR. If the light ray undergoes TIR then the incident angle will be the critical angle. We know that the critical angle is given by:
\[\Rightarrow \theta_{crit} = \sin^{-1} (\frac{n_{2}}{n_{1}})\]
Where,
\[n_{1}\]-The refractive index of the incident medium
\[n_{2}\]-The refractive index of the refracted medium
Substituting, the values of n1 and n2in the above equation and simplify we get:
\[\Rightarrow \theta_{crit} = \sin^{-1} (\frac{1.3}{{1.52}}) = \sin^{-1} (0.8552)\]
\[\Rightarrow \theta_{i} = 58.78^{0}\]
Therefore, the angle of incidence for the given light ray is \[58.78^{0}\]
FAQs on Critical Angle Formula
1. What is the critical angle?
Critical angle is one of the important necessary conditions to achieve total internal reflection. Now, the critical angle can be defined as the angle of incidence of light at which the angle of refraction will be \[90^{0}\]
2. How do we calculate the refractive index by using the critical angle?
The refractive index of the material can be found using the critical angle formula given by:
\[\Rightarrow n = \frac{\sin r}{\sin i}\]
We know that the critical angle is the angle of incidence at which the angle of refraction is 90. Therefore we get:
\[\Rightarrow n = \frac{\sin 90}{\sin \theta_{critical}} = \frac{1}{\sin \theta_{c}}\]
