Question

A narrow beam of light with wavelengths from 450 nm to 700 nm is incident perpendicular to one face of a prism made of crown glass, for which the index of refraction ranges from n = 1.533 to n = 1.517 for those wavelengths. The light strikes the opposite side of the prism at an angle of 40.0°. What is the angular spread of the beam as it leaves the prism? Express your answer in degrees.

Answer 1

Given data, to 450 nm 700 nm Wavelength Refractive Index of crown glass For 450 nm n = 1.533 For 700 nm, n₂ = 1.517 Angle of Incidence inside the prism = 40.00 In order to find the angular spread, we need to find the angle of refraction corresponding to above mentioned wavelengths Angle of Refraction for 450nm : Using Snell's Law: n, sin o, R, - Refractive index of the Crown glass for 450 nm O, - Angle of incidence inside n2 Sin O₂ the prism M2 Refractive index of ais O₂ - Angle of Refraction

we need to find a : n, sino, ₂ Sin O₂ sin n. sina) п, Sint 1.533 2 Sin 40 40) O 80-195 0 2 - Angle of Refraction for 700nm n, sino 1₂ Sinoz here, - 1.517 40 nz 2 Substituting the values, n, sin o, ₂ Sinoz 0 2 Sin- sino n₂ O2 Sint 1.517 Sin 40 40) 11 77.1890 2

Given data, Wavelength 450 nm Angulas Spread of the beam as it leaves the prism Difference between O for 450nm and O₂ for 700 nm = 0 - @ O 80. 195 77.1899 3.006° = 3.006 Angular spread of the beam