Question

QUESTIONS 22 - 24: Three flat, 4 cm thick plates of differing refractive indices are stacked as shown. The letters A-D indicate the surfaces/interfaces between the different layers. air n1 1.2 n2-1.4 22. Monochromatic light is incident on surface A from above at an angle of 30° with the normal. The angle (with the normal) at which the ray emerges from the bottom of the stack wil be n, = 1.6 (a) 30° air (b) 60° (c) 18o (d) no emergence, due to total internal reflection (e) none of these

Answer 1

In, total internal reflection, light is completely reflected from denser to rarer medium when angle of incidence is greater than the critical angle. Consider the first surface, when light is incident at the air and surface A boundary. Use Snell law and find the angle of reflection. n, sin 30º = (1.2) sin r; sin rį = _ (1) sin 30° 1.2 sin r; = 0.4167 This angle of reflection becomes the incidence angle at the boundary of surface A and B. (1.2)sin r; = (1.4)sin r; sin r; = 1.2(0.4167) sin r2 = 1.4 sin r, = 0.357 For the boundary of surfaces C and D, (1.4)sin rı =1.6 sin r . 1.4(0.357) sin r == 1.6 sin rz = 0.3123 rz = 18.2°

This is the angle of incidence for the surface when light goes from surface D to air i.e. rarer to denser medium. The critical angle for this surface is, 0c = sin- () = 26° Thus, the angle of incidence is less than he critical angle which implies that TIR does not occur. For surface D to air, 1.6 sin r; = 1sinr " 29.979 -30° Thus, the correct option is (A).