Answer
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Hint:In optics critical angle is that angle of incidence at which the angle of refraction is \[{90^0}\]. The direction of light bends towards the normal if light enters a denser medium from a rarer medium. From Snell’s law, we can find a refractive index in the given medium.
Formula used:
The critical angle is given as:
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)\]
Where C is the critical angle
\[\mu {\rm{ is refractive index}}\]
Complete step by step solution:
Given the refractive index of the material of the prism for red colour,\[{\mu _R}\] =1.27
The refractive index of the material of the prism for green colour, \[{\mu _G}\] =1.42
The refractive index of the material of the prism for the blue colour, \[{\mu _B}\] =1.49
Image: In right angled triangle PQR, shows \[\angle PRQ = {45^0}\]
For Total Internal Reflection, at the face PR,
Incident angle, \[i = {45^0} = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)\]
As critical angle is given as:
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)\]
If the rays which are not passing through the face PR, must follow the condition,
i > C or \[{45^0} > C\]
And if the rays which are pass through the face PR, must follow the condition,
i < C or \[{45^0} < C\]
For blue colour,
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{{1.49}}} \right)\]
\[\Rightarrow {\mu _B} = 1.49 > \mu = \sqrt 2 = 1.414\]
Thus it shows i > C, blue not passing through the face PR.
For green colour,
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{{1.42}}} \right)\]
\[\Rightarrow {\mu _G} = 1.42 > \mu = \sqrt 2 = 1.414\]
Thus it shows i > C, green not passing through the face PR.
For red colour,
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{{1.27}}} \right)\]
\[\Rightarrow {\mu _R} = 1.27 < \mu = \sqrt 2 = 1.414\]
Thus it shows i < C, red pass through the face PR.
Here μG and μB are more than μ.
∴ only red will come out.
Hence option C is the correct answer.
Note: The light is a form of energy that exhibits various phenomena like reflection, refraction, dispersion, total internal reflection(TIR). If the light rays pass from a denser medium to a less dense medium then this phenomenon is known as total internal reflection.
Formula used:
The critical angle is given as:
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)\]
Where C is the critical angle
\[\mu {\rm{ is refractive index}}\]
Complete step by step solution:
Given the refractive index of the material of the prism for red colour,\[{\mu _R}\] =1.27
The refractive index of the material of the prism for green colour, \[{\mu _G}\] =1.42
The refractive index of the material of the prism for the blue colour, \[{\mu _B}\] =1.49
Image: In right angled triangle PQR, shows \[\angle PRQ = {45^0}\]
For Total Internal Reflection, at the face PR,
Incident angle, \[i = {45^0} = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)\]
As critical angle is given as:
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)\]
If the rays which are not passing through the face PR, must follow the condition,
i > C or \[{45^0} > C\]
And if the rays which are pass through the face PR, must follow the condition,
i < C or \[{45^0} < C\]
For blue colour,
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{{1.49}}} \right)\]
\[\Rightarrow {\mu _B} = 1.49 > \mu = \sqrt 2 = 1.414\]
Thus it shows i > C, blue not passing through the face PR.
For green colour,
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{{1.42}}} \right)\]
\[\Rightarrow {\mu _G} = 1.42 > \mu = \sqrt 2 = 1.414\]
Thus it shows i > C, green not passing through the face PR.
For red colour,
\[C = {\sin ^{ - 1}}\left( {\dfrac{1}{{1.27}}} \right)\]
\[\Rightarrow {\mu _R} = 1.27 < \mu = \sqrt 2 = 1.414\]
Thus it shows i < C, red pass through the face PR.
Here μG and μB are more than μ.
∴ only red will come out.
Hence option C is the correct answer.
Note: The light is a form of energy that exhibits various phenomena like reflection, refraction, dispersion, total internal reflection(TIR). If the light rays pass from a denser medium to a less dense medium then this phenomenon is known as total internal reflection.
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