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 green light without changing the angle of incidence, the beam of green light would bend away from the normal as it gets refracted through the second medium.
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 medium1 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)〗〗_g<〖〖(sin〗〖θ_c)〗〗_y
So, from the equation (ii) we get ‘n’ for green is more than that for yellow.
Hence, the critical angle for green is less than that for yellow.
Therefore, green will show total internal reflection if it is incident at an angle equal to the critical angle for yellow. Therefore, the statement that the green light will get refracted is wrong
Final Answer: No.
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 medium1 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)〗〗_g<〖〖(sin〗〖θ_c)〗〗_y
So, from the equation (ii) we get ‘n’ for green is more than that for yellow.
Hence, the critical angle for green is less than that for yellow.
Therefore, green will show total internal reflection if it is incident at an angle equal to the critical angle for yellow. Therefore, the statement that the green light will get refracted is wrong
Final Answer: No.
