The angle of incidence in the denser medium at which angle of reflection is 90° is called critical Angle.
[caption id="attachment_22114" align="aligncenter" width="500"]
Formation of critical angle[/caption]
Let see in case of total internal reflection
When a ray of light passes from denser to rarer medium, the refracted ray bends away from the normal. Hence, the angle of refraction is greater than the angle of incidence. Gradually if the angle of incidence increases, angle of refraction also increases. At a particular angle of incidence, the refracted ray just grazes the refracting surface making the angle of refraction is equal to 90°. This angle of incidence in the denser medium is called 'critical angle'. If the angle of incidence is further increased, the ray does not undergo refraction, it gets reflected into the same optically denser medium. This is known as total internal reflection.
A ray PQ refracting from denser to rare medium[/caption]
From Snell's Law,
$^1μ_2 = \dfrac{sin i}{sin r}$ → eq (1)
But i = c and r = 90°
$^1μ_2 = \dfrac{sin c}{sin 90}$
$^1μ_2 = \dfrac{sin c}{1}$
$(∴ sin 90 = 1) ⇒ \dfrac{μ_2}{μ_1}= sin\sqrt{C}$
[$^1μ_2 = \dfrac{μ_2}{μ_1}$ from relative reflective index]
If the second medium is air, then, $μ_2 = 1 and μ_1 = μ ⇒ μ = \dfrac{1}{sin}.$
Formation of critical angle[/caption]
Let see in case of total internal reflection
When a ray of light passes from denser to rarer medium, the refracted ray bends away from the normal. Hence, the angle of refraction is greater than the angle of incidence. Gradually if the angle of incidence increases, angle of refraction also increases. At a particular angle of incidence, the refracted ray just grazes the refracting surface making the angle of refraction is equal to 90°. This angle of incidence in the denser medium is called 'critical angle'. If the angle of incidence is further increased, the ray does not undergo refraction, it gets reflected into the same optically denser medium. This is known as total internal reflection.
Relation between Critical Angle and Reflective Index
Consider a ray of light PQ travelling from denser medium to rarer medium. Let the ray be incident at the critical angle. the angle of refraction will be equal to 90°. [caption id="attachment_22116" align="aligncenter" width="500"]
A ray PQ refracting from denser to rare medium[/caption]
From Snell's Law,
$^1μ_2 = \dfrac{sin i}{sin r}$ → eq (1)
But i = c and r = 90°
$^1μ_2 = \dfrac{sin c}{sin 90}$
$^1μ_2 = \dfrac{sin c}{1}$
$(∴ sin 90 = 1) ⇒ \dfrac{μ_2}{μ_1}= sin\sqrt{C}$
[$^1μ_2 = \dfrac{μ_2}{μ_1}$ from relative reflective index]
If the second medium is air, then, $μ_2 = 1 and μ_1 = μ ⇒ μ = \dfrac{1}{sin}.$
Comments
Post a Comment