Question

Prism-shaped Air Cavity in Glass: Rays A and B approach the equilateral prism, and the prism is air surrounded by glass with indexes of refraction shown. Following the rules of geometrical optics, trace each ray as it reflects and refracts at the prism's surfaces. If a ray strikes a surface and undergoes both reflection and refraction, then only trace the refraction. Continue tracing each ray until the ray is moving away from the prism. As each ray strikes a surface, record the incident angle, angle of reflection, and the angle of refraction in a table. n = 1.5 na= 1

Answer 1

Solution : Prism-shaped Air cavity in Glass : KEY ------ Yay A - - -Tay B , refracted say 1B2 / 1,=105 derya refracted ray 60-TBT incident - 4 Yays 60- in -1-5 incident rays 60° LA = incidence angle of say A at p. TAL = re fracked angle of say A at P. 8A2 = refracted angle of ray A at Q. 60°- tag - incident angle' of xay A at Q. ip - incident xay Bungle of ray B at P & Bg = refracted angle of say B at P. 862 = refracted angle of ray B at Q. 60°- 884 = incident angle of ray B at Q, Let's, take, [in = 30 By Shell's law, som inne Sin ใน sin 300 = 1 sinr 1.5 » = 48.59°). 60-8A1 = 11,41° .:

By shell's law , at 8: sin (60 – Al) sin YAZ ne sin 11.410 Sin VAZ S A2 = 7.578° Similarly, for say B. let's take lib = 350] » By Shell's law, simio = n2 sin rol sin 35 sin VBI *B, = 59.359 = 60° - YB, = 0.650 By applying Snell's at R, sin (60°– VBI ) Sin VB2 = sin 0.65° 105 sin & B2 no 8B2=0.433 ray A ray B. angle of incidint angle of eflectim angle of | refraction At point p angle of angle of angle of incident I reflection / refraction / 30° 30° 48.590 11:42 11.419 | 70578° 350 | 350 54.350 | 0.650 0.65 At point 8 At point R 0.433°

The given equilateral triangle is the inside of the prism having air as the medium, the outside of the triangle is the glass medium. The angle of reflection at each of the points for each of the rays is same as the corresponding angle of incidence at those points.