You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90∘. State whether the following statement is correct or not. If the source of yellow light is replaced with that of blue light without changing the angle of incidence, the beam of blue light would undergo total internal reflection.
As we know, when the light goes from optically denser medium to optically rarer medium it bents away from the normal.
Let the light move from medium 1 to medium 2. Let the refractive index of 1 with respect to 2 be n_12.
Refractive index of medium1, (n_1 )=n
Refractive index of medium 2, (n_2 )=1
By Snell's law,
n_2/n_1 =sin(θ_i )/(sin(θ_r ) ) …(i)
By using equation (i) we get,
1/n=sin(θ_i )/(sin(θ_r ) )
nsin(θ_i )=sin(θ_r )
where
θ_i is the angle of incidence
and θ_r is the angle of refraction.
The θ_i at which θ_r= 〖90〗^∘ is known as the critical angle(θ_c).
Substitute θ_i=θ_c and θ_r=〖90〗^∘,we get
sinθ_c=1/n …(ii)
If the angle of incidence larger than critical angle, there is no refraction but only reflection. This is known as total internal reflection.
Since n∝1/λ …(iii)
〖〖(sin〗〖θ_c)〗〗_b<〖〖(sin〗〖θ_c)〗〗_y
So, from the equation (iii),we get ‘n’ for blue is more than that for yellow.
Hence, the critical angle for blue is less than that for yellow.
Therefore, blue will show total internal reflection if it is incident at an angle equal to the critical angle for yellow. Therefore, the statement is correct.
Final Answer: Correct.
As we know, when the light goes from optically denser medium to optically rarer medium it bents away from the normal.
Let the light move from medium 1 to medium 2. Let the refractive index of 1 with respect to 2 be n_12.
Refractive index of medium1, (n_1 )=n
Refractive index of medium 2, (n_2 )=1
By Snell's law,
n_2/n_1 =sin(θ_i )/(sin(θ_r ) ) …(i)
By using equation (i) we get,
1/n=sin(θ_i )/(sin(θ_r ) )
nsin(θ_i )=sin(θ_r )
where
θ_i is the angle of incidence
and θ_r is the angle of refraction.
The θ_i at which θ_r= 〖90〗^∘ is known as the critical angle(θ_c).
Substitute θ_i=θ_c and θ_r=〖90〗^∘,we get
sinθ_c=1/n …(ii)
If the angle of incidence larger than critical angle, there is no refraction but only reflection. This is known as total internal reflection.
Since n∝1/λ …(iii)
〖〖(sin〗〖θ_c)〗〗_b<〖〖(sin〗〖θ_c)〗〗_y
So, from the equation (iii),we get ‘n’ for blue is more than that for yellow.
Hence, the critical angle for blue is less than that for yellow.
Therefore, blue will show total internal reflection if it is incident at an angle equal to the critical angle for yellow. Therefore, the statement is correct.
Final Answer: Correct.
