In optical physics, various phenomena are occurring as refraction, reflection, etc. The refraction event is the event for a change in direction of propagation of light due to a change in its transmission medium. If light enters a denser medium from a comparatively rarer medium, then the direction of light changes and the light ray bends towards the normal. This normal line is an imaginary perpendicular line drawn at the point of contact of light. In this article one important measurement in this regard. It is a critical angle. Let us learn the critical angle formula with examples.
Critical Angle Formula
Concept of Critical Angle
The critical angle of a matter when passing a light ray from that matter to vacuum is very important as you must use it to determine whether total internal reflection will occur. It is also important for your physics test.
The critical angle in optics refers to a specific angle of incidence. Beyond this angle, the total internal reflection of light will occur. The trajectory 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 angle of exit of the ray is always greater than the angle of incidence. This type of reflection is called internal reflection.
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Whenever light travels from a medium with a higher refractive index i.e. \(n_1\) to a medium with a lower refractive index i.e. \(n_2\). Here, the angle of refraction is more than the angle of incidence. Due to this difference in the refractive index, the ray bends towards the surface.
Therefore, the critical angle is defined as the angle of incidence which provides a 90 degree angle of refraction. It must be noted here that the critical angle is an angle of incidence value. For the water-to-air limit, the critical angle is 48.6 degrees. For the boundary between the glass and crown water, the critical angle is 61.0 degrees. The actual value of the critical angle is depending on the combination of materials present on each side of the boundary.
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The Formula for Critical Angle
Let us consider two different media. The critical angle is that of \( θ_{cric}\) which gives a value of exactly 90 degrees. If these values are substituted in the Snell’s Law equation, we will get a generic equation that will be used to predict the critical angle.
Therefore,
The critical angle = the inverse function of the sine of \( (\frac{refraction inde}{incident index}) \)
The equation is:
\( \theta_{cric} = \sin^{-1}\frac{n_r}{n_i} \)
Where,
\(\theta_{cric}\) | The critical angle. |
\(n_r\) | Refraction index. |
\(n_i \) | Incident index. |
Solved Examples on Critical AngleÂ
Q.1: What must be the angle of incidence for there to be a total internal reflection of a ray going from the water with \( n_w \)= 1.3 to glass with \( n_g \)= 1.52?
Solution: Given the indices for the means by which the ray passes,
we use the formula
\( \theta_{cric} = \sin^{-1}\frac{n_r}{n_i}\)
\( \theta_{cric} = \sin^{-1}\frac{1.3}{1.52}\)
So,
\( \theta_{cric} = 1.064 rad \)
Critical angle is 1.064 rad.
Example-2: A ray of light strikes from a medium with n = 1.67 on a surface of separation with the air with n = 1. Calculate the value of a critical angle.
Answer: Given the indices for both the means.
We know the formula,
\( \theta_{cric} = \sin^{-1}\frac{n_r}{n_i}\)
\( \theta_{cric} = \sin^{-1}\frac{1}{1.67}\)
Therefore,
\(\theta_{cric} = 0.064 rad \)
Critical angle is 0.064 rad.
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
Interesting studies
It is already correct f= ma by second newton formula…