Question

ABC represents an equiangular prism. Light of a given frequency falls on face AB. gets refracted andtravels parallel to the base of prism. While emerging from face AC, the ray deviates by an angle 30Refractive index​

Answer 1

Answer:

•Given: A ray of light undergoes deviation of 30 ∘when incident on an equilateral prism of refractive index 2.

To find the angle made by the ray inside the prism with the base of the prism.

To find the angle made by the ray inside the prism with the base of the prism.Solution:

As per the given criteria,angle of deviation, δ=30 ∘refractive index of the prism, μ= 2

As per the given criteria,angle of deviation, δ=30 ∘refractive index of the prism, μ= 2As the prism is equilateral, the refractive angle, A=60 ∘And we know,

As per the given criteria,angle of deviation, δ=30 ∘refractive index of the prism, μ= 2As the prism is equilateral, the refractive angle, A=60 ∘And we know,μ= sin 2Asin( 2A+δ m )Substituting the values, we get

sin( 2A+δ m )Substituting the values, we get2=sin 260sin( 260+δ) m⟹ 2= sin30sin( 260+δ m )⟹2× 2 =sin( 260+δ m )⟹ 260+δ m =sin −1 ( 21 )⟹ 260+δ m =45⟹δ m=90−60=30°Hence the minimum deviation angle is equal to the deviation angle, hence at minimum deviation condition the ray travels parallel to the base inside the prism.

Hence the minimum deviation angle is equal to the deviation angle, hence at minimum deviation condition the ray travels parallel to the base inside the prism.The angle between the ray and base inside the prism is zero.