Answer to Question #200377 in Optics for Piolo

Question

The path of a light beam in air goes from an angle of incidence of 40° to an angle of refraction of 21° when it enters a rectangular block of plastic. What is the index of refraction of the plastic?

answer to 2 decimal places

Answer 1

• The Principle
• Once we know the angle of incidence and the angle of refraction, as well as the refractive index of the incident beam, we can calculate the refractive index of a material in which a refractive ray passes by using Snell’s law: n1 sin (θ_i) = n2 sin (θ_r)
• In which n1 is the refractive index for the incoming light beam medium, n2 is the refractive index for the outgoing light beam medium. Θi is the incident angle between the ingoing light beam and the normal to the refractive surface i.e. surface at which the incident refracts, Θr is the angle made by the outgoing beam and the normal to the refractive surface.
• N2 can be given by
• n2=[n1sin(θ_i)/sin (θ_r)]-----------------------------------------------------------------------------------------------------1
• n2=[n1sin(θ_i)$divide by$ sin (θ_r)]
• In our case, incident light goes through air with a refractive index of 1.000293 hence n1=1.000293
• The angle of incidence is 40 degrees and the angle of refraction is 21 degrees.
• Hence, we can know the refractive index of the refractive beam medium by using equation 1 as follows.
• n2 = [1.000293 sin (40)/sin(21)] = 1.794178.
• n2 = [1.000293 sin (40)$divide by$ sin(21)] = 1.794178.
• We round this off to 2 decimal places to get 1.79
• Ans=1.79